Atoms, Molecules and Moles Lecture Notes PDF

Atoms, Molecules and Moles Lecture 1  What's an atom? ?Element All Matter in universe is composed of Atoms (An atom is the smallest particle of an element which exhibits the physical and chemical characteristics of that element.)  Elements are composed of only 1 type...

Atoms, Molecules and Moles Lecture 1  What's an atom? ?Element All Matter in universe is composed of Atoms (An atom is the smallest particle of an element which exhibits the physical and chemical characteristics of that element.)  Elements are composed of only 1 type of atom. (An element is a substance which contains only one kind of atom: lead (Pb), silver (Ag), hydrogen (H2), oxygen (O2). 2  ATOMIC PARTICLES  Atoms consist of three subatomic particles:  electrons  which are very small  negatively charged  have a negligible mass (mass = 0)  protons  protons are positively charged particles  have a mass =1  The number of protons in an atom’s nucleus determines what element it is  neutrons  neutrons have no charge and are therefore called Neutral  have a mass = 1 3  ATOMIC STRUCTURE  Atoms are mostly empty space  Electrons move in orbits around the center of the atom - in relatively distinct areas called Energy Levels.( Orbits or shells) 4  Electrons was discovered by British physicist J.J. Thomson  Protons were discovered by Rutherford in 1919.  Neutrons were discovered by James Chadwick in 1932. 5  ELECTRONS  Small negatively charged particle  Orbit, circle, around the nucleus  Have no mass  Number of electrons = number of protons 6  PROTONS  small, positively charged particles  reside in the nucleus  along with the neutron, make up most of the mass of the atom  the number of protons is what defines the type of a particular atom  Number of protons = number of electrons 7 8  NEUTRONS  small particles with no charge  reside in the nucleus  along with the proton, make up most of the mass of the atom  a differing number of neutrons is what defines an "isotope" of an atom 9  Properties Particle Mass (k g) # Mass (amu) *Charge Electron x 10 -31 9.10939 0 = 0.00055 1- Proton x 10 -27 1.67262 1 = 1.00728 1+ Neutron x 10 -27 1.67493 1 = 1.00866 0 10 Atomic Theory of Matter  The theory that atoms are the fundamental building blocks of matter reemerged in the early 19th century, championed by John Dalton. 11  Dalton’s Postulates  Each element is composed of extremely small particles called atoms. 12  Dalton’s Postulates  All atoms of a given element are identical to one another in mass and other properties, but the atoms of one element are different from the atoms of all other elements. 13  Dalton’s Postulates  Atoms of an element are not changed into atoms of a different element by chemical reactions;  atoms are neither created nor destroyed in chemical reactions. 14  Dalton’s Postulates  Compounds are formed when atoms of more than one element combine;  a given compound always has the same relative number and kind of atoms. 15  Dalton’s Atomic Theory (1808) 1. Elements are composed of extremely small particles called atoms. All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of one element are different from the atoms of all other elements. 2. Compounds are composed of atoms of more than one element. The relative number of atoms of each element in a given compound is always the same. 3. Chemical reactions only involve the rearrangement of atoms. Atoms are not created or destroyed in chemical reactions. 2.1 16 16 X + 8Y 8 X2Y 2.1 17 Law of Conservation of Mass  The total mass of substances present at the end of a chemical process is the same as the mass of substances present before the process took place 18  Law of Constant (Definite) Composition Joseph Proust (1754–1826)  Also known as the law of definite proportions  The elemental composition of a pure substance never varies.  The law of definite composition states that “Compounds always contain the same elements in a constant proportion by mass”.  Sodium chloride is always 39.3% sodium and 60.7% chlorine by mass, no matter what its source.  Water is always 11.2% hydrogen and 88.8% oxygen by mass. 19  Law of Definite Composition A drop of water, a glass of water, and a lake of water all contain hydrogen and oxygen in the same percent by mass. 20 2 21 2.1 What is Molecule 22 A molecule is an aggregate of two or more atoms in a definite arrangement held together by chemical bonds H2 H2O NH3 CH4  A diatomic molecule contains only two atoms Ex: H2, N2, O2, Br2, HCl, CO  A polyatomic molecule contains more than two atoms Ex: O3, H2O, NH3, CH4 2.5 23 24  Symbols of Elements Atomic number (Z) = number of protons in nucleus (the atomic number defines a specific type of atom since each different type of atom (representing each element) will have a different number of protons in the nucleus ) Mass number (A) = number of protons + number of neutrons = atomic number (Z) + number of neutrons Mass Number A ZX Element Symbol Atomic Number 2.3 25  Isotopes : same element (X) with Isotopes are atoms of the different numbers of neutrons in their nuclei 1 2 3 1H 1H (D) 1H (T) 235 238 92 U 92 U 26 Isotopes 2.3 27  Do You Understand Isotopes? 14 How many protons, neutrons, and electrons are in 6 C ? 6 protons, 8 (14 - 6) neutrons, 6 electrons 11 How many protons, neutrons, and electrons are in 6 C ? 6 protons, 5 (11 - 6) neutrons, 6 electrons 28 2.3  2.6 29  The Mole Concept  The mole (mol) is a unit of measure for an amount of a chemical substance.  ex. Its just like the use of dozen to describe 12 of something  Avogadro’s Number (symbol N) is the number of atoms in 12.01 grams of carbon.  A mole is Avogadro’s number of particles, that is 6.02 × 1023 particles.  1 mol = Avogadro’s Number = 6.022 × 1023 particles of anything  We can use the mole relationship to convert between the number of particles and the mass of a substance. 30  The Mole One mole = amount of any substance that contains as many particles of that substance as there are atoms of C in 12 g of pure carbon-12 1 mole = 6.022× 1023 particles = Avogadro’s Number 1 mole of Fe = 6.022× 1023 atoms of Fe 1 mole of H2O = 6.022× 1023 molecules of H2O Abbreviation for mole is mol 31  Mole Calculations I How many sodium atoms are in 0.120 mol Na? Step 1: we want atoms of Na Step 2: we have 0.120 mol Na Step 3: 1 mole Na = 6.02 × 1023 atoms Na 6.02 × 1023 atoms Na 0.120 mol Na × = 7.22 × 1022 atoms Na 1 mol Na 32  Mole Calculations I How many moles of potassium are in 1.25 × 1021 atoms K? Step 1: we want moles K Step 2: we have 1.25 × 1021 atoms K Step 3: 1 mole K = 6.02 × 1023 atoms K 1 mol K 1.25 × 10 atoms K × 21 = 2.08 × 10-3 mol K 6.02 × 1023 atoms K 33  Molar Mass  The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance. (number of grams of that substance equal to the sum of atomic masses of the element in the formula)  The atomic mass of iron is 55.85 amu.  Therefore, the molar mass of iron is 55.85 g/mol.  Since oxygen occurs naturally as a diatomic, O2, the molar mass of oxygen gas is 2 times 16.00 g or 32.00 g/mol. 34  Calculating Molar Mass  The molar mass of a substance is the sum of the molar masses of each element.  What is the molar mass of magnesium nitrate, Mg(NO3)2?  The sum of the atomic masses is: 24.31 + 2(14.01 + 16.00 + 16.00 + 16.00) = 24.31 + 2(62.01) = 148.33 amu  The molar mass for Mg(NO3)2 is 148.33 g/mol. 35  Atomic Mass Atomic Mass (Atomic Weight) = number of grams in 1 mole of that element ( 55.85 = number of grams in mol of Fe) Examples: 1 mole of C = 12.011 grams C 1 mole of Na = 22.99 grams Na So now: 6.022× 1023 atoms of Fe = 1 mol of Fe = 55.85 g Fe 36 Problem: How many Na atoms are in 10 g Na ? 1 mole Na = 6.022× 1023 atoms Na = 22.99 g Na ↦ 10 g Na Solution: 10 g Na x 6.022× 1023 atoms Na = 2.62 x 1023 atoms Na 22.99 g Na 37  Mass-Mole Calculations What is the mass of 1.33 moles of titanium, Ti? We want grams, we have 1.33 moles of titanium. Use the molar mass of Ti: 1 mol Ti = 47.88 g Ti 47.88 g Ti 1.33 mole Ti × = 63.7 g Ti 1 mole Ti 38  Mole Calculations II What is the mass of 2.55 × 1023 atoms of lead? We want grams, we have atoms of lead. Use Avogadro’s number and the molar mass of Pb 1 mol Pb 207.2 g Pb 2.55 × 10 atoms Pb × 23 × 6.02×10 atoms Pb 23 1 mole Pb = 87.8 g Pb 39  Mole Calculations II How many O2 molecules are present in 0.470 g of oxygen gas? We want molecules O2, we have grams O2. Use Avogadro’s number and the molar mass of O2 1 mol O2 6.02×1023 molecules O2 0.470 g O2 × × 32.00 g O2 1 mole O2 8.84 × 1021 molecules O2 40  Gas Density  The density of gases is much less than that of liquids.  We can calculate the density of any gas at STP easily.  The formula for gas density at STP is: molar mass in grams = density, g/L molar volume in liters 41  Calculating Gas Density  What is the density of ammonia gas, NH3, at STP?  First we need the molar mass for ammonia; 14.01 + 3(1.01) = 17.04 g/mol  The molar volume NH3 at STP is 22.4 L/mol.  Density is mass/volume: 17.04 g/mol = 0.761 g/L 22.4 L/mol 42  Molar Mass of a Gas  We can also use molar volume to calculate the molar mass of an unknown gas.  1.96 g of an unknown gas occupies 1.00L at STP. What is the molar mass?  We want g/mol, we have g/L. 1.96 g 22.4 L × = 43.9 g/mol 1.00 L 1 mole 43  Mole Unit Factors  We now have three interpretations for the mole: 1 mol = 6.02 × 1023 particles 1 mol = molar mass 1 mol = 22.4 L at STP for a gas  This gives us 3 unit factors to use to convert between moles, particles, mass, and volume. 44  Mole-Volume Calculation  A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present?  We want moles, we have volume.  Use molar volume of a gas: 1 mol = 22.4 L 1 mol CH4 4.50 L CH4 × = 0.201 mol CH4 22.4 L CH4 45  Mass-Volume Calculation  What is the mass of 3.36 L of ozone gas, O3, at STP?  We want mass O3, we have 3.36 L O3.  Convert volume to moles then moles to mass: 1 mol O3 48.00 g O3 3.36 L O3 × × 22.4 L O3 1 mol O3 = 7.20 g O3 46  Molecule-Volume Calculation  How many molecules of hydrogen gas, H2, occupy 0.500 L at STP?  We want molecules H2, we have 0.500 L H2.  Convert volume to moles and then moles to molecules: 1 mol H2 6.02×1023 molecules H2 0.500 L H2 × × 22.4 L H2 1 mole H2 = 1.34 × 1022 molecules H2 47  Chemical Formulas  A particle composed of two or more nonmetal atoms is a molecule.  A chemical formula expresses the number and types of atoms in a molecule.  The chemical formula of sulfuric acid is H2SO4. 48  Writing Chemical Formulas  The number of each type of atom in a molecule is indicated with a subscript in a chemical formula.  If there is only one atom of a certain type, no ‘1’ is used.  A molecule of the vitamin niacin has 6 carbon atoms, 6 hydrogen atoms, 2 nitrogen atoms, and 1 oxygen atom. What is the chemical formula? C6H6N2O 49  Percent Composition  The percent composition of a compound lists the mass percent of each element.  For example, the percent composition of water, H2O is: 11% hydrogen and 89% oxygen  All water contains 11% hydrogen and 89% oxygen by mass. 50  If a formula of a compound is known, its percent composition by mass of each element can be found and vice versa 51 Calculating Percent Composition Calculate the percent composition of the elements in ammonium nitrate. Given formula= NH4NO3 Formula mass = 80.0 g/mol Solution: 3 elements present, what is the % N, % O , % H % N = mass of N x 100 mass of compound = 2 x 14 g/mol x 100 = 35 % N 80 g/mol % H = 5.04 % H % O = 60 % O 35 % N +5.04 % H +60 % O = 100 %....this makes a good check 52  Percent Composition Problem  TNT (trinitrotoluene) is a white crystalline substance that explodes at 240°C. Calculate the percent composition of TNT, C7H5(NO2)3. 7(12.01 g C) + 5(1.01 g H) + 3 (14.01 g N + 32.00 g O) = g C7H5(NO2)3  84.07 g C + 5.05 g H + 42.03 g N + 96.00 g O = 227.15 g C7H5(NO2)3. 53 Percent Composition of TNT 84.07 g C × 100% = 37.01% C 227.15 g TNT 5.05g H × 100% = 2.22% H 227.15 g TNT 42.03 g N × 100% = 18.50% N 227.15 g TNT 96.00 g O × 100% = 42.26% O 227.15 g TNT 54 A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance An empirical formula shows the simplest whole-number ratio of the atoms in a substance molecular empirical H2O H2O C6H12O6 CH2O O3 O N2H4 NH2 55  Empirical Formulas  The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule.  The molecular formula of benzene is C6H6 The empirical formula of benzene is CH.  The molecular formula of octane is C8H18 The empirical formula of octane is C4H9. 56  Empirical Formulas from Percent Composition  We can use percent composition data to calculate empirical formulas.  Assume that you have 100 grams of sample.  Benzene is 92.2% carbon and 7.83% hydrogen, what is the empirical formula.  If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen. 57  Empirical Formulas from Percent Composition Calculate the moles of each element: 1 mol C 92.2 g C × = 7.68 mol C 12.01 g C 1 mol H 7.83 g H × = 7.75 mol H 1.01 g H The ratio of elements in benzene is C7.68H7.75. Divide by the smallest number to get the formula. 7.68 7.75 C 7.68 H 7.68 = C1.00H1.01 = CH 58  Molecular Formulas  The empirical formula for benzene is CH. This represents the ratio of C to H atoms of benzene.  The actual molecular formula is some multiple of the empirical formula, (CH)n.  Benzene has a molar mass of 78 g/mol. Find n to find the molecular formula. (CH)n 78 g/mol n = 6 and the molecular = CH 13 g/mol formula is C6H6. 59  Problem A compound is found to contain 64.80 % carbon, 13.62 % hydrogen, and 21.58 % oxygen by weight. What is the empirical formula for this compound? The molecular weight for this compound is 74.14 g/mol. What is its molecular formula? 60  :Solution 1) Assume 100 g of the compound is present. This changes the percent to grams: C ⇒ 64.80 g H ⇒ 13.62 g O ⇒ 21.58 g 2) Convert the masses to moles: C ⇒ 64.80 g / 12 = 5.4 mol H ⇒ 13.62 g / 1 = 13.62 mol O ⇒ 21.58 g / 16 = 1.349 mol 3) Divide by the lowest, seeking the smallest whole-number ratio: C ⇒ 5.4 / 1.349 = 4 H ⇒ 13.62 / 1.349 = 10 O ⇒ 1.349 / 1.349 = 1 4) Write the empirical formula: C4H10O Determine the molecular formula:  Molecular mass of C4H10O ⇒ 48+10+16 = 74 g/mol  So (CHO)n = 74.14 ⇒ n=1 CHO 74  molecular formula = C4H10O 61