Calculations and Key Notes in Chemistry PDF

**Chapter 1** Atoms, molecules, ions and radicals **Learning Outcome** When you have completed this chapter, you will be able to: - Understand the concepts of the following terms: atoms, molecules, ions and radicals. - Know the components of an atom; its subatomic particle - Determine the number of protons, electrons and neutrons of an atom. If you already feel confident about these chapter why not try the Quiz over the page? **Practice Questions** 1. For the noble gases (The group 8A elements) 2. Indicate the number of protons, electrons and neutrons in each of the following species 3. Write the appropriate symbol for each of the following atoms 4. Determine the number of electrons and neutron in Mg^2+^ 5. An atom has atomic number 92 and contains 143 neutrons in its nucleus. What is the atomic mass of the atom and how many proton does it contain in its nucleus? **1. Atom** What is an Atom: An atom is the smallest particle of an element that can take part in a chemical reaction. An atom is made up of three sub-atomic particle which I call P E N of an atom P for proton \[which is positively charged\] E for electron \[which is negatively charged\] N for neutron \[which is neutral\] Now let us understand the meaning of the word ATOM. *Example 1* The Symbol O: should be pronounced as Oxygen atom The symbol O -- O i.e. O~2~: should be pronounced as oxygen molecule. This means that a molecule is made up of at least two atoms so two atoms of an element can form a molecule, three atoms of an element can form a molecule, four atoms of an element can form a molecule as the case may be. *Example 2* The Symbol N: should be pronounced as Nitrogen atom. The Symbol N -- N i.e. N~2~ should be pronounced as Nitrogen molecule. *Example 3* The Symbol P: should be pronounced as Phosphorus atom. The symbol P -- P -- P -- P i.e. P~4~: should be pronounced as Phosphorus molecule. **2. Determination of the Number of Protons, Electrons, Neutron of an atom** Let us examine an unknown atom X A Z Where X = unknown atom A = Atomic mass or mass number or nucleon number Z = Atomic number or proton number To determine the number of proton, electrons and neutron of an atom, it will be in two CASES. **CASE 1: For a NEUTRAL ATOM \[Atom without a charge\]** P = is same as atomic number E = is same as proton number \[i.e. same as atomic number\] N = is calculated by Mass Number -- Atomic Number Where A = mass number Z = atomic number N = A -- Z **CASE 2: For a CHARGED ATOM \[atom with a charged wither positive or negative charge\]** P = is same as atomic number *N.B:* For a CHARGED ATOM, the Electron Number is not equal to the proton number E = Proton Number -- Charge N = is calculated by Mass Number -- Atomic Number Where A = mass number Z = atomic number N = A -- Z **Solved Questions with Detailed Explanation** 1. Determine the number of proton, electron and neutron in Na P = is same as atomic number = 11 E = is same as proton number = 11 N = Mass Number -- Atomic Number Atomic number = Z = 11 2. Determine the number of proton, electron and neutron in Na^+1^ P = is same as atomic number = 11 E = Proton Number -- Charge Where Proton Number = 11 Charge = + 1 E = 11 -- (+ 1) E = 11 -- 1 E = 10 N = Mass Number -- Atomic Number Where Mass number = A = 23 Atomic number = Z = 11 **3. Molecule** What is Molecule? A molecule is the smallest particle of a substance that can exist alone or independently and still retain that same chemical identity of that same atom. **4. Ions** What is an Ion? An ion is a charged atom, when an ion is positively charged, it is called a CATION. But when it is negatively charged, it is called an ANION. *Example:* Na^+1^, K^+1^, Ca^2+^, Al^3+^. These are Cations *Example:* Cl^−1^, I^−1^, O^2−^, S^2−^. These are Anions **5. Radicals** What are Radicals? Radicals are group of atoms behaving as a single charge unit i.e. a particular charge is holding all the atoms together. *Example:* N : O : O : O^−1^ = NO \[This is called the Nitrate radical\] C : O : O : O^2−^ = CO \[This is called the carbonate radical\] S : O : O : O^2−^ = SO \[This is called the sulphate radical\] **Chapter 2** Isotopy and Relative Atomic Mass Calculations **Learning Outcome** When you have completed this chapter, you will be able to: - Understand the word called isotopy. - Know how to calculate the relative atomic mass dealing with different cases and other related calculations involved. - Understand the meaning of some basic terms like ISOBAR, ISOTONES and ISO-DIAPHERS. If you already feel confident about this chapter, why not try the Quiz over the page. **Practice Questions** 1. An element X exists in two isotopic form X and X in the ratio 3:2 respectively. What is the relative atomic mass of X? 2. An element X with relative atomic mass 16.2 contains two isotopes. X with relative abundance of 90% and X with relative abundance of 10%. What is the value of m? 3. Rubidium has two naturally occurring isotope Rb with a mass of 84.9118 a.m.u and Rb with a mass of 86.9092 amu. If the atomic mass of rubidium is 85.47. What is the relative abundance of Rb? 4. Lithium exist as Li and Li in the ratio 2:25. Calculate the relative atomic mass of Lithium. 5. Two isotopes of Z with mass number 18 and 20 are in the ratio 1:2. Determine the relative atomic mass of Z. **1. ISOTOPY** What is Isotopy: Isotopy is a phenomenon where by atoms of same element have DIFFERENT MASS NUMBER but same ATOMIC NUMBER. For Example: Cl and Cl where 35 and 37 for chlorine are the MASS NUMBER which is DIFFERENT and 17 for both atoms are the ATOMIC NUMBER which is the SAME. **2. Relative Atomic Mass** What is relative atomic mass? The relative atomic mass of an element is the AVRAGE MASS of its atoms, compared to 1/12^th^ the mass of carbon-12. **3. Determination of relative atomic mass of an atom** To determine Relative Atomic Mass (RAM) of an element, it can be in different cases. **CASE 1: Relative atomic mass calculations dealing two isotopes.** *Formula:* R.A.M = Where %~1~ = Percentage abundance of the first isotope M~1~ = Mass number of the first isotope \%~2~ = Percentage abundance of the second isotope M~2~ = Mass number of the second isotope **Solved Questions with Detailed Explanations on CASE 1** 1\) Oxygen is a mixture of two isotope O and O with percentage abundance of 90% and 10% respectively. The relative atomic mass of oxygen is (a) 16 (b) 16.2 (c) 17 (d) 18 *Soln* ***N.B:*** The relative atomic mass of element are not whole numbers. So with this hint, a smart student might not solve this question before choosing the correct answer. From the hint 1 just said, the answer to the question will be option (B) because it is 16.2 and the only option without a whole number But let us solve together Parameter First isotope = O = 90% Second isotope = O = [10%.] [100% ] \[The percentage abundance in a question should be equal to 100%\] \%~1~ = Percentage abundance of the first isotope = 90% M~1~ = Mass number of the first isotope = 16 \%~2~ = Percentage abundance of the second isotope = 10% M~2~ = Mass number of the second isotope = 18 Let us input all the parameters into the formula I gave initially. RAM = RAM = RAM = 14.4 + 1.8 RAM = 16.2 2\) Calculate the relative atomic mass of chlorine which contains 75% Cl and 25% Cl *Soln* First isotope = Cl = 75% Second isotope = Cl = [25%] [100% ] \[The percentage abundance in a question should be equal to 100%\] \%~1~ = Percentage abundance of the first isotope = 75% M~1~ = Mass number of the first isotope = 35 \%~2~ = Percentage abundance of the second isotope = 25% M~2~ = Mass number of the second isotope = 37 Let us input all the parameters into the formula I gave initially. RAM = RAM = RAM = 26.25 + 9.25 RAM = 35.5 **\ ** **CASE 2: Relative atomic mass calculations dealing with three isotopes** *Formula:* R.A.M = Where %~1~ = Percentage abundance of the first isotope M~1~ = Mass number of the first isotope \%~2~ = Percentage abundance of the second isotope M~2~ = Mass number of the second isotope \%~3~ = Percentage abundance of the third isotope M~3~ = Mass number of the third isotope **Solved Question with Detailed Explanations on CASE 2** 1\) Calculate the relative atomic mass of oxygen atom with the following percentage abundances O (99.757%), O (0.037%) and O (0.204%) ↓ ↓ ↓ 15.995amu 16.999amu 17.9992 *Soln* First isotope = O = 99.757% Second isotope = O = 0.037% Third isotope = O = [0.204%] [100%\_\_ ] \[The percentage abundance in a question should be equal to 100%\] \%~1~ = 99.757% M~1~ = 16 \[The exact value is 15.995\] \%~2~ = 0.037% M~2~ = 17 \[The exact value is 16.9991\] \%~3~ = 0.204% M~3~ = 18 \[The exact value is 17.9992\] **CASE 3: Relative atomic mass calculations dealing with Atoms** *Formula:* R.A.M = **Solved Questions with Detailed Explanation** 1\) If two atoms of an element X contains 70 atoms of ^9^X and 30 atoms of ^11^X. Calculate the relative atomic mass of X. *Soln* First isotope = ^9^X = 70 atoms Second isotope = ^11^X = [30 atoms] [100 atoms] No of atom~1~ = 70 atoms M~1~ = 9 No of atom~2~ = 30 atoms M~2~ = 11 Let us input all the parameters into the formula I gave initially R.A.M = R.A.M = R.A.M = 6.3 + 3.3 = 9.6 **CASE 4: Relative atomic mass calculations dealing with ratio** **Solved Questions with Detailed Explanations** 1\) An atom has two isotopes X and X in the ratio 1:3 respectively. Calculate the relative atomic mass of X *Soln* Second isotope = X = [75%] [100%] \[The percentage abundance in a question should be equal to 100%\] \%~1~ = 25% M~1~ = 20 \%~2~ = 75% M~2~ = 22 *Formula* RAM = RAM = RAM = 5 + 16.5 RAM = 21.5 **CASE 5: General calculations on relative atomic mass determination.** **Solved Questions with Detailed Explanations** 1\) Chlorine has two isotopes Cl and Cl. If the relative atomic mass of chlorine is 35.5, what is the percentage abundance of both isotopes Cl and Cl? *Solution* RAM = \%~1~ = Cl \%~2~ = Cl \%~1~ = *x* M~1~ = 35 \%~2~ = 100 -- *x* M~2~ = 37 R.A.M = 35.5 Let us input all the parameters into the formual I gave initially RAM = 35.5 = 35.5 = L.C.M 35.5 × 100 = 35*x* + 3700 -- 37*x* Collect like term 3550 -- 3700 = 35*x* -- 37*x* − 150 = − 2*x* Cancel minus sign *x* = 75% Recall, *x* was %~1~ ∴ *x* = 75% of Cl \%~2~ = 100 -- *x* \%~2~ = 100 -- 75 \%~2~ = 25% → Cl The percentage abundance of Cl = 75% The percentage abundance of Cl = 25% 2\) The relative atomic mass of antimony is 121.8. antimony exist as two isotopes. antimony 121 and antimony 123. Calculate the percentage abundance of both isotopes. The chemical symbol of antimony is 5b We are still to use the CASE 1 formula RAM = \%~1~ = ^121^Sb \%~2~ = ^123^Sb \%~1~ = *x* M~1~ = 121 \%~2~ = 100 -- *x* M~2~ = 123 R.A.M = 121.8 Let us input all the parameters into the formual I gave initially RAM = 121.8 = 121.8 = L.C.M 121.8 × 100 = 121*x* + 12300 -- 123*x* 12180 = 121*x* + 12300 -- 123*x* Collect like term 12180 -- 12300 = 121*x* -- 123*x* − 120 = − 2*x* Cancel minus sign = 60% Recall, *x* was %~1~ ∴ *x* = 60% of ^121^Sb \%~2~ = 100 -- *x* \%~2~ = 100 -- 60 \%~2~ = 40% → ^123^Sb The percentage abundance of ^121^Sb = 60% The percentage abundance of ^123^Sb = 40% 3\) Two isotopes X and X with 80% and 20% of the percentage abundance respectively with relative atomic mass of 64.2. What is A? *Soln* \%~1~ = 80% M~1~ = A \%~2~ = 20% M~2~ = 65 R.A.M = 64.2 *Formula* RAM = 64.2 = 64.2 = 64.2 = 0.8A + 13 Collect like term 64.2 -- 13 = 0.8A A = 64 **4. Basic terms you must know** ISOBARS: are atoms with DIFFERENT ELEMENT which have SAME MASS NUMBER but DIFFERENT ATOMIC NUMBER. *Example*: C and N You can see from the example above, the mass number of the two atoms are the same "ISOBARIC EFFECT". ISOTONES are atoms with DIFFERENT ATOMIC NUMBER, DIFFERENT MASS NUMBER but THE SAME NEUTRON NUMBER *Example* N and O Now let calculate the neutron number for each atom For N = Mass number -- Atomic number = 14 -- 7 = 7 For O = Mass number -- Atomic number = 15 -- 8 = 7 Wow!!! They are truly isotones to each other. **ISO-ELECTRONIC SERIES:** These are atoms with same number of electron. *Example* Mg^2+^ and Neon (Ne) For Mg^2+^ Electron number = proton number -- charge Electron number = 12 -- (+ 2) Electron number = 12 -- 2 Electron number = 10 electrons For Ne Neon atom has 10 electrons **ISO-DIAPHERS:** They are nuclide (atoms) with the same number of neutron defect. *For example:* U and Th Neutron defect = Neutron number -- Proton number For U (Uranium) Neutron number = Mass number -- Atomic number Neutron number = 238 -- 92 = 146 Neutron defect = neutron number -- proton number Where proton number = 92 Neutron defect = 146 -- 92 Neutron defect = 54 For Th (Thorium) Neutron number = Mass number -- Atomic number Neutron number = 234 -- 90 Neutron number = 144 Neutron defect = neutron number -- proton number Where proton number = 90 Neutron defect = 144 -- 90 Neutron defect = 54 So you can see that both atoms have the same neutron defect (ISO-DIAPHER'S) Chapter 3 Neil Bohr Theory of Hydrogen Spectrum and Quantum Number **Learning Outcome** When you have completed this chapter, you will be able to: - Understand the theory of Neil Bohr - Solve calculations involved in hydrogen spectrum - Solve calculation questions on De-Broglie wavelength - Understand formulas relating energy, frequency, wavelength and emitted photons. - Understand the concept of Quantum number - Solve calculations involved in Quantum number If you already feel confident about this chapter, why not try the quiz over the page? **Practice Question** 1. Calculate the wavelength and frequency of the first line in the Balmer series when n = 3 and 5. 2. Calculate the frequency of the second line in the Balmer series. 3. How many electrons can exist with principal quantum number n = 4 and l = 3? 4. Calculate the De-broglie wavelength O ten electron with a velocity of 1.00 × 10^7^m/s mass of electron (kg) = 9.11 × 10^−31^kg h = 6.626 × 10^−34^kg m^2^/s 5. Write out the values for l when n = 3 **1. Neil Bohr Theory of Hydrogen Spectrum** Neil Bohr in the year 1913 successfully accounted for the line seen in an hydrogen spectrum. **WHAT IS AN HYDROGEN SPECTRUM?** Hydrogen spectrum is a series of line that is seen when light is emitted from a discharge tube filled with hydrogen and it viewed with a SPECTROSCOPE. **2. Calculations involved in hydrogen spectrum** There are various line seen in the hydrogen spectrum. These lines are arranged serially which are +-----------------------+-----------------------+-----------------------+ | *Spectral line* | *n~1~* | *n~2~* | +=======================+=======================+=======================+ | Lyman series | 1 | 2, 3, 4 | | | | | | Balmer series | 2 | 3, 4, 5 | | | | | | Paschen series | 3 | 4, 5, 6 | | | | | | Bracket series | 4 | 5, 6, 7 | | | | | | Pfund series | 5 | 6, 7, 8 | | | | | | Humphrey series | 6 | 7, 8, 9 | +-----------------------+-----------------------+-----------------------+ *N.B:* The first line is gotten from the n~2~ first value So for the Lyman series: The first line = 2 The second line = 3 The third line = 4 *N.B:* It also applies to all series To calculate the wavelength of these spectral lines, it is done by an Equation which was described by JOHANN BALMER EQUATION. The Equation: Where: = inverse of wavelength = wave number Rh = Rydberg constant for hydrogen atom Most times in Question, we will be asked to convert the wavelength we are to get in a particular question in nanometer so it is best to solve with the constant Rh = 109678cm^−1^. At the end of the calculation you simply convert your answer to nanometer (nm) by \[Answer × 10^−2^ × 10^9^\] = Answer will then be in nanometer (nm). **Solved Questions with Detailed Explanations** 1\) What is the wavelength in nanometers of a photon emitted during a transition from n = 5 state to n = 2 state in the hydrogen atom? *Soln* Using Johann Balmer's Equation The equation is Where \[They are both same\] ∴ Parameter given *N.B:* The lesser value goes to n~1~ and the larger value goes to n~2~ ∴ = 109678cm^-1^ = 109678cm^-1^ = 109678cm^-1^ \[0.21\] = λ = Let us convert to nanometer Wavelength = λ λ in nm = λ in cm x 10^-2^ x 10^9^ λ in nm = 4.342 x 10^-5^cm x 10^-2^ x 10^9^ = 434.2nm 2\) Calculate the wavelength of the first and second line in the Balmerino series in nanometer *Soln* --------------- ------ --------- Spectral line n~1~ n~2~ Balmer series 2 3, 4, 5 --------------- ------ --------- Where 3 is the value for the first line, and 4 is the value for the second line The second line value for n~2~ is 4. So that\'s how it is gotten. The third line value is \_\_\_\_\_\_\_\_\_. Try it. Now let\'s solve together Remember the formula to use The equation is: *N.B:* The lesser value goes to n~1~ and the larger value goes to n~2~ = 109678cm^-1^ = 109678cm^-1^ = 109678cm^-1^ \[0.13889\] = λ = Let us convert to nanometer Wavelength = λ λ in nm = λ in cm x 10^-2^ x 10^9^ λ in nm = 6.565 x 10^-5^cm x 10^-2^ x 10^9^ = 656.5nm **3. De-Broglie Wavelength** **Calculations on De-Broglie Wavelength** To solve questions on this aspect, a formula must be noted. De-Broglie Wavelength = (λ) Where h = planks constant = 6.626 x 10^-34^kg.m^2^/s De-Broglie wavelength (λ) = **Solved Questions with Detailed Explanations** 1\) Calculate the De-Broglie wavelength of an electron with a velocity of 1.00 x 10^6^m/s (electron mass = 9.11 x 10^-31^kg) \[h = 6626 x 10^-34^kg.m^2^/s\] *Soln* De-Broglie wavelength (λ) = Where P is momentum = mass x velocity De-Broglie wavelength (λ) = Parameters given h = 6.626 x 10^-34^kg.m^2^/s m = 9.11 x 10^-31^kg V = 1.00 x 10^6^m/s De-Broglie wavelength (λ) = De-Broglie wavelength (λ) = 7.27 x 10^-10^m 2\) Proton can be accelerated to a speed of light in a particle accelerator. Estimate the wavelength of such a proton moving with at 2.90 x 10^8^m/s \[mass of proton = 1.673 x 10^-27^kg\] *Soln* De-Broglie wavelength (λ) = Where P is momentum = mass x velocity De-Broglie wavelength (λ) = Parameters given h = 6.626 × 10^-34^kg.m^2^/s m = 1.673 × 10^-27^kg V = 2.90 × 10^8^m/s De-Broglie wavelength (λ) = De-Broglie wavelength (λ) = 1.37 × 10^-15^m 3\) What is the De-Broglie wavelength in (cm) of 13.24kg parrot flying at 1.93 x 10^2^km/hr? *Soln* λ = λ = = 9.3349 x 10^-37^m But remember the question says De-Broglie wavelength in (cm) Recall, 1m = 100cm 9.3349 x 10^-37^m = xcm **4) The Concept of Energy, Frequency and Wavelength of emitted photons relationship** There is a formula that relate all of these parameter above which is E = Where **5) Energy and Johann Balmer equation relationship** E = Rhch Where Rh = 109678 x 10^2^ E = 2.18 x 10^-18^ **Solved Questions with Detailed Explanations** 1\) In an hydrogen atom, an electron jumps from the third orbit to the first orbit. What is the frequency of the Spectral line? *Soln* We are asked to determine frequency Recall, E = hF Where h = 6.626 x 10^-34^kgm^2^/s Using the formula below E = 2.18 x 10^-18^ I believe you can still remember how 2.18 x 10^-18^ is gotten *N.B:* The lesser value goes to n~1~ and the larger value goes to n~2~ **6) Energy associated with a mole of photon** To solve question of this type E = n~p~hF Where E = Energy n~p~ = number of photons h = planks constant = 6.626 x 10^-34^kgm^2^/s F = Frequency **Solved Questions with Detailed Explanations** 1\) A laser emits light that has a frequency of 4.73 x 10^12^ herz. Calculate the number of photons in a pulse that emits 1.6 x 10^-3^\] *Soln* Recall, when dealing with number of photon, we use Making np subject of formula Where E = 1.6 x 10^-3^J n~p~ = = 5.105 x 1017 photons **7. Quantum Numbers** There are four Quantum numbers, which are - Principal Quantum number (n) - Azimuthal Quantum number (l) - Magnetic Quantum number (ml) - Spin Quantum number (s) **Principal Quantum Number** **Quick Note** - The principal Quantum number is denoted with n. - The principal Quantum number describes MAIN ENERGY LEVEL of an atom - The principal Quantum number describes the SIZE OF AN ORBITAL. - It also describe shells. **What are Shells?** They are K shell L shell M shell M shell O shell *NOTE:* These shells have numbers which are 1, 2, 3, 4, 5 respectively +-----------------------------------+-----------------------------------+ | *SHELLS* | *NUMBERS (n)* | +-----------------------------------+-----------------------------------+ | K | 1 | | | | | L | 2 | | | | | M | 3 | | | | | N | 4 | | | | | O | 5 | +-----------------------------------+-----------------------------------+ The table above will be very important when we start solving questions. n = 1 2 3 4 5 *N.B:* For principal Quantum number = K L M N O **Azimuthal Quantum number** **Quick Note** - The Azimuthal Quantum number is denoted with L. - The Azimuthal Quantum number is also called subsidiary Quantum number or secondary Quantum number. - The Azimuthal Quantum number describes the number of sub energy level. - It also describes the shape of an orbital. *NOTE:* These orbitals have numbers which are 0, 1, 2, 3, 4 respectively. +-----------------------------------+-----------------------------------+ | *ORBITALS* | *NUMBER (l)* | +-----------------------------------+-----------------------------------+ | s | 0 | | | | | p | 1 | | | | | d | 2 | | | | | f | 3 | | | | | g | 4 | +-----------------------------------+-----------------------------------+ The table above will be very important when we start solving questions. *N.B:* These orbitals have names and respective shapes +-----------------------+-----------------------+-----------------------+ | *Orbitals* | *Name* | *Shape* | +-----------------------+-----------------------+-----------------------+ | s | spin | Spherical | | | | | | p | principal | Dumbbell | | | | | | d | diffuse | Double dumbbell | | | | | | f | fundamental | Complex | | | | | | g | gravitation | Complex | | | | | | h | height | Complex | +-----------------------+-----------------------+-----------------------+ *N.B:* The P orbital is three dimensional which are Px Py Pz l = 0 1 2 3 4 For Azimuthal Quantum number s p d f g **Magnetic Quantum number** **Quick Note** - The magnetic Quantum number is denoted with (ml) - The magnetic Quantum number describes the number of orbitals in a given sub-shell (Degree of degeneracy) - It describes the orientation of orbitals in space. - It ranges from - l to + l i.e. ml = - l to + l **Spin Quantum Number** **Quick Note:** - The spin Quantum number is denoted with (s) - The spin Quantum number describes the direction of electron either clockwise or anticlockwise - The spin Quantum is S is + and - ½ **Relationship between Principal Quantum Number and Azimuthal Quantum Number** These two Quantum number are related by an equation which is l = n - 1 Where l = Azimuthal Quantum number n = Principal Quantum number *Alert* The s, p, d, f, g, h orbitals occupies a maximum number of electron which are 2, 6, 10, 14, 18, 22 respectively. The K, L, M, N, O shells occupies a maximum of electrons which are 2, 8, 18, 32, 50 electrons respectively. **QUICK FORMULAS TO NOTE FOR CALCULATIONS** - To determine the maximum number of electron in a given shell (which can be K, L, M, N, O) with principle Quantum number (n) = 2n^2^ - To determine the maximum number of sub-shell in a given shell with principal Quantum number (n) = n - To determine the maximum number of orbitals in a given shell with principal Quantum number (n) = n^2^. - To determine the maximum number of orbitals in a given sub-shell with azimuthal quantum number (l) = 2l + 1 When we start solving questions, you will understand the concept of quantum number so well. **Solved Questions with Detailed Explanation** 1\) What value of angular momentum quantum are allowed for a principal Quantum number n = 3. *Soln* Remember (n) = 1 2 \[3\] 4 5 For principal Quantum number = K L M N O \(l) = 0 1 \[2\] 3 4 For Azimuthal Quantum number = s p d f g 2\) Give the name, magnetic quantum number and number of orbital for the substance with the following Quantum number n = 3, l = 2. *Soln* If n = 3 l = 2 Let analyze n = 3 Recall, n = 1 2 \[3\] For principal Quantum number = K L M N O 4 means that 4 is in the M shell Now lets analyze l = 2 Recall, l = 0 1 \[2\] 3 4 For azimuthal Quantum number = s p d f g It means that it is in the d orbital Counting the value of ml, it shows that they are five (5). 1 2 3 4 5 ml = − 2, − 1, 0, + 1, + 2 So it means that the d orbital have five degenerate orbital d orbital ↓ ANSWERS BECOMES - - - 3\) For the Azimuthal Quantum number l = 3, write out the values for ml *Soln* Since l = 3 ml = − l to + l ml = − 3 to + 3 There are values from -- 3 to + 3 which are ml = − 3, − 2, − 1, 0, + 1, + 2, + 3. 4\) Consider the M -- Shell \(i) What is the maximum number of election it can take? \(ii) How many sub-shell can it hold, name them? \(iii) How many orbitals are in the shell? *Soln* M -- Shell Principal Quantum number (n) for M shell = 3 Recall, n = 1 2 3 4 5 Principal Quantum number = K L M N O ∴ n = 3 for the M shell i. Maximum number of electron = 2n^2^ ii. Number of subshell in a shell = n iii. Number of orbitals in a shell = n^2^ Chapter 4 Relative Molecular Mass Calculations **Learning Outcome** When you have completed this chapter, you will be able to - know what relative molecular mass is - determine the different compound and also compound with water of crystallization (H~2~O) - If you feel already confident about these chapter, why not try the Quiz over the page? **Practice Questions** 1. Calculate the relative molecular mass of sodium tetraoxosulphate (VI) salt Na~2~SO~4~ \[Na = 23, S = 32, O = 16\]. 2. Calculate the relative molecular mass of Ca~3~(PO~4~)~2~ \[Ca = 40, P = 31, O = 16\]. 3. Calculate the relative molecular mass of washing soda. \[Na = 23, C = 12, O = 16, H = 1\]. **1. Relative Molecular Mass** What is Relative Molecular Mass? The relative molecular mass is the mass of an element or a compound is the number of times the average mass of one molecules of it is heavier than one -- twelfth the mass of one atom of carbon -- 12. To be able to calculate the relative molecular mass of compound, the atomic mass of atleast the first 20 elements must be known +-----------------------+-----------------------+-----------------------+ | *Atomic Number* | *Element* | *Atomic Mass* | +=======================+=======================+=======================+ | 1 | Hydrogen | 1.008 | | | | | | 2 | Helium | 4.0026 | | | | | | 3 | Lithium | 6.941 | | | | | | 4 | Beryllium | 9.0122 | | | | | | 5 | Boron | 10.81 | | | | | | 6 | Carbon | 12.011 | | | | | | 7 | Nitrogen | 14.0067 | | | | | | 8 | Oxygen | 15.9994 | | | | | | 9 | Fluorine | 18.9984 | | | | | | 10 | Neon | 20.183 | | | | | | 11 | Sodium | 22.9898 | | | | | | 12 | Magnesium | 24.305 | | | | | | 13 | Aluminum | 26.9815 | | | | | | 14 | Silicon | 28.086 | | | | | | 15 | Phosphorus | 30.9738 | | | | | | 16 | Sulphur | 32.06 | | | | | | 17 | Chlorine | 35.453 | | | | | | 18 | Argon | 39.948 | | | | | | 19 | Potassium | 39.102 | | | | | | 20 | calcium | 40.008 | +-----------------------+-----------------------+-----------------------+ **Solved Questions with Detailed Explanations** 1\) Calculate the relative molecular mass of CaCO~3~ \[Ca = 40, C = 12, O = 16\] *Soln* Atomic mass of Ca = 40 Atomic mass of C = 12 Atomic mass of O = 16 Ca C O~3~ Relative molecular mass = 40 + 12 + 16 × 3 Relative molecular mass = 40 + 12 + 48 = 100 2\) Calculate the relative molecular mass of soda ash \[Na~2~CO~3~\] \[Na = 23, C = 12, O = 16\] *Soln* Atomic mass of Na = 23 Atomic mass of C = 12 Atomic mass of O = 16 Na~2~ C O~3~ Relative molecular mass = 23 × 2 + 12 + 16 × 3 Relative molecular mass = 46 + 12 + 48 3\) Calculate the relative molecular mass of H~2~SO~4~ \[H = 1, S = 32, O = 16\] *Soln* Atomic mass of H = 1 Atomic mass of S = 32 Atomic mass of O = 16 H~2~ S O~4~ Relative molecular mass = 1 × 2 + 32 + 16 × 4 Relative molecular mass = 2 + 32 + 64 Relative molecular mass = 98 4\) Calculate the relative molecular mass of Lead (II) Trioxonitrate (IV) \[Pb = 207, N = 14, O = 16\] *Soln* First lead (II) trioxonitrate (IV) is a compound *Question:* What is the chemical formula of the compound? Lead (II) Trioxonitrate (V) Pb^2+^ NO~3~^−1^ Pb~1~(NO~3~)~2~ or Pb(NO~3~)~2~ This is the chemical formula of the compound Pb(NO~3~)~2~ = PbN~2~O~6~ Atomic mass of Pb = 207 Atomic mass of N = 14 Atomic mass of O = 16 Pb N~2~ O~6~ Relative molecular mass = 207 + 14 × 2 + 16 × 6 Relative molecular mass = 207 + 28 + 96 Relative molecular mass = 331 5\) Calculate the relative molecular mass of Al~2~O~3~.2H~2~O \[Al = 27, O = 16, H = 1\] *Soln* Atomic mass of Al = 27 Atomic mass of O = 16 Atomic mass of H = 1 \+ Al~2~ O~3~ ↑ 2 H~2~ O Relative molecular mass = 27 × 2 + 16 × 3 + 2\[1 × 2 + 16\] Relative molecular mass = 54 + 48 + 2\[2 + 16\] Relative molecular mass = 102 + 2\[18\] Relative molecular mass = 102 + 36 Relative molecular mass = 138 6\) Calculate the relative molecular mass of potash alum \[K = 39, Al = 27, S = 32, O = 16, H = 1\] *Soln* Potash alum has the chemical formula to be KAl(SO~4~)~2~. 12H~2~O Atomic mass of k = 39 Atomic mass of Al = 27 Atomic mass of S = 32 Atomic mass of O = 16 Atomic mass of H = 1 \+ K Al S~2~ O~8~ ↑ 12H~2~ O Relative molecular mass = 39 + 27 + 32 × 2 + 16 × 8 + 12\[1 × 2 + 16\] Relative molecular mass = 39 + 27 + 64 + 128 + 12\[2 + 16\] Relative molecular mass = 258 + 12\[18\] Relative molecular mass = 258 + 216 Chapter 5 Percentage composition of each elements and water of crystallization in a compound determination **Learning Outcome** When you have completed this chapter, you will be able to: - Determine the percentage of each elements in a compound. - Determine the percentage of water of crystallization. If you already feel confident about these chapters, why not try the quiz over the page? **Practice Questions** 1. Calculate the percentage composition of Nitrogen in NH~4~NO~3~ \[H = 1, N = 14, O = 16\] 2. What is the percentage composition of water (H~2~O) in CuSO~4~. 5H~2~O? 3. What is the percentage composition of oxygen in Al~2~(SO~4~)~3~. 2H~2~O \[Al = 27, S = 32, H = 1, O = 16\]? 4. Calculate the percentage composition of each elements in Na~2~CO~3~ \[Na = 23, C = 12, O = 16\]. 5. What is the percentage composition of sodium in sodium hydroxide \[Na = 23, O = 16, H = 1\]? **Percentage Composition** In this chapter, solving questions on percentage composition will be made so easy. Percentage composition = % In this chapter, determination of percentage composition will be in two CASES. **CASE 1: Determination of percentage composition of each element in a compound.** \% = **Solved Questions with Detailed Explanations** 1\) Calculate the percentage composition of each element in H~2~SO~4~ *Soln* The compound = H~2~SO~4~ *Elements present Atomic mass No of atom* H (Hydrogen) = 1 2 atoms S (Sulphur) = 32 1 atom O (Oxygen) = 16 4 atoms Now to get the no of atoms of each elements present in a compound H~2~SO~4~ Believe it is clear now Molar mass of the compound H~2~SO~4~ H~2~SO~4~ Molar mass = 1 × 2 + 32 + 16 × 4 Molar mass = 2 + 32 + 64 = 98g/mol Now lets determine the percentage composition of each element in H~2~SO~4~ Recall, the formula \% = \%H = = 2.041% \%S = = 32.65% \%O = = [65.31%] [100%] 2\) Calculate the percentage of each element of Iron (III) oxide *Soln* The compound is Iron (III) oxide Iron (III) oxide Fe^3+^ O^2−^ The chemical formula = Fe~2~O~3~ *Elements present Atomic mass No of atom* Fe (Iron) = 56 2 atoms O (Oxygen) = 16 3 atoms Now to get the no of atoms of each elements present in a compound is Fe~2~O~3~ Believe it is clear now Molar mass of the compound Fe~2~O~3~ Fe~2~O~3~ Molar mass = 56 × 2 + 16 × 3 Molar mass = 112 + 48 = 160g/mol Now lets determine the percentage composition of each element in Fe~2~O~3~ Recall, the formula \% = \%Fe = = 70% \%O = = [30%] [100%] 3\) Calculate the percentage composition of Oxygen is Sulphur (IV) oxide *Soln* The compound is Sulphur (IV) oxide The chemical formula = SO~2~ *Elements present Atomic mass No of atom* S (Sulphur) = 32 1 atoms O (Oxygen) = 16 2 atoms Now to get the number of atoms of each elements present in a compound is SO~2~ Believe it is clear now Molar mass of the compound SO~2~ SO~2~ Molar mass = 32 + 16 × 2 Molar mass = 32 + 32 = 64g/mol Recall, the formula \% = \%O = = 50% **CASE 2: Determination of Percentage Composition of Water of Crystallization (H~2~O)** Now to solve question of this such the initial formula will be altered. The part of the formula that has: Atomic mass of element × Number of atom will be changed to molar mass of water of crystallization. So with this change it becomes, \% = So, when question on this CASE is asked the formula above should be used. **Solved Questions with Detailed Explanations** 1\) Calculate the percentage composition of water of crystallization in the compound Na~2~CO~3~. 10H~2~O \[Na = 23, C = 12, O = 16, H = 1\] *Soln* The formula to apply is the formula under CASE 2 \% = The compound is Na~2~CO~3~. \[10H~2~O\] ↓ The water of crystallization part Let us determine the molar mass of the compound Molar mass of Na~2~CO~3~. 10H~2~O 23 × 2 + 12 + 16 × 3. 10\[1 × 2 + 16\] 46 + 12 + 48. 10\[2 + 16\] 46 + 12 + 48. 10\[18\] 106 + 180 = 286g/mol The molar mass of water of crystallization Molar mass of IO0H~2~O 10\[1 × 2 + 16\] 10\[2 + 16\] 10\[18\] Molar mass of water of crystallization = 180g/mol 2\) Calculate the percentage composition of all element in Na~2~CO~3~. 10H~2~O\] *Soln* The compound is Na~2~CO~3~. IOH~2~O *Elements present Atomic mass No of atom* Na (Sodium) = 23 2 atom C (Carbon) = 12 1 atom O (Oxygen) = 16 13 atom H (Hydrogen) = 1 20 atom Now to get the number of atoms of each element present in a compound is Na~2~CO~3~. IOH~2~O Believe it is clear now Molar mass of the compound \[Na~2~CO~3~. IOH~2~O\] Molar mass of Na~2~CO~3~. IOH~2~O 23 × 2 + 12 + 16 × 8. 10\[1 × 2 + 16\] 46 + 12 + 48. 10\[2 + 16\] 106 + 10\[18\] Molar mass = 106 + 180 = 286g/mol Now let solve the percentage composition of each element in Na~2~CO~3~. IOH~2~O From the formula, I gave for CASE 1. [100%] Chapter 6 Empirical and Molecular Formula **Learning Outcome** When you have completed this chapter, you will be able to: - Know the meaning of empirical and molecular formula of a compound. - Solve calculations on empirical formula - Solve calculations on molecular formula If you already feel confident about these chapter, why not try the Quiz over the page? **Practice Questions** 1. An organic compound contains 72%, 12% H and 16% O by mass. What is the empirical formula of the compound \[H = 1, C = 12, O = 16\]. 2. Find the empirical formula of the following compounds from their percentage composition by mass a. b. 3. 6g of metal M react completely with 23.66g of chlorine to form 29.66g of metallic chloride. Determine the empirical formula of the metallic chloride \[M = 27, Cl = 35.5\]. 4. If 0.5 mole of a monoalkanoic acid weighs 44g, determine the molecular formula and name the acid \[H = 1, C = 12, O = 16\]. 5. An organic compound has the empirical formula CH~2~. If its molecular mass is 42g/mol, what is the molecular formula \[H = 1, C = 12\]. 6. A 4.25g sample of a compound that contains only carbon, hydrogen and oxygen was burnt in an atmosphere of pure O~2~. This produced 9.34g CO~2~ and 5.09g H~2~O. What is the empirical formula of the compound? **1. Empirical Formula** What is empirical formula? Is the sample of formula of a compound that tell which elements are present and the simplest whole number ratio of their atoms but not necessarily the actual number of atom in a given molecules. **2. Molecular Formula** What is Molecular Formula? Is the true formula of a compound which expresses the actual number of atoms of each elements present in a molecules in the compound. Let us take an example to understand this concept better, +-----------------------+-----------------------+-----------------------+ | *Example of compound* | *Molecular formula* | *Empirical formula* | +=======================+=======================+=======================+ | Benzene | C~6~H~6~ | CH | | | | | | | divide through by 6 | | +-----------------------+-----------------------+-----------------------+ | Ethane | C~2~H~6~ | CH~3~ | | | | | | | divide through by 2 | | +-----------------------+-----------------------+-----------------------+ | Ethyne | C~2~H~2~ | CH | | | | | | | divide through by 2 | | +-----------------------+-----------------------+-----------------------+ **3. Calculations in Empirical Formula** To solve questions on this aspect, it will be in 3 CASES. **CASE 1: Empirical formula determination percentage given** **Solved Questions with Detailed Explanations** 1\) A compound contains 92.31% carbon and 7.69% hydrogen. Find the empirical formula of the compound \[C = 12, H = 1\] *Soln* Elements Carbon Hydrogen Element symbol C H Percentage of element [92.31%] [7.69%] Divide by atomic mass 12 1 [7.69 7.69] Divide by the smallest value 7 Mole ratio 1 1 ∴ The empirical formula of the compound is CH. 2\) Analysis of a sample of an organic compound showed H to contain 39.9%, 6.9% hydrogen and 53.2% oxygen. Calculate the empirical formula \[C = 12, H = 1, O = 16\] +-----------------+-----------------+-----------------+-----------------+ | Element | Carbon | Hydrogen | Oxygen | | | | | | | Element symbol | C | H | O | | | | | | | Percentage of | [39.9%]{.underl | [6.9%]{.underli | [53.2%]{.underl | | element | ine} | ne} | ine} | | | | | | | Divide by | 12 | 1 | 16 | | atomic mass | | | | +=================+=================+=================+=================+ | | [3.33 6.9 | | | | | 3.33]{.underlin | | | | | e} | | | +-----------------+-----------------+-----------------+-----------------+ | Divide by the | 1 | 3.33 | 1 | | smallest | | | | | | | 2.10 | | | Mole ratio | | | | +-----------------+-----------------+-----------------+-----------------+ 3\) What is the empirical formula of cyrolite if it contains 32.85% Na, 12.85% Al and 54.30% F by mass? \[Na = 23, Al = 27, F = 19\] +-----------------+-----------------+-----------------+-----------------+ | Element | Sodium | Aluminium | Fluorine | | | | | | | Element symbol | Na | Al | F | | | | | | | Percentage of | [32.85%]{.under | [12.85%]{.under | [54.30%]{.under | | element | line} | line} | line} | | | | | | | Divide by | 23 | 27 | 19 | | atomic mass | | | | +=================+=================+=================+=================+ | | [1.428 0.476 | | | | | 2.858]{.underli | | | | | ne} | | | +-----------------+-----------------+-----------------+-----------------+ | Divide by the | 3 | 0.476 | 6 | | smallest | | | | | | | 1 | | | Mole ratio | | | | +-----------------+-----------------+-----------------+-----------------+ **CASE 2: Empirical formula determination with masses given** **Solved Questions with Detailed Explanations** 1\) An organic compound is made up of 0.72g of carbon, 0.16g of hydrogen and 0.32g of oxygen. Calculate the empirical formula of the organic compound \[C = 12, H = 1, O = 12\] *Soln* +-----------------+-----------------+-----------------+-----------------+ | Element | Carbon | Hydrogen | Oxygen | | | | | | | Element symbol | C | H | O | | | | | | | Percentage of | [0.72%]{.underl | [0.16%]{.underl | [0.32%]{.underl | | element | ine} | ine} | ine} | | | | | | | Divide by | 12 | 1 | 16 | | atomic mass | | | | +=================+=================+=================+=================+ | | [0.06 0.16 | | | | | 0.02]{.underlin | | | | | e} | | | +-----------------+-----------------+-----------------+-----------------+ | Divide by the | 3 | 0.02 | 1 | | smallest | | | | | | | 8 | | | Mole ratio | | | | +-----------------+-----------------+-----------------+-----------------+ The empirical formula of the compound is C~3~H~8~O 2\) If 6.75g of an oxide of lead was reduced to 6.12g of metal. Calculate the empirical formula of the oxide \[Pb = 207, O = 12\]. *Soln* The compound is an oxide of Lead. So the compound consist of Lead (Pb) and Oxygen (O). Mass of the Oxide \[Pb, O\] = 6.75g Mass of Lead \[Pb\] = 6.12g To get the mass of oxygen, subtract the mass of the oxide from the mass of Lead (Pb). Pb + O = Mass of Lead Oxide Pb + O = 6.75g Mass of Lead = 6.12g 6.12g + O = 6.75g O = 6.75g -- 6.12g O = 0.63g of Oxygen \[Mass of Oxygen\] +-----------------+-----------------+-----------------+-----------------+ | Element | Lead | Oxygen | | | | | | | | Element symbol | Pb | O | | | | | | | | Percentage of | [6.12%]{.underl | [0.63%]{.underl | | | element | ine} | ine} | | | | | | | | Divide by | 207 | 16 | | | atomic mass | | | | +=================+=================+=================+=================+ | | [0.02957 | | | | | 0.03938]{.under | | | | | line} | | | +-----------------+-----------------+-----------------+-----------------+ | Divide by the | 0.02957 | 1.33 | | | smallest | | | | | | 1 | | | | Mole ratio | | | | +-----------------+-----------------+-----------------+-----------------+ 3\) A 0.10g sample of alcohol containing C, H, and O was burnt completely in Oxygen to form CO~2~ and H~2~O. These were trapped and weight to give 0.1919g and 0.1172g of CO~2~ and H~2~O respectively. What is the empirical formula? \[C = 12, H = 1\] *Soln* Mass of sample = 0.10g Mass of CO~2~ = 0.1919g Mass of H~2~O = 0.1172g There is a step that must be followed when solving question of this such. *Step:* Extract the mass of Carbon (C), Hydrogen (H) from CO~2~ and H~2~O respectively. For Carbon \[from CO~2~\] Mass of Carbon = Molar mass of Carbon (C) = 12g/mol Molar mass of CO~2~ = 12 + 16 × 2 = 44g/mol Mass of Carbon = Mass of Carbon = 0.05234g of Carbon For Hydrogen \[from H~2~O\] Mass of Carbon = Molar mass of Hydrogen (H~2~) = 1 × 2 = 2g/mol Molar mass of H~2~O = 1 × 2 + 16 = 18g/mol Mass of Hydrogen = Mass of Hydrogen = 0.01302g of Hydrogen Let us add the mass of C and H together. Mass of Carbon + Mass of Hydrogen 0.05234g + 0.01302g = 0.06536g Now, you can see that the mass is not up to that of the original mass (0.10g) sample. This means that Oxygen mass is not yet gotten. So let us calculate for it. Mass of Sample = Mass of C + Mass of H + Mass of O 0.10g = 0.05234g + 0.01302g + Mass of O 0.10g = 0.06536g + Mass of O Mass of O = 0.10g -- 0.06536g Mass of O = 0.03464g of Oxygen ∴ The mass of each element in the compound is determined Mass of Carbon = 0.05234g Mass of Hydrogen = 0.01302g Mass of Oxygen = 0.03464g So let determine the empirical formula of the sample. +-----------------+-----------------+-----------------+-----------------+ | Element | Carbon | Hydrogen | Oxygen | | | | | | | Element symbol | C | H | O | | | | | | | Percentage of | [0.05234%]{.und | [0.01302%]{.und | [0.03464%]{.und | | element | erline} | erline} | erline} | | | | | | | Divide by | 12 | 1 | 16 | | atomic mass | | | | +=================+=================+=================+=================+ | | [0.004362 | | | | | 0.01302 | | | | | 0.002165]{.unde | | | | | rline} | | | +-----------------+-----------------+-----------------+-----------------+ | Divide by the | 2 | 0.002165 | 1 | | smallest | | | | | | | 6 | | | Mole ratio | | | | +-----------------+-----------------+-----------------+-----------------+ **CASE 3: Empirical formula determination with volume is given** **Solved Question with Detailed Explanations** 1\) 20cm^3^ of a gaseous hydrocarbon were mixed with 10cm^3^ of oxygen, and the mixture exploded. At room temperature, 60cm^3^ of the gas were left, 40cm^3^ of the gas \[Carbon (IV) Oxide\] were absorbed by sodium hydroxide, leaving 20cm^3^ of oxygen. Find the empirical formula of the hydrocarbon. *Soln* *N.B:* Since they are asking for the empirical formula of the hydrocarbon, what will be in your mind is carbon and hydrogen. C*~x~*H~y~ + O~2~ → CO~2~ + H~2~O Equation of combustion Volume of hydrocarbon = 20cm^3^ Volume of oxygen left = 20cm^3^ Volume of carbon (IV) oxide = 40cm^3^ Volume of H~2~O = 60cm^3^ From mole concept n = No of mole n = n = at standard temperature and pressure (S.T.P) ∴ so *1 Step:* is to calculate the mass of CO~2~. So that the mass of carbon can be extracted. *2 Step:* is to calculate the mass of H~2~O, so that the mass of Hydrogen can be gotten. From the formula aforementioned Remember that the volume must be in the same S.I units. ALERT: 1dm^3^ = 1000cm^3^ ∴ For 40cm^3^ of CO~2~ 1 dm^3^ = 1000cm^3^ x = 40cm3 x = = 0.04dm^3^ Then mass = Where CO~2~ = 12 + (16 × 2) = 44gmol^−1^ Mass = Mass = 0.07857g of CO~2~ Mass of Carbon is Then, × Mass of CO~2~ × 0.07857 = 0.02143g of carbon For H~2~O (Water) Alert: Water is also referred to as OXIDANE Recall: Mass = For 60cm^3^ of H~2~O 1 dm^3^ = 1000cm^3^ x = 40cm3 x = x = 0.06dm^3^ Then mass becomes = Where H~2~O = (1 × 2) + 16 = 18gmol^−1^ Mass = Mass = 0.0482143g of H~2~O Mass of Hydrogen is Then, × Mass of H~2~O × 0.04821439 = 0.0053571g of Hydrogen Application of the concept of empirical formula we get +-----------------+-----------------+-----------------+-----------------+ | Element | Carbon | Oxygen | | | | | | | | Element symbol | C | H | | | | | | | | Divide by the | 0.021430g | 0.0053571g | | | molar mass | | | | | | 0.021430 | 0.0053571g | | | Divide by the | | | | | smallest | [0.021430]{.und | 0.0053571 | | | | erline} | | | | | | 1 | | | | 12 | | | | | | 0.0053571 | | | | 0.0017900 | | | +=================+=================+=================+=================+ | | [0.0017900 | | | | | 0.0053571]{.und | | | | | erline} | | | +-----------------+-----------------+-----------------+-----------------+ | Mole ratio | 0.01790 | 3 | | | | | | | | | 1 | | | +-----------------+-----------------+-----------------+-----------------+ The empirical formula is CH~3~ 2\) On heating under suitable conditions, 1 litre of amono atomic gas × combines with 1½ litre of oxygen to form an oxide. What is the formula of the oxide? *Soln* **X O** Volume that reacted 1L 1½ litres Mole of molecules that reacted 2 3 Since X is monoatomic and oxygen is diatomic The atoms of X and O combine in the ratio 2 to 6 i.e. **X O** Mole of atoms 2 6 Mole ratio 1 3 The formula of the oxide is XO~3~ **Molecular Formula Determination** **Solved Questions with Detailed Explanations** 1\) Analysis of a sample of an organic compound showed it to contain 39.9% carbon, 6.9% hydrogen and 53.2% oxygen. \(a) Calculate the empirical formula *Soln* +-----------------+-----------------+-----------------+-----------------+ | Element | Carbon | Hydrogen | Oxygen | | | | | | | Element symbol | C | H | O | | | | | | | Percentage of | [39.9%]{.underl | [6.9%]{.underli | [53.2%]{.underl | | element | ine} | ne} | ine} | | | | | | | Divide by | 12 | 1 | 16 | | atomic mass | | | | +=================+=================+=================+=================+ | | [3.3 6.9 | | | | | 3.33]{.underlin | | | | | e} | | | +-----------------+-----------------+-----------------+-----------------+ | Divide by the | 1 | 3.33 | 1 | | smallest | | | | | | | 2 | | | Mole ratio | | | | +-----------------+-----------------+-----------------+-----------------+ The empirical formula of the sample is CH~2~O \(b) The molecular formula: Since the relative molecular mass was given to be 60 Simply use the formula below \[Empirical Formula\]~n~ = \[Molecular Formula\] \[CH~2~O\]n = \[Molecular Formula\] ∴ \[Empirical Mass\]n = \[Molecular mass\] Empirical mass of CH~2~O is 12 + 1 × 2 + 16 = 30 Molecular mass = 60 \[30\]n = \[60\] n = ∴ Recall; \[Empirical formula\]n = \[Molecular formula\] \[CH~2~O\]~2~ = C~2~H~4~O~2~ ∴ The molecular formula of the sample is C~2~H~4~O~2~ Chapter 7 Gas Law **Learning Outcome** When you have completed this chapter, you will be able to: - Understand the variables used to describe Gases - Solve calculation questions on the Gas Laws which include Boyle's law, Charles's law, Amonton's law, General gas law, Dalton's law of partial pressure, Avogatro's Hypothesis, Ideal gas law, Graham's law of diffusion and Gray-Lussac's law of combining volume. If you already feel confident about these chapter, why not try the Quiz over the page? **Practice Questions** 1. The partial pressure of oxygen in a sample of air is 425mmHg and the total pressure is 780mmHg. What is the molefraction of oxygen gas? 2. A given amount of gas occupies 10dm^3^ at 4atm and 273^o^C, the number of mole of gas present is \_\_\_\_\_\_\_\_ 3. If 30cm^3^ of oxygen gas diffuses through porous plug in 7 secs, how long will it take 60cm^3^ of chlorine to diffuse through the same plug? 4. If 60cm^3^ of a gas is heated from 27^o^C to 50^o^C, what is the new volume of the gas at constant pressure? 5. If 11g of a gas occupies 5.6dm^3^ at stp, calculate its vapour density. 6. The mass of 800cm^3^ of a gas x at stp is 1g. What is the molar mass of X? 7. 1500cm^3^ of carbon dioxide escaped through a porous pot in 6 minutes if 800cm^3^ of propane is put in the same porous pot, how long will it take to escape from the pot? 8. The densities of two gases X and Y are 0.5g/dm^3^ and 2.0g/dm^3^ respectively. What is the rate of diffusion of X relative to Y \[Hint i.e. R~x~ : R~y~ or R~x~/R~y~\]. 9. What volume was previously occupied by a given amount of chlorine in gas jar that occupied 85cm^3^ when heated from room temperature to 45^o^C? 10. Calculate the volume that will be occupied by 208cm^3^ of propane if its pressure changes from 24atm to 48atm. **1. Variables for Describing Gas Behaviour** The behaviour of a gas is usually described in terms of four major variables which are Volume (V) Pressure (P) Temperature (T) Number of mole (n) Now before we learn all the Gas Laws properly, we need to be familiar with all these aforementioned variables. **VARIABLES FOR DESCRIBING GAS BEHAVIOUR** **1) Volume (V)** Now volume can be expressed in various S.I units *N.B:* The volume of an gas is the volume of the container in which it is placed or put. The units of volume are L, mL, cm^3^, dm^3^, m^3^ Where L = litres mL = milliliters cm^3^ = centimeter cube dm^3^ = decimeter cube m^3^ = metre cube All these volume metrics (units) have their respective relationships when it comes to conversion. *N.B:* 1L = 1000mL = 1000cm3 = 1dm3 = 1 × 10^−3^m^3^ So the volume of a particular gas can be expressed in any of these S.I. units. It solely depends on which is used in the calculation question. **2) Pressure (P)** Pressure can be expressed in various S.I. units *N.B:* The pressure of a gas is the total force exerted by a gas per unit on the wall of its container. The units of pressure are Atm, mmHg, torr, N/m^2^, pascal Where atm = atmosphere mmHg = millimeter mercury torr = torriceli N/m^2^ = Newton per metre squared Pascal = pascal All these units of pressure have their respective relationship when it comes to conversion. N.B.: 1 atm = 760mmHg = 760 torr = 101325N/m^2^ = 101325 pascal So the pressure of a particular gas can be expressed in any of these S.I. units, it solely depends on which is used in the calculation question. **3) Temperature (T)** In Gas Law calculations, Temperature is usually expressed in Kelvin. *N.B:* But most times the temperature will be in degree Celsius. So how do you convert to kelvin; it is very easy to do so. Simply use the formula below ^o^C + 273 = Kelvin **4) Number of Mole (n)** Number of mole is expressed with a symbol which is "n" (small letter n not capital letter) The number of mole of a substance is the quantity of a gas that contains 6.02 × 10^23^ molecules. Ok let us stop here for now. The fourth variable will be explained exhaustively in the next chapter. **2. The Gas Law** **BOYLE'S LAW** The Boyle's law is also called Manotte law. This law was discovered by a man called Robert Boyle's. The law states that the volume of a forced mass of gas is inversely proportional to its pressure provided temperature is kept constant. *N.B:* Since this law works when temperature is kept constant, it means that when solving calculation question on GAS LAW, we often hear in most question "temperature is held constant". Our mind should just go to the Boyle's law equation. Mathematically, V \[This expression means that volume is inversely proportional to pressure\] ∴ We introduce a constant k V ; Making k subject of formula k = PV this is why the Boyle's law equation is P~1~V~1~ =P~2~V~2~ Where P~1~ = Initial pressure V~1~ = Initial volume P~2~ = Final pressure V~2~ = Final volume Note this below For instance, we are solving a question on Boyle's law and the first volume (V~1~) is 200cm^3^ and second volume (V~2~) is 0.8dm^3^. **What do we do?** We are to simply make sure that both volumes (V~1~ and V~2~) in the question should be in the same S.I units. This also applies to pressure and also when solving questions on other laws to be discussed as we proceed. *N.B:* Boyle's law is not only represented mathematically but it is also represented graphically. **GRAPHICALLY REPRESENTATION OF BOYLE'S LAW** \(a) (b) PV P \(c) P (d) V V \(e) P V All the graphs above represent Boyle's law. In each case Pressure (P) is inversely proportional to Volume (V). *N.B:* As Pressure (P) increases, Volume (V) decrease which is inverse proportion. To calculate questions on Boyle's Law, it will be in two cases. **CASE 1: When the parameters of volume and pressure are in same S.I unit.** **Solved Questions with Detailed Explanation** \(1) A given mass of gas occupies a volume of 400cm^3^ at 950mmHg. Calculate the volume occupied by the same gas at 750mmHg if the temperature is held constant. *Solution* *Step 1:* Check for what is kept constant as you can see in the question above temperature is held constant then it involves using the Boyle's law equation. *Step 2:* Bring out the parameters involved in the question with their S.I units. *Step 3:* Calculate with the formula P~1~V~1~ = P~2~V~2~ Where P~1~ = 950mmHg V~1~ = 400cm^3^ P~2~ = 750mmHg V~2~ = ?cm^3^ Now you can see that the two pressure provided in the question are of same S.I units which is mmHg. Applying the formula P~1~V~1~ = P~2~V~2~ From the question we are to make V~2~ subject of formula P~1~V~1~ = P~2~V~2~ Divide both sides by the coefficient of V~2~ which is P~2~. V~2~ = V~2~ = V~2~ = 506.67cm^3^ \(2) Calculate the volume that will be occupied by 210cm3 of propane if its pressure changes from 25atm to 50atm. *Soln* V~2~ = V~2~ = V~2~ = 105cm^3^ **CASE 2: When the parameters of volume and pressure are not in same S.I units** **Solved Questions with Detailed Explanations** 1\) 290cm^3^ of a gas in a gas jar was observed to exert a pressure of 700mmHg. The volume occupied by the gas was increased to 420cm^3^. Determine the new pressure it will exert \[in atmosphere \[atm\]\] *Soln* From the question, it started the new pressure which is P~2~, we are to calculate for should be in atmosphere i.e. atm. Do not make this mistake to solve the question like that without checking the instructions given by the examiner. Parameters provided P~2~ = P~2~ = = 0.64atm The question can still be solved in another way. P~2~ = P~2~ = = 483.33mmHg Now the P~2~ value we just got is 483.33mmHg. Let us convert it to atm. Recall the conversion factor to use 760mmHg = 1atm 483.33mmHg = *x*atm *x*atm = *x*atm = 0.64atm So use the one that suits you. **CHARLES' LAW** This law discovered by a man called Jacques Charles. This law states that the volume of a fixed mass of gas is directly proportional to its temperature provided pressure is kept constant. Mathematically, ∴We introduce a constant k Making k subject of formula This is why the Charles' law equation is Where *N.B:* Charles' law is not only presented mathematically, it is also represented graphically. **GRAPHICAL REPRESENTATION OF CHARLES' LAW** \(a) V (b) V -273 0 T (^o^c) T(k) \(c) (d) T(k) V or T \(e) V V~T~(k^−1^) All the graphs above represent Charles' law. In each cases, temperature is directly proportional to Volume (V) *N.B:* As volume (V) increase, temperature (T) increases which is direct proportion. **Solved Questions with Detailed Explanation** 1\) Convert 180^o^C to Kelvin temperature *Soln* Recall; ^o^C + 273 = Kelvin ^o^C = 180^o^C 180^o^C + 273 = 453 kelvin (k) Parameter provided ^o^C = 45^o^C 3\) 600cm^3^ of a certain gas at constant pressure had a temperature of 49^o^C. At what temperature in Kelvin will it occupy 800cm^3^? *Soln* Parameters provided V~1~ = 600cm^3^ V~2~ = 800cm^3^ T~1~ = 49^o^C \[*N.B:* The temperature should be converted to Kelvin before applying the formula\]. To convert ^o^C to Kelvin temperature ^o^C + 273 = Kelvin temperature 49^o^C + 273 = 322 Kelvin T~2~ = ? Now let us calculate using the formula Making T~2~ subject of formula T~2~ = 429.33 kelvin 4\) The temperature of a gas is 37^o^C. At what temperature would its original volume by halved if the pressure is kept constant? *Soln* Parameters provided T~1~ = 37^o^C \[cause that is the first temperature mentioned\] Remember the temperature should be converted to Kelvin temperature before applying the formula. ^o^C + 273 = Kelvin 37^o^C + 273 = 310 Kelvin T~2~ = *x* V~1~ = *x* \[V~1~ is the original volume and also it is unknown\] Formula to apply Making T~2~ subject of formula Now let us input the parameters into the formula T~2~ = Now let us re-structure the equation above T~2~ = T~2~ = T~2~ = T~2~ = 155 kelvin **AMONTON'S LAW** Amonton's law is also called pressure law. This law was discovered by a man called Guillaume Amontons. We have previously discussed that for Boyle's law temperature is the parameter that is kept constant, while for Charles' law pressure is the parameter that is kept constant. Now for the pressure law or Amonton's law volume is kept constant; so here we relate pressure and temperature. The pressure law states that the pressure of a fixed volume of a gas is directly proportional to its temperature. *N.B:* Since this law works when volume is kept constant, it means that when solving calculations questions. On this law, we will most times hear in the question "if volume is held constant" our mind should just tell us to pressure law equation. Mathematically, P α T \[This expression means that pressure is directly proportional to temperature\] ∴ We introduce constant k P = KT ; Making K subject of formula K = This is why the pressure law equation is Where **Solved Questions with Detailed Explanations** 1\) At 40^o^C, a given quantity of a gas was found to exhibit a pressure of 480mmHg. What pressure will it exact at 74^o^C? *Soln*\ Parameters provided T~1~ = 40^o^C \[This is the first temperature mentioned\] P~1~ = 480mmHg \[This is the first pressure mentioned\] P~2~ = *x* \[This is what we are to calculate for\] T~2~ = 74^o^C \[This is the second temperature mentioned\] As I earlier said in Charles' law, before solving any equation on the gas law that requires temperature as a parameter, do well to check the S.I unit of the temperature. Most time it is given in ^o^C. Simply convert it to Kelvin. Recall, the formula ^o^C + 273 = Kelvin For T~1~ = 40^o^C 40^o^C + 273 = 313 Kelvin For T~2~ 74^o^C + 273 = 347 Kelvin Parameters rewritten T~1~ = 40^o^C = 313K P~1~ = 480mmHg P~2~ = *x* T~2~ = 74^o^C = 347K P~2~ = P~2~ = 532.141mmHg **GENERAL GAS LAW** The general gas law is deduced from two very important laws discussed previously. The laws are the Boyle's law and Charles' law. Recall, the mathematically expression for both law Boyle's law= V Charles' law = V α T Bringing both laws expression together, we get V α T Introducing a constant K V = Make K subject of formula K = This is why the general gas law equation is Where **Solved Questions with Detailed Explanations** 1\) A gas occupies a volume of 5cm^3^ at 180^o^C and 0.9atm. Calculate the final volume of the gas if it occupies 1.0atm at 4^o^C. *Soln* Parameters provided V~1~ = 5cm^3^ \[This is the first volume mentioned in the question\] T~1~ = 18^o^C \[This is the first temperature mentioned in the question\] ^o^C + 273 = Kelvin 18^o^C + 273 = 291 Kelvin P~1~ = 0.9atm \[This is the first pressure mentioned in the question\] V~2~ = x \[This is what we are to calculate for\] T~2~ = 4^o^C \[This is the second temperature mentioned in the question\] ^o^C + 273 = Kelvin 18^o^C + 273 = 277 Kelvin P~2~ = 1atm \[This is the second pressure mentioned in the question\] Applying the formula Making V~2~ subject of formula V~2~ = V~2~ = = 4.3cm^3^ 2\) 300cm^3^ of a gas X excerceted a pressure of 700mmHg at 33^o^C. What volume will the gas occupy at stp? *Soln* Parameters provided V~1~ = 300cm^3^ \[This is the first volume mentioned in the question\] T~1~ = 33^o^C \[This is the first temperature mentioned in the question\] ^o^C + 273 = Kelvin 33^o^C + 273 = 306 Kelvin P~1~ = 700mmHg \[This is the first pressure mentioned in the question\] *Final condition* This condition was at stp \[standard temperature and pressure\] V~2~ = *x*cm^3^ T~2~ = standard temperature = 273K P~2~ = standard pressure = 760mmHg Applying with the formula Making V~2~ subject of formula V~2~ = V~2~ = = 246.52cm^3^ **DALTON'S LAW OF PARTIAL PRESSURE** This law states that the total pressure of a mixture of gas is equal to the sum of their partial pressure provided the gases do not react with each other. Now from the law stated above Let me explain!!! Let us assume we have a container that contains three different gases. A B C This container above has a total pressure which is called P~T~ and each gases inside the containers have their specific partial pressures. *N.B:* The sum of the partial pressure of the gases forms the total partial pressure. Mathematically, P~T~ = P~A~ + P~B~ + P~C~ Where, P~T~ = Total pressure of the Gas mixture P~A~ = Partial pressure of Gas A P~B~ = Partial pressure of Gas B P~C~ = Partial pressure of Gas C *N.B:* It can have PD, PE and above dependent on how many gases were present in the container. *Question* What is the partial pressure of a gas? This is the pressure which that gas in a mixture would exact if it were alone in the container. *Another Question* How do we calculate the partial pressure of the gases in the container (P~A~, P~B~ and P~C~). For partial pressure of gas A (P~A~) P~A~ = mole fraction ~A~ × P~T~ Where mole fraction ~A~ = P~A~ = × P~T~ For partial pressure of gas B (P~B~) P~B~ = mole fraction ~B~ × P~T~ Where mole fraction ~B~ = P~B~ = × P~T~ For partial pressure of gas C (P~C~) P~C~ = mole fraction ~C~ × P~T~ Where mole fraction ~C~ = P~C~ = × P~T~ **Solved Questions with Detailed Explanations** 1\) A gases mixture is made up of 3 moles of Neon, 2 moles of Argon and 1 mole of Helium. If the gas mixture exerts a pressure of 18 atm i\) What is the partial pressure of each gas? ii\) Show that P~T~ = P~Ne~ + P~Ar~ + P~He~ *Soln* Total pressure of the gas mixture (P~T~) = 18atm P~Ar~ = 6 atm *For Helium* P~He~ = 3 atm ii\) Show that P~T~ = P~He~ + P~Ar~ + P~He~ Where P~T~ = 18atm P~He~ = 9atm P~Ar~ = 6atm P~He~ = 3atm P~T~ = 9atm + 6atm + 3atm P~T~ = 18atm There are other aspect we have to take note of under the Dalton's law of partial pressure; it is called COLLECTION OF A GAS OVER WATER. When a gas is collected over water, the total pressure (P~T~) exerted by the gas is actually the sum of the partial pressure of the gas (P gas) and the partial pressure of the water vapour (P). From this words above, a mathematical expression will be deduced. P~T~ = P~gas~ + P **Solved Question with Detailed Explanations** 1\) What is the partial pressure of oxygen gas collected over water at standard atmospheric pressure and 25^o^C. If the saturation vapour pressure of water is 23mmHg at 25^o^C. *Soln* So this is what the question says. "What is the partial pressure of oxygen gas that means they are asking us to get the partial pressure of gas (P~gas~). *Parameters provided* P~T~ = 760mmHg \[The standard atmospheric pressure\] P~gas~ = ? P = 23mmHg Recall; P~T~ = P~gas~ + P P~gas~ = P~T~ - P P~gas~ = 760mmHg -- 23mmHg P~gas~ = 737mmHg **AVOGADRO'S HYPOTHESIS** This hypothesis was brought by a man called Amedeo Avogadro. He explained that: 1 mole of any Gas at stp contains a volume of 22.4dm^3^ and 6.02 × 10^23^ molecules. From this Explanation i. ii. iii. **Solved Question with Detailed Explanation** 1\) Which of the following contain the same number of molecule as 25cm^3^ of Cl~2~ \[Cl = 35.5\] \(a) 20cm^3^ of HCl(g) \(b) 22.4dm^3^ of H~2~(g) \(c) 21cm^3^ of N~2~(g) \(d) 25cm^3^ of CH~4~ gas \(e) 25cm^3^ of HCl (aq) *ANS: Option D* Option e is wrong because HCl is not in gaseous state but rather in aqueous state. **IDEAL GAS LAW** The ideal gas law is a combination of three laws which are Boyle's law, Charles's law and Avogadro's Law. Recall; Boyle's Law = V α Charles' Law = V α T Avogadro's Law = V α n Bringing these laws together, we get V α α T α n Introducing a constant V = × T × n PV = nkT Change k to R PV = nRT Where P = Pressure V = Volume T = Temperature n = Number of mole k = molar mass gas constant or ideal gas constant For O, level examinations and A Level Examination like 100 level and 200 level chemistry exams. Two important constants for R must be noted. R = 0.0821atm. L/mol/k The S.I unit is pronounced as atmosphere. Litre per mole per Kelvin temperature. The / mean per R = 8.314J/deg/mol/k The S.I unit is pronounced as Joules per mole per Kelvin temperature. We use 0.0821 atm. L/deg/mol/k When pressure = atm and when volume = dm^3^ or L We use 8.314J/deg/mol/k When pressure = N/m^2^ or Pascal and volume = m^3^ How to determine the constant? For R = 0.0821 As earlier said pressure should be in atm and volume should be in dm^3^ or L. Recall the Formula DV = nRT Lets make R (molar gas constant) R = Where P = 1atm \[This is a standard value\] V = 22.4dm^3^ \[This is a constant for volume at standard condition\] n = 1mol T = 273 kelvin R = R = 0.0821 atm.L/mol/k or atm.dm^3^/mol/k for R = 8.314 As I earlier said, pressure should be in N/m^2^ or Pascal and volume should be in m^3^. Do not forget the formula PV = nRT Let make R (molar mass constant) subject of formula R = P = 101325N/m^2^ or 101325 Pascal \[This is a standard value\] V = 22.4dm^3^ or 22.4L but now, the volume should be expressed in m^3^. So we simply convert dm^3^ to m^3^ Very Easy!!! 1dm^3^ = 1 × 10^−3^m^3^ 22.4dm^3^ = xm^3^ xm^3^ = xm^3^ = 0.0224dm^3^ or 2.4 × 10^−3^m^3^ So this value just gotten is what is to be written in the formula above. n = 1 mole T = 273 kelvin R = R = 8.314N/m^2^. m^3^/mol/Kelvin or simply put as R = 8.314J/mol/Kelvin **Solved Questions with Detailed Explanations** 1\) A sample of an ideal gas consists of 0.0176 mole and occupies 8.64L at a pressure of 0.432 atm. What is the temperature of the gas in ^o^C? *Soln* Formula to use PV = ∩RT Parameters provided Pressure = 0.432atm Volume = 8.64L or 8.64dm3 Number of mole (n) = 0.176mole R = 0.0821atm. dm^3^/mol/Kelvin Temperature = ? *Quick note:* I can decide to use any of the constant; I prefer using this one (0.0821) but if the question specified the one to use, then I will use the specified one as started by the question. *NOTE:* You can see that the molar gas constant (R) S.I unit is atm.L/mol/Kelvin. So you can see Kelvin there, it simply means that the temperature we are to solve for will be in Kelvin scale but at the end of the question, it was specified we should convert the answer to temperature in degree Celsius scale (^o^C) which is very easy!!! So let us solve PV = nRT Making T subject of formula T = T = T = 258.5 Kelvin Let us convert to Celsius scale To do it, we simply use a formula which is ^o^C + 273 = Kelvin ^o^C = Kelvin -- 273 ^o^C = 258.5 kelvin -- 273 = − 14.5^o^C 2\) 350cm^3^ of an ideal gas exert a pressure of 1.05 atm at 37^o^C. What amount in mole (n) of gas is present? *Soln* Formula to use PV = nRT Parameter provided Pressure = 1.05atm Volume = 350cm^3^ \[It must be converted to dm^3^ or L\] Recall; 1dm^3^ = 1000cm^3^ xdm^3^ = 350cm^3^ xdm^3^ = xdm^3^ = 0.35dm^3^ Temperature = 37^o^C \[It must be converted to Kelvin\] ^o^C + 273 = Kelvin 37^o^C + 273 = 310 kelvin R = 0.0821 atm.dm^3^/mole/Kelvin PV = nRT Make number of mole (n) subject of formula n = n = = 0.01445 mole **GRAHAM'S LAW OF DIFFUSION** Graham's law of diffusion state the rate of diffusion of a gas is inversely proportional to the square root of its density or molecular mass. Mathematically, R α Or R α So from this expressions above, it means that when rate of diffusion (R) increases, density (d) or molecular mass (m) decreases and vice versa. From deduction, the basic formulas involved in the grahams law of diffusion are Where R~1~ = rate of diffusion 1 R~2~ = rate of diffusion 2 d~2~ = Density of gas 2 d~1~ = Density of gas 1 Where R~1~ = rate of diffusion 1 R~2~ = rate of diffusion 2 M~2~ = molecular mass of gas 2 M~1~ = molecular mass of gas 1 This two formula above are the basic formulas that must be noted on the Graham's law of diffusion. So first of all "what is rate of diffusion"? This is a term that must be noted. Mathematically, it is Rate of diffusion = The unit of volume can be expressed as cm^3^, dm^3^, L or mL or even m^3^ as seen in a particular question. The unit of time is simply expressed majorly in seconds. The sole aim of saying this above is to deduce the S.I unit of rate of diffusion. Because it can be asked in exam condition. For example, I wan to use cm^3^ as the unit for volume and seconds as the unit for time. Recall; Rate of diffusion = Now, you can see how it works so this expression above which can also be expressed as cm^3^/seconds. **How is it pronounced?** cm^3^/seconds is pronounced as cm^3^ per seconds. Let get back to the rate of diffusion formulas which is Rate of diffusion = From the formula above, it means that rate of diffusion is directly proportional to volume of gas (V) and inversely proportional to time taken (t) to diffuse. Mathematically, Rate of diffusion (R) α Volume of gas (V) R α V Mathematically Rate of diffusion (R) α R α Formulas to note on this aspect 1) This formula relate rate of diffusion and density 2) This formula relate rate of diffusion and molecular mass of gas. 3\) Rate of diffusion = 4) This formula relate rate of diffusion, volume and molecular mass 5) This formula relate rate of diffusion, volume and density of gas. 6) This formula relate rate of diffusion, time and molecular mass. 7) This formula relate rate of diffusion, time and density. 8) This formula relate volume and molecular mass of gas. 9) This formula relate volume and density 10) This formula relate time and molecular mass of gas 11) This formula relate time and density 12) This formula relate volume, time and molecular mass 13) This formula relate volume, time and density So looking at the formula above, you should not be scared cause it is very easy to understand. Meanwhile knowing the formula will be very helpful to solve various on this law which we are to see below. **Solved Questions with Detailed Explanations** 1\) Given that the rate of diffusion of a gas X is 200cm^3^/seconds. What volume of X will take 1440 min to diffuse through an orifice? *Soln* Parameters provided Rate of diffusion of the gas = 200cm^3^/seconds Time taken = 1440 min Volume of X = ? Remember time should be in seconds Let us convert 1440 min to seconds Recall; 60 seconds = 1 min x seconds = 1440 min x seconds = = 86,400 seconds Recall; the formula *N.B:* 1000cm^3^ = 1dm^3^ 17280000cm^3^ = xdm^3^ xdm^3^ = = 17280dm^3^ How do you convert from cm^3^ to m^3^ 1000cm^3^ = 1 × 10^−3^m^3^ 17280000cm^3^ = xm^3^ xdm^3^ = xdm^3^ = 17.28m^3^ Believe it is very clear!!! 2\) If 465cm^3^ of sulphur (IV) oxide, can diffuse through a partition in 30 seconds, how long will an equal volume of hydrogen sulphide diffuse through same partition \[S = 32, O = 16, H = 1\] *Soln* Parameters provided V~1~ = 465cm^3^ of SO~2~ T~1~ = 30 seconds of SO~2~ V~2~ = 465cm^3^ of H~2~S T~2~ = x seconds M~1~ = SO~2~ 32 + 16 × 2 32 + 32 = 64g/mol M~2~ = H~2~S K2 + 32 2 + 32 = 34g/mol Let make x seconds subject of formula 3\) If SO~2(g)~ and CH~4~ are released at the same time at the opposite end of a tube, the rate of diffusion of SO~2~ and CH~4~ will be in the ratio \[S = 32, O = 16, C = 12\] \(a) 2:1 (b) 4:1 (c) 1:4 (d) 1:2 *Soln* Now for this question first let us calculate the molecular mass of both compound SO~2~ and CH~4~ Molecular mass of SO~2~ 32 + 16 x 2 = 64g/mol Molecular mass of CH~4~ 12 + 1 x 4 = 16g/mol From the question SO~2~ was mentioned first so it becomes M~1~ and CH~4~ becomes M~2~. From the question, we are to relate rate of diffusion and molecular mass of the gas. We simply use one of the formula in the explanation previously. The formula to use is Where M~1~ = 64g/mol → SO~2~ **GAY-LUSSAC'S LAW OF COMBINING VOLUME** Gay-Lussac's law state that when gases react, they do so in volume which are in simple ratios to one another and to the volume of the product if gaseous provided temperature and pressure remains constant. *For example* When hydrogen gas and oxygen are combined together to form gaseous water. The volume of the reactant and products are related to each other in a simple way if they are all measured to the same temperature and pressure. Using a reaction to explain this phrase above 2H~2(g)~ + O~2(g)~ → 2H~2~O~(g)~ 2 mole of hydrogen + 1 mole of oxygen → 2 mole of gaseous water Since we are on the Gay-Lussac's law of combining volume 2 volumes of hydrogen + 1 volume of oxygen → 2 volume of water *N.B:* 1 volume of oxygen is better still called a volume of oxygen. *Another Example* When Nitrogen gas and hydrogen gas are combined to form Ammonia. Equation to explain better N~2(g)~ + 3H~2(g)~ → 2NH~3(g)~ 1 mole of Nitrogen gas + 3 mole of hydrogen → 2 mole of Ammonia On Gay-Lussac's law A volume of Nitrogen gas + 3 volumes of hydrogen gas → 2 volume of Ammonia Chapter 8 Mole Concept **Learning Outcome** When you have completed this chapter, you will be able to: - Understand strongly the Basic concept of solving questions on mole concept. - Understand formulas on mole concept. - Solve questions on stoichiometry If you already feel confident about these chapters, why not try the quiz over the page? **Practice Questions** 1. What volume will 0.5g H~2~ occupy at stp? \[H = 1, 1 mole of a gas occupy 22.4dm^3^ at stp\]. 2. What volume of hydrogen gas would be produced from 6.0g of the magnesium \[H = 1, 1 mole of a gas occupy 22.4dm^3^ at stp\]? 3. Determine the volume of carbon (IV) oxide measured at stp that would be produced by the thermal decomposition of 10g of calcium trioxoxide carbonate (iv) (CaCO~3~) 4. It is known that carbon has an allotropic force called fullerence, containing molecules of formular C~66~. Calculate the mass of one mole of these molecules (C = 12) 5. Calculate the mass of chlorine gas which occupies a volume of 1.12dm^3^ at stp \[Cl = 35.5, 1 mole of a gas occupies 22.4dm^3^ at stp\]. 6. Determine the mass of Sulphur (IV) oxide obtained when 91g of oxygen reacts completely with Sulphur according to the following equation \[S = 32, O = 16\] 7. Calculate the mass of ZnSO~4~ produced when excess of ZnCO~3~ is added to 50cm^3^ of 4.00mol/dm^3^ H~2~SO~4~. The equation for the reaction **1. Mole** What is a mole: A mole is the amount of a pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of C-12 (i.e. 6.02 × 10^23^). The concept of mole is all about stoichiometry and stiochiometry is mole. **FORMULAS TO NOTE ON MOLE CONCEPT** n [=] Number of mole n = Knowing all these formulas above starts to build your knowledge about the mole concept. *N.B:* n = Concentration × Volume (dm^3^) You can see that the volume in the question is in dm^3^ so don't divide by 1000 but if the volume in the question is in cm^3^, you can divide by 1000. *N.B:* n = *Quick Note:* Number of atoms can also be called number of molecules or number of ions or number of particles. **Common parameters on mole concept** Mass = g (grams) Molar mass = g/mol (grams per mole) Concentration = mol/dm^3^ (mole per dm^3^) Volume = cm^3^, dm^3^, L, mL, m^3^ Pressure = atm, mmHg, torr, N/m^2^, Pascal T = temperature = ^o^C, Kelvin To solve questions on the mole concept, it will be in different CASES. **CASE 1: Question dealing with number of mole, mass and molar mass** **Solved Question with Detailed Explanation** 1\) How many mole of calcium trioxocarbonate (IV) are there in 2.5g of calcium trioxocarbonate (IV) \[Ca = 40, O = 16, C = 12\] 2\) What is the mass of 2.3 mole of sodium \[Na = 23\] *Soln* Parameter provided 3\) How many mole of Iron (III) oxide are contained in 1kg of the compound \[Fe = 56, O = 16\] *Soln* The chemical formula of Iron (III) oxide Fe^3+^ O^2−^ = Fe~2~O~3~ Parameters Provided Mass = 1kg \[Mass should be in g\] *N.B:* 1kg = 1000g Mass = 1000g Compound = Fe~2~O~3~ Molar mass = 56 × 2 + 16 × 3 Molar mass = 112 + 48 = 160g/mol Number of mole (n) = ? n = 6.25 mole of Fe~2~O~3~ 4\) One mole of a compound MHCO~3~ has a mass of 84g. Calculate the relative atomic mass of M \[H = 1, C = 12, O =10\] *Soln* Parameters provided Number of mole = 1 mole Mass = 84g Molar mass = MHCO~3~ Recall; **CASE 2: Questions dealing with number of mole, concentration and volume** **Solved Questions with Detailed Explanation** 1\) Calculate the number of mole of element present in 200cm^3^ of a 0.2mol/dm^3^ H~2~SO~4~ solution. *Soln* Parameters provided Volume (cm3) = 200cm^3^ Concentration = 0.2mol/dm^3^ Number of mole = ? Recall; n = n = n = 0.04mole **CASE 3: Questions dealing with number of mole, pressure, volume, temperature and molar gas constant** **Solved Question with Detailed Explanations** 1\) Calculate the number of mole of element present in 250cm^3^ of an ideal gas exerting a pressure of 700mmHg at 27^o^C Recall; n = n = n = 0.009354 mole **CASE 4: Questions dealing with number of mole, volume at stp \[standard, temperature and pressure\]** **Solved Questions with Detailed Explanation** 1\) What is the number of mole of 5.6dm^3^ of SO~2~ gas at stp \[S = 32, O = 16\] *Soln* Parameters provided Volume (dm^3^) = 5.6dm^3^ at stp Number of mole = n = ? Standard volume = 22.4dm^3^ Recall; n = n = n = 0.25 mole **CASE 5: Question dealing with number of mole, number of atoms and Avogadro's constant** **Solved Questions with Detailed Explanation** 1\) Calculate the number of moles present in 2.06 × 10^23^ atm of Cl~2~. Number of mole = ? Number of atoms = 2.06 × 10^23^ Avogadro's constant = 6.02 × 10^23^ Recall; n = **CASE 6: Question dealing with relationship between two formulas from the formulas on mole concept** **Solved Questions with Detailed Explanations** 1\) What is the mass of potassium hydroxide present in 500cm^3^ of 2mol/dm^3^ solution of the compound? \[H = 39, O = 16, H = 1\] *Soln* Parameter provided Mass = ? Volume = 500cm^3^ Concentration = 2mol/dm^3^ Potassium hydroxide KOH 39 + 16 + 1 = 56g/mol Formulas that relate all the parameters Making Mass subject of formula Mass = Mass = Mass = 56g 2\) What is the mass of 2.6dm^3^ of SO~2~ gas at stp? *Soln* Parameter provided Mass = ? Volume = 2.6dm^3^ at stp SO~2~ 32 + 16 x 2 = 64g/mol 3\) Upon heating 1.25g of a solid 280cm^3^ of a gas measured at stp were evolved and a residue of 0.7g was left. Calculate the molar mass of the gas \[molar volume of a gas at stp = 22.4dm^3^\] *Soln* Mass of gas = 1.25g -- 0.7g Mass of gas = 0.55g Volume at stp = 280cm^3^ Molar mass = n = Making molar mass subject of formula Molar mass = Molar mass = Molar mass = 44g/mol **CASE 7: General calculations in stoichiometry** 1\) Calculate the number of mole of oxygen provided from the decomposition of 2 mole of KClO~3~ *Soln* Equation of reaction = 2KClO~3~ → 2KCl + 3O~2~(g) Mole ratio 2 mole 2 mole 3 mole From the balanced equation 2 mole → 3 mole From question 2 mole → x mole 2\) How many mole of HCl will be required to react with 5g of NaOH \[Na = 23, O = 16, H = 1, Cl = 35.5\] *Soln* Equation of reaction: HCl + NaOH → NaCl + H~2~O Mole ratio 1 mole 1 mole 1 mole 1 mole Let us convert 5g of NaOH to mole Recall; n = Mass = 5g Molar mass = NaOH n = = 0.125.mole of NaOH From balanced equation 1 mole → 1 mole From question x mole → 0.125 mole 3\) Fe(s) + 2HCl → FeCl~2~(aq) + H~2~ What is the mass of iron that would be required to react with 250cm^3^ of 0.1mol/dm^3^ HCl? *Soln* Equation of Reaction: Fe(s) + 2HCl → FeCl~2~(aq) + H~2~ Mole ratio 1 mole 2 mole 1 mole 1 mole Let us covert 250cm^3^ of 0.1mol/dm^3^ HCl to mole Recall; n = Volume = 250cm^3^ Concentration = 0.1mol/dm^3^ From balanced equation 1 mole → 2 mole From question x mole → 0.025mole Recall; n = Mass = n x molar mass n = 0.0125.mole of Fe Molar mass = 56g/mol Mass = 0.0125 mole x 56g/mol Mass = 0.7g of Iron (Fe) 4\) 2C~4~H~10~ + 13O~2~ \--\> 8CO~2~ + 10H~2~O From the equation above, what volume of oxygen at stp is required to burn 100cm^3^ of butane \[molar volume of a gas at stp = 22.4dm^3^\] *Soln* Equation of reaction 2C~4~H~10~ + 13O~2~ → 8CO~2~ + 10H~2~O Mole ratio 2 mole 13 mole 8 mole 10 mole Let us convert 100cm^3^ of butane at stp to mole Recall; n = Volume = 100cm^3^ From balanced equation 2 mole → 13 mole From question 0.0045mole → x mole Let us convert the answer (mole) to volume Recall: n = Making volume (cm^3^) subject of formula Volume (cm^3^) = n × 22400cm^3^ Volume (cm^3^) = 0.02925 mole x 22400cm^3^ Volume (cm^3^) = 655.2cm^3^ 5\) 5.00g of a mixture of CaCO~3~ and CaO liberated 1.32g of carbon (IV) oxide (CO~2~) on strong heating. What is the percentage of CaO in the mixture \[Ca = 40, C = 12, O = 16\] *Soln* *N.B:* Only CaCO~3~ portion of the mixture will decompose upon heating. Equation of reaction: CaCO~3~ → CaO + CO~2~ Let us convert 1.32g of CO~2~ to mole Recall; n = From balanced equation 1 mole → 1 mole From question x mole → 0.03 mole Let us convert the answer (mole) to grams Recall; n = Mass of mixture = 5g \[CaCO~3~ and CaO\] Mass of CaCO~3~ = 3g Mass of CaO Mass of mixture = Mass of CaCO~3~ + Mass of mixture Mass of mixture = 3g -- *x*g Percentage of CaO in the mixture = Percentage of CaO in the mixture = Percentage of CaO in the mixture = 40% 6\) 2g of limestone was treated with excess 0.5mol/dm^3^ HCl. The acid left at the end of the reaction required 32cm^3^ of 0.02mol/dm^3^ Na~2~CO~3~ solution for neutralisation. What is the original volume of the acid? \[Na = 23, Ca = 40, C = 12, O = 16, Cl = 35.5\] *Soln* Equation of reaction: CaCO~3~ + 2HCl → CaCl~2~ + H~2~O + CO~2~ Let us convert 2g of Limestone (CaCO~3~) to mole Recall; n = Mass = 2g Molar mass = CaCO~3~ 40 + 12 + 16 × 3 = 100g/mol n = = 0.02 mole We are to relate CaCO~3~ and 2HCl CaCO~3~ → 2HCl From balanced equation 1mole → 2 mole From question 0.02mol → x mole x mole = x mole = 0.04 mole of HCl We were told at the end of the reaction the acid left reacted with Na~2~CO~3~ solution. Equation of reaction becomes: 2HCl + Na~2~CO~3~ → 2NaCl + H~2~O + CO~2~ Let us convert 32cm^3^ of 0.02mol/dm^3^ of Na~2~CO~3~ to mole Recall; n = Concentration = 0.02mol/dm^3^ Volume (cm^3^) = 32cm^3^ n = n = 0.00064 mole of Na~2~CO~3~ We are to relate HCl and Na~2~CO~3~ 2HCl → Na~2~CO~3~ From the balanced equation 2 mole → 1 mole From question x mole → 0.00064 mole x mole = x mole = 0.00128 mole of HCl Total number of mole of HCl in the two reactions Reaction 1 + Reaction 2 n~HCl~ Reaction 1 + n~HCl~ Reaction 2 0.04 mole + 0.00128 mole = 0.04128 mole To calculate the original volume of the acid \[HCl\] Recall; n~T~ = Making volume (cm^3^) subject of formula Volume (cm^3^) = n~T~ = 0.04128 mole Concentration = 0.5mol/dm^3^ of HCl \[This is gotten from the question\] nT = Total number of mole Volume = Volume = 82.56dm^3^ = 80dm^3^ 7\) The volume of 0.20mol/dm^3^ H~2~SO~4~ that will exactly neutralization 25cm^3^ of 0.05mol/dm^3^ NaOH solution. *Soln* From the word "neutralize". The neutralization reaction must be written. Equation of Reaction: H~2~SO~4~ + 2NaOH → Na~2~SO~4~ + 2H~2~O From the question, we are asked to find the volume of acid \[H~2~SO~4~\] From the titration equation Where C~A~ = Concentration in mol/dm^3^ of Acid V~A~ = Volume in mol/dm^3^ of Acid C~B~ = Concentration in mol/dm^3^ of Base V~B~ = Volume in mol/dm^3^ of Base N~A~ = Number of mole of Acid N~B~ = Number of mole of Base Making VA subject of formula V~A~ = Where CB = 0.05mol/dm^3^ VB = 25cm^3^ NA = 1 mole \[1H~2~SO~4~\] NB = 2 mole \[2NaOH\] CA = 0.2mol/dm^3^ V~A~ = 8\) How many atoms of carbon are in 4.0 × 10^5^g propane \[C~3~H~8~\] \[C = 12, H = 1\] *Soln* *N.B:* The question asked for number of atom of carbon 9\) Calculate the Density of NH~3~ \[Molar mass = 17.03g/mol\] at 752mmHg and 55^o^C \[R = 0.0821\] *Soln* Remember: Let us make molar mass subject of formula Molar mass = Where = density Let us make density subject of formula \[This is what we are asked to calculate for in the question\] *N.B:* Pressure should be in atm. 1atm = 760mmHg xatm = 752mmHg xatm = xatm = 0.99atm *N.B:* Temperature should be in Kelvin scale ^o^C + 273 = Kelvin ^o^C = 55 55^o^C + 273 = 328K Parameters provided Molar mass = 17.03g/mol Pressure = 0.99atm R = 0.0821 Temperature = 328k Density = = 0.626g/L 10\) A scientist has a synthesized greenish-yellow gelatinous compound of chlorine and oxygen whose dentists is 7.7g/L at 36^o^C and 2.88atm. Calculate the molar mass of compound *Soln* Apply same concept in the above question Formula; Density = From the question, we are asked to calculate the molar mass; Let make molar mass subject of formula Molar mass = Pressure = 2.88atm Density = 7.7g/L Temperature = 36^o^C oC + 273 = Kelvin 36oC + 273 = 309K R = 0.0821 Molar mass = Molar mass = 68g/mol Chapter 9 pH Concept **Learning Outcome** When you have completed this chapter, you will be able to: - Understand the concept of pH. - Solve calculations involved in the pH concept. - Understand the concept of basicity of an acid and acidity of a base. If you feel confident about these chapters, why not try the quiz over the page? **Practice Question** 1. The POH of 0.25mol/dm^3^ of HCl is \_\_\_\_\_\_\_\_\_ 2. Which of the following salt solution will have a pH greater than 7? (a) NaCl (b) Na~2~CO~3~ (c) Na2SO~4~ (d) NaHSO~4~ 3. What is the pH of a 0.001M solution of NaOH? 4. What is the hydroxyl ion concentration \[OH^−^\] in a solution of sodium hydroxide of pH 10.0? 5. What is the hydrogen ion concentration \

Calculations and Key Notes in Chemistry PDF

Document Details

Tags

Related

Summary

Full Transcript

Upgrade to continue