General Inorganic Chemistry PDF

General Inorganic Chemistry Dr. Azza Hassoon Chem 101  Chapter 1 The periodic table The history of the periodic table  periodic Table 18 Groups 7 periods Types of Elements Important properties of the periodic Table 1. Atomic Size This is the d...

General Inorganic Chemistry Dr. Azza Hassoon Chem 101  Chapter 1 The periodic table The history of the periodic table  periodic Table 18 Groups 7 periods Types of Elements Important properties of the periodic Table 1. Atomic Size This is the distance between the nucleus and the outer energy level  As general, the atomic size increases on descending a group, but decreases across a period from left to right. Atomic size decrease Atomic size increase In a group  in a period? 3 Li: 2 1s ,2s1 4Be: 2 2 1s ,2s + e + e e 2. Ionization energy This is the energy needed to remove an outer electron from a free atom of the element. (units of ionization energy are kJ mol-1) For example: As general, the ionization energies decrease on descending a group, but increase a cross a period. Although not quite regularly. In a group In a period 3. Electron affinity It is the change in energy (in kJ/mole) when an electron is added to the neutral atom in the gaseous state to form a negative ion. A(g) + e- → A-(g) What is the difference between stable atom and unstable atom???? In case of stable element (As: 7N) The element has no ability to gain or loss electrons So, when we add an electron, the element must absorp energy (Endothermic process) N + e- → N- ………….> E.A.= -7 KJ/mole (end.) 7N: 1S2 2S2 2P3 Half filled orbital In case of unstable element (as: 5B) The element has an ability to gain or loss electrons So, when we add an electron, the element must be release energy (Exothermic process) B + e - → B- ………….> E.A. = 27 KJ/mole (exo.) 5B: 1S2 2S2 2P1 Another example Electron Affinity Trend 4. Electronegativity It is the ability of an atom in a compound to attract electron towards itself Electronegativity of Cl > electronegativity of H 3.6 2.2  Electronegativity Trend 1A 2A 3B 4B 5B 6B 7B 8B 1B 2B 3A 4A 5A 6A 7A 8A 1 2 H Electronegativity increases He 2.1 3 4 5 6 7 8 9 10 Electronegativity decreases Li Be B C N O F Ne 1.0 1.5 2.0 2.5 3.0 3.5 4.0 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar 0.9 1.2 1.5 1.8 2.1 2.5 3.0 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 0.85 1.0 1.3 1.5 1.6 1.6 1.5 1.8 1.9 1.9 1.9 1.6 1.6 1.8 2.0 2.4 2.8 3.0 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 0.82 1.0 1.2 1.4 1.6 1.8 1.9 2.2 2.2 2.2 1.9 1.7 1.7 1.8 1.9 2.1 2.5 2.6 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 0.79 0.9 1.1 1.3 1.5 1.7 1.9 2.2 2.2 2.4 1.9 1.8 1.9 1.9 2.0 2.2 2.4 87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Uuu Uub Uut Uuq Uup Uuh 0.7 0.9 1.1  In a group In a period Why is oxygen more electronegative than Why is oxygen more electronegative than sulfur? nitrogen?  Oxygen has 2 energy levels, sulfur has 3 +7 P  The bonding electrons in sulfur are further +8 P away from the nucleus of the atom  When bonding electrons are further from the nucleus of the atom, there is less attraction  Oxygen has 8 protons in the nucleus whereas from the nucleus nitrogen only has 7  The bonding pair of electrons in oxygen will  A bonding pair of electrons will experience experience more attraction from its nucleus more attraction from the oxygen’s nucleus than sulfur’s bonding electrons that from nitrogen’s, thus the Hence oxygen is a more electronegative atom electronegativity of oxygen is greater. What is the Most Electronegative Element? Answer: Fluorine is the most electronegative element. Fluorine has an electronegativity of 4 9F: s2, s2, p5 1 2 2 ↑↓ ↑↓ ↑ 4. Electropositivity The opposite of electronegativity is electropositivity: a measure of an element's ability to donate electrons. Electropositivity trend  In a group Number of electron shells increase Electropositivity increases less attraction between the nucleus and the valence electrons due to this and valence electrons are easily lost. Decrease Ionization energy Electropositivity increases 5. lattice energy the energy required to convert one mole of an ionic solid into gaseous ionic constituents. NaCl(s) Na+(g) + Cl-(g) Or It is the energy released when oppositely charged ions in the gas(g) phase come together to form a solid (s). For example The lattice energy of NaCl Na+(g) + Cl-(g) NaCl(s) Ho = -787.3 kJ/mol How is lattice energy estimated using Born-Haber cycle? Na(s) + 1/2 Cl2(g) → NaCl(s) Hf (formation energy) Na(s) → Na(g) ∆Hsub (sublimation energy) Na (g) → Na+(g) + e ∆HIE (ionization energy) 1/2Cl2(l) → Cl(g) ∆Hdiss (dissociation energy) Cl (g) + 2 e → Cl- (g) ∆HEA (electron affinity) Add all the above equations leading to Na+(g) + Cl-(g) → NaCl(s) -788 kJ/mol = Ecryst Hf (formation energy )= Hsub + ∆HIE + ∆½Hdiss – ∆HEA  Orbital Filling & Hund's Rule 1. Electron configuration: is the arrangement of the electrons in an atom 2. Hund's Rule: “Electrons are distributed among the orbitals of a subshell singly then paried. Ex. 8O : 1s2, 2s2, 2p4 ↑↓ ↑↓ ↑↓ ↑ ↑ Outermost shell is called valence shell Electrons in valence shell are called valence electrons 3. Pauli exclusion principle: The max. Number of electrons in any orbit equal two electrons & these two electrons spin in opposite direction. ↑↑ ↑↓ X √  Electronic configuration Aufbau principle The statement that the ground state configuration of an atom is generated by filling in levels from the lowest (energy-wise) to the highest with electrons observing the maximum for each of these levels  Magnetic Measurements to calculate number of unpaired electrons Paramagnetic substances: Diamagnetic substances:  Contains unpaired electrons  All electrons are paired  Drawn to magnetic field  Not attracted to magnetic field  Magnetic moment increase as no. of unpaired electrons increase  Chapter 2 Electromagnetic radiation  Question? What is the sound? What is the light?  Wave Basics  Waves are energy carriers that propagate through space or a medium, transporting energy without a net movement of matter. Water waves Sun  They can be categorized into two main types: 1- mechanical waves, which require a medium to travel through 2- electromagnetic waves, which can propagate through a vacuum.  1- Mechanical waves - require a medium to travel through Mechanical waves include two types of waves: 1- longitudinal waves, cause the medium to move parallel to the direction of the wave. (e.g., sound waves). 2- Transverse waves, cause the medium to move perpendicular to the direction of the wave. (e.g., light waves). Types of electromagnetic radiation Characterization of electromagnetic radiation - All waves share common fundamental properties. - The properties are wavelength (the distance between consecutive wave crests or troughs), frequency (the number of oscillations per unit time), and amplitude (the maximum displacement from the equilibrium position).  Frequency  Wavelength  Amplitude 1. Frequency (υ): is the number of waves that pass through definite point in the second and is measured in hertz or S-1 unit of frequency 2. Wavelength (λ): is the distance between two adjacent crests or troughs. Medium Wavelength  c = υλ Frequency is inversely proportional to wavelength, according to the equation: υ α 1/λ υ = c/λ Where: υ: is the frequency in Hz or sec-1 c: is the speed of the light 3x108 m/sec. λ: is the wavelength in m, cm, nm, Ao.  The differences among EMR’s is in their wavelengths or frequency Wavelength decreases Frequency increases Energy increases Electromagnetic spectrum 3. Amplitude (a): is the height of electromagnetic waves.  What Is Quantum Theory?  The theory basically explains the nature and behaviour of matter and energy on the atomic level. Photoelectric effect 10 of the most influential figures in the history of quantum mechanics. Left to right: Max Planck, the founding father of quantum theory Albert Einstein, one of the pioneers of quantum theory Niels Bohr, Louis de Broglie, Max Born, Paul Dirac, Werner Heisenberg, Wolfgang Pauli, Erwin Schrödinger, Richard Feynman.  History of Quantum theory - In the early 1900s, a German physicist named Max Planck stated his quantum hypothesis, where he explained that radiation from a sparkling body changed its shades from red to orange to blue when the temperature was increased. This phenomenon was also known as black body radiation. - He proposed that all material systems can absorb or emit electromagnetic radiation only in “chunks” of energy, quanta E, and that these are proportional to the frequency of that radiation E = hν h: Planck's constant ν: frequency  Later 1905, Albert Einstein proposed that the quanta of light might be regarded as real particles called photon. Photoelectric effect  A photon has an energy, E, proportional to its frequency, υ, by this equation: E = hυ = hc/λ Where: h is Planck,s constant (6.626x10-34 J.s) λ is the wavelength c is the speed of light (3x108 m/sec.) ► Bohr,s Theory of Hydrogen Atom and its Spectrum: 1913 Bohr explained how an electron can orbit in Hydrogen atom. 1- Electrons moves in circular orbits around the nucleus 2- Electrons have only a fixed set of allowed orbits, called stationary states ► When an electron jumps from one orbit to another, it absorb / emit energy. The Emission of Light by a Hydrogen Atom in an Excited State. (a) Light is emitted when the electron undergoes a transition from an orbit with a higher value of n (at a higher energy) to an orbit with a lower value of n (at lower energy). (b) The Balmer series of emission lines is due to transitions from orbits with n ≥ 3 to the orbit with n = 2. The differences in energy between these levels corresponds to light in the visible portion of the electromagnetic spectrum.  Excited state Ground state Electron Transitions Responsible for the Various Series of Lines Observed in the Emission Spectrum of Hydrogen. The Lyman series of lines is due to transitions from higher-energy orbits to the lowest-energy orbit (n = 1); these transitions release a great deal of energy, corresponding to radiation in the ultraviolet portion of the electromagnetic spectrum. The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation. Series ni nf Lyman 1 2-∞ Balmer 2 3-∞ Paschen 3 4-∞ Bracket 4 5-∞ Pfund 5 6-∞ ∆E = Ei - Ef = A (1/nf2 – 1/ ni2) ……………………….. (1) Where: nf is the final energy level, ni is the initial energy level, A is constant equal 2.18x10-18 J Since the energy of a photon is E = hc/λ ……………………………………… (2) From equs. 1, 2 the wavelength of the photon is given by: 1/ λ = R(1/nf2 – 1/ ni2) where R is empirical constant called Rydberg's Constant (1.097 x10 7m -1). 1. Problem Calculate the wave length of the first line in the lyman series for hydrogen atom? ►Solution: 1/ λ = R(1/nf2 – 1/ ni2) 1/ λ = 1.097 x10 7 (1/22 – 1/ 12)  2. Problem Calculate the change in energy & frequency and wavelength of an electron transfer from first energy level to fifth energy level; If you know that : h = 6.626x10-34 J.s c = 3x108 m/sec. A = 2.18x10-18 J ►Solution: ∆E = A (1/nf2 – 1/ ni2) ∆E = 2.18x10-18 (1/52 – 1/ 12) E = hυ = hc/λ  “Dual Nature” ►Louis de Broglie (1924): He postulated that light has dual character, which it act as a particle (known as photon) and as a wave. E = h n …………… (1) E = mc2.…………... (2) \ m c2 = h n \ m c2 = hc/ λ \ m c = h/ λ \ λ = h/mc \ λ = h/mv \ λ = h/p De-Broglie as the correlation equation between particle (momentum p =mv) and wave (wave length l) characteristics. ►Heisenberg's uncertainty principle (1926): “It is impossible to determine simultaneously the exact position and exact momentum of a body as small as the electron” applications of the quantum theory Quantum optics Quantum computing Light-emitting diodes Superconducting magnets Optical amplifiers and lasers Transistors Semiconductors Magnetic resonance imaging Electron microscopy  Chapter 3 Quantum numbers  ‫‪Quantum Numbers‬‬ ‫أعذاد الكم التي تستخذم في ‪:‬‬ ‫طاقة االلكترون‬ ‫وصف إلكترون في ررة أو أيون‬ ‫شكل االوربتال‪ ,‬حجم االوربتال‪ ,‬اتجاه االوربتال‬ ‫وصف االوربتاالت‬  Quantum Numbers 1. Principal quantum number 2. Angular quantum number 3. Magnetic quantum number 4. Spin quantum number ‫‪‬‬ ‫‪Principal quantum number (n):‬‬ ‫عدد الكم الرئيسى‬ ‫يحدد مستويات الطاقة الرئيسية فى الذرة ) يصف طاقة المدار وبعد‬ ‫‪‬‬ ‫االلكترون عن النواة (‬ ‫‪‬‬ ‫‪describes the size and energy of the orbital.‬‬ ‫]∞‪[n = 1, 2, 3, 4,...............‬‬ ‫‪K, L, M, N, O, P, Q‬‬ ‫‪n‬‬ ‫‪1s‬‬ ‫‪2s‬‬ ‫‪3s‬‬  ‫هل منطقى‬ ‫كلما زاد عدد الكم الرئيسى كلما زاد حجم االوربتال؟‬ ‫كلما زاد العدد الكم الرئيسى‬ ‫‪n‬‬ ‫كلما زادت طاقة االلكترون‬ ‫‪1s‬‬ ‫‪2s‬‬ ‫‪3s‬‬ ‫كلما كانت حركته أكبر‬ ‫يشغل حيز من الفراغ أكبر‬  Angular quantum number (l): ‫عدد الكم الثانوى او المدارى‬ ‫يحدد مستويات الطاقة الفرعية الموجوده فى مستويات الطاقة الرئيسيه‬  s p d f l= 0 1 2 3 describes the shape of the orbital. The s orbitals are spherical (l = 0). The p orbitals are polar (l = 1). The d orbitals are clover-leaf shaped (l = 2). spherical dumbbell-shaped cloverleaf shaped  Magnetic quantum number (m ): ‫عدد الكم المغناطيسى‬ x, y, z ‫ يحدد اتجاهات االوربتاالت فى االبعاد الثالثة‬  describes the orientation in space of a particular orbital. m = -l,…………..+l ►If (l=0)............(m=0) ►If (l=1)............(m=-1, 0, +1) ►If (l=2)............(m=-2, -1, 0, +1, +2) Orbital Structures  Spin quantum number (S): ‫عدد الكم المغزلى‬ ‫يصف اتجاه دوران االلكترون حول محوره‬   describes spin state of the electron can be -½ or +½. ↑ ↑↓ -½ +½  Calculate four quantum number of: 11Na: 1s2, s2, p6, s1 2 2 3 ↑ n = 3, l = 0, m = 0, S = -1/2 + Na : 2 2 1s , 2s ,2p 6 ↑↓ ↑↓ ↑↓ -1 0 +1 n = 2, l = 1, m = +1, S = +1/2 20Ca: 2 2 6 2 6 2 1s , 2s ,2p , 3s , 3p , 4s n = 4, l = 0, m = 0, S = +1/2 ↑↓ Ca 2+ 2+ Ca : 2 2 6 2 1s , 2s ,2p , 3s , 3p 6 ↑↓ ↑↓ ↑↓ -1 0 +1 n = 3, l = 1, m = +1, S = +1/2 20Ca: 2 2 6 2 6 2 1s , 2s ,2p , 3s , 3p , 4s n = 4, l = 0, m = 0, S = +1/2 ↑↓ Ca2+: 1s2, 2s2,2p6, 3s2, 3p6 ↑↓ ↑↓ ↑↓ -1 0 +1 n = 3, l = 1, m = +1, S = +1/2 9F: 2 2 1s , 2s ,2p 5 ↑↓ ↑↓ ↑ -1 0 +1 n = 2, l = 1, m = 0, S = +1/2 - F: 1s 2, s 2 2 2, p6 ↑↓ ↑↓ ↑↓ -1 0 +1 n = 2, l = 1, m = +1, S = +1/2 23V: 2 2 6 2 6 2 3 1s , 2s ,2p , 3S , 3P , 4S , 3d ↑ ↑ ↑ -2 -1 0 +1 +2 n = 3, l = 2, m = 0, S = -1/2 24Cr: 2 2 6 2 6 2 4 1s , 2s ,2p , 3S , 3P , 4S , 3d 2 2 6 2 6 1 5 1s , 2s ,2p , 3S , 3P , 4S , 3d ↑ ↑ ↑ ↑ ↑ -2 -1 0 +1 +2 n = 3, l = 2, m = +2, S = -1/2 29Cu: 2 2 6 2 6 2 9 1s , 2s ,2p , 3S , 3P , 4S , 3d 2 2 6 2 6 1 10 1s , 2s ,2p , 3S , 3P , 4S , 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ -2 -1 0 +1 +2 n = 3, l = 2, m = +2, S = +1/2  Chapter 4 Stoichiometry and Chemical Equations  “Stoichiometry: Chemical Arithmetic”  Stoichiometry: The relative proportions in which elements form compounds or in which substances react. aA + bB cC + dD The mole is a unit of measurement used in chemistry to express amounts of a chemical substance. Amount of a substance that contains elementary particles (molecules, atoms, ions, electrons) equal to Avogadro’s number (NA = 6.022×1023 mol-1) Molar mass of a substance is the mass in grams of one mole of the compound. The standard unit for this is g mol−1  Molecular mass or Molecular weight The molecular mass of a substance is the mass of one molecule of that substance (g/mol) & it is equal to the sum of atomic weights of all atoms in the molecule. Number of moles (n) = weight/Molecular weight 1 mol C = 12.0 g C  Example What is the mass of one atom of calcium? Solution: 1 mol Ca = 6.022 × 1023 atoms 1 mol Ca = 40 g Ca 1 atom Ca = 40/ 6.022 × 1023 = 6.66 × 10-23 Mass of 1 atom Ca = atomic weight / NA  Steps to Calculate Stoichiometric Problems 1. Correctly balance the equation. 2. Convert the given amount into moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit.  Mole ratio Reaction between magnesium and oxygen to form magnesium oxide. Mg(s) + O2(g) MgO(s) Mole Ratios:  Mole ratio Reaction between magnesium and oxygen to form magnesium oxide. 2 Mg(s) + O2(g) 2 MgO(s) Mole Ratios: 2 : 1 2  Mole to Mole conversions  How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2Al2O3 4Al + 3O2 2mole : 3mole 3.34 : ? No. of moles = 3 x 3.34 / 2 = 5.01 mole of O2  A can of butane lighter fluid contains 1.20 moles of butane (C4H10). Calculate the number of moles of carbon dioxide given off when this butane is burned. Equation of reaction 2C4H10 + 13O2 8CO2 + 10H2O Mole ratio 2C4H10 : 8CO2 C4H10 4CO2 1 : 4 1.2 : ? By cross-multiplication X = 1.2 x 4 / 1 = 4.8 moles of CO2 given off Problem : 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced? [k = 39, Cl = 35.5, O = 16] 2 KClO3 2 KCl + 3 O2 2 : 3 1.50 : X X = 3 x 1.50 / 2 = 2.25 mole Convert to mass 2.25 mol x 32.0 g/mol = 72.0 grams  We want to produce 2.75 mol of KCl. How many grams of KClO3 would be required? [k = 39, Cl = 35.5, O = 16] Soln: 2 KClO3 2 KCl + 3 O2 2 : 2 X : 2.75 X = 2 X 2.75 / 2 = 2.75 mol In mass: 2.75 mol X 122.55 g/mol = 337 grams  Note: The mass of the reactants must equal the mass of the products. 2H2 + O2 2H2O 2x2 g H2 + 32O2 = 36.00 g H2O 36.00 grams reactant = 36.00 grams product  Practice: 2C2H2 + 5 O2 4CO2 + 2 H2O If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? (9.6 mol) How many moles of C2H2 are needed to produce 8.95 mole of H2O? (8.95 mol) If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed? (4.94 mol)  Example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu 2x56 : 3x63.5 10.1 : ? Number of grams of Cu = 3 x 63.55 x 10.1 / 2 x 56 = 17.24 g Cu  Volume-Volume Calculations: How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ? CH4 + 2O2 CO2 + 2H2O 1 : 2 x : 17.5 x = 1 x 17.5 / 2 = 8.75 L CH4  Percentage composition % atom = (mass of atom/total mass) × 100 e.g. CHCl3 % C = (mass C/mass CHCl3) × 100 % C = (12/119.5) × 100 = 10.04 % % H = (mass H/mass CHCl3) × 100 % H = (1/119.5) × 100 = 1.195 % % Cl = (mass Cl/mass CHCl3) × 100 % Cl = (3x35.5/119.5) × 100 = 89.12 %  Mass of an element in a sample of compound Calculate mass of iron in a 10.0 g sample of rust, Fe2O3 1 mol Fe2O3 2 mol Fe 2 x 56 + 3 x 16 : 2 x 56 160 g/mol : 112 g/mol 10 g : X X = (10 × 112) / 160 = 7.0 g Fe  Chemical formulas: Indicates the elements and their proportions in a compound 1. Empirical formula: Is the simplest formula that gives the relative number of atoms present in the compound. 2. Molecular formula: Is the formula that gives the actual number of each kind of atom in a molecule. 3. Structural formula: Is the formula that represent the chemical bonds between the atoms in the molecule.  Examples ►Acetic acid: E. F. M. F. S. F. CH2O C2H4O2 Problem: oxalic acid , n- hexane, benzene Proplem A sample of a brown- colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the simplest formula of the compound? 1 mole of N 14.0 ? 2.34 Number of mole of N = 1 x 2.34 / 14.0 = 0.167 mole 1 mole of O 16.0 ? 5.34 Number of mole of O = 1 x 5.34 / 16.0 = 0.334 mole N : O 0.167 : 0.334 1 : 2 NO2 Proplem : Colored gas is used in rocket engins , with empirical formula is NO2 and has a molecular weight of 92.0 gm. What is the molecular formula ? The no. of times of the E. F. weight = Mol. Wt. / E. F. Wt. The number of E. F. = 92 / 14 + 2 x 16 = 92 / 46 = 2 Molecular Formula = n x empirical formula The M. F. = 2 x NO2= N2O4  Limiting Reactant Is the reactant that disappear firstly from the reaction. Proplem Zinc & sulfur react to form zinc sulfid, a substance used in phosphors that coat the inner surfaces of TV picture tubes. The equation for the reaction is: Zn + S → ZnS 1. How many grams of ZnS can be formed when 12.0 gm of Zn are allowed to react with 6.5 gm of S ? 2. Which is the limiting reactant ? 3. How much of which elements will remain unreacted ? Solution 1 mole of Zn → 65.4 gm of Zn ? → 12.0 gm of Zn No. of mole of Zn = 1 x 12 / 64.4 = 0. 183 mole 1 mole of S → 32.1 gm of S ? → 6.5 gm of S No. of mole of S = 1 x 6.5 / 32.1 = 0. 202 mole Zn + S → ZnS 0.183 0.183 0.183 1 mole ZnS → 97.5 gm ZnS 0.183 mole → ? No. of gm of ZnS = 0.183 x 97.5 / 1 = 17.8 gm  Zn is the limiting reactant because it is disappear firstly from the reaction. 1 mole of S → 32.1 gm 0.202 – 0.183 → ? 0.019 → ? No. of grams of unreacted = 0.019 x 32.1 /1 = 0.16 gm  The Concept of: A little different type of yield than you had in Driver’s Education class.  ►What is Yield? Yield is the amount of product made in a chemical reaction. There are three types: 1. Actual yield- what you actually get in the lab when the chemicals are mixed 2. Theoretical yield- what the balanced equation tells should be made x 100 3. Percent yield = Actual / Theoretical Theoritical yield calculation -Molar concentration: It is defined as: the number of gram molecular weight. M = Wt. / Mol. Wt. x V (in Liter) OR It is defined as: the number of moles of substance in liter of solution. Problem: 2 gm of NaOH was dissolved in 200 ml of solution, what is the molar concentration ? M == No. M 2 x 1000 /40 xx200 of moles = Liter) V (in 0.25 M  Dilution -Problem: How much water must be added to 25.0 cm3 of 0.5 M KOH solution to produce a solution whose concentration is 0.350 M ? Mi. Vi = Mf. Vf 25 x 0.5 = 0.35 x Vf Vf = 35.7 ml