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CHAPTER 3: Quantitative Relationships in Chemical Reactions Stoichiometry: Greek for “measure elements” Stoichiometry involves calculations based on chemical formulas and chemical equations (reactions) – quantitative. Law of conservation of mass is central to stoichiometry. Atoms are neither created nor destroyed in a chemical reaction. Atoms are simply rearranged to yield new substances. The following is a common teaching laboratory reaction: Zn + HCl ZnCl2 + H2 Reactants Products This reaction is not balanced: reactant atoms ≠ product atoms Zn + HCl ZnCl2 + H2 Reactants Products 1 Zn 1 Zn 1 H 2 H 1 Cl 2 Cl Procedure for balancing a reaction: Balance the elements that appear in only one reactant and one product first. Zn - Already balanced Balance H and Cl; need 2 H atoms on each side of the reaction Putt a coefficient P ffi i nt of f 2 in front f nt of f th the HCl molecule on the reactant side. Also results in balancing Cl's. Balance this reaction by the method of inspection. Balancing Chemical Reactions • Formulas of reactants and products can’t be changed (i.e., cannot change subscripts) • Only amounts of reactants & products can be changed to reach atom balance – put coefficients (multipliers) in front of the substances in the reaction Zn + 2 HCl ZnCl2 + H2 Now, the reaction is balanced: Reactants Products 1 Zn 1 Zn 2 H 2 H 2 Cl 2 Cl reactant atoms = product atoms 1 Another example: C4H10 + O2 CO2 + H2O Designate physical states of each species: Balance C's and H's first: – only y appear pp in one reactant and one product p each – O's balanced last; appears in two products Balance C's, multiply the CO2 product by 4. C4H10 + O2 4 CO2 + H2O Next, balance H's: multiply the H2O product by 5 C4H10 + O2 4 CO2 + 5 H2O 13 O’s Balance the O's; multiply the O2 reactant by 13/2 / to give 13 O atoms: C4H10 + 13/2 O2 4 CO2 + 5 H2O Reactants Products 4 C 4 C 10 H 10 H 13 O 13 O BALANCING CHEMICAL REACTIONS • Typically, want the smallest whole-number coefficients in a balanced reaction. • Hence, fractional coefficient is removed by multiplying the entire equation by 2. 2 2 C4H10 + 13 O2 Often, the physical state of each substance is specified. 8 CO2 + 10 H2O (g) – gas (l) - liquid (s) - solid (aq) - aqueous (water) solution Chemical equation with physical states specified: 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(g) Combustion reactions: Involve O2 as a reactant, typically form CO2 and H2O as products. Examples: CH4 + 2 O2 2 C2 H 2 + 5 O 2 2 C8H18 + 25 O2 CO2 + 2 H2O 4 CO2 + 2 H2O 16 CO2 + 18 H2O Examples: Balancing Reactions __NO + __Cl2 __Fe Fe2O3 + __CO CO __C3H6 + __O2 __NOCl __Fe Fe + __CO CO2 __CO + __H2O Balanced chemical reactions are called chemical equations. 2 ATOMIC AND MOLECULAR MASSES • It is rather easy to determine the relative masses of atoms experimentally. • For example, C has a mass 12 times that of H. • O a mass that is 16 times that of H or 1.33 times that of C. • Cl a mass that is 35.4 times that of H or 2.95 times that of C. RELATIVE ATOMIC MASS (WEIGHT) SCALE H 100 1 200 H2(g) + 2 Li(s) 2 LiH(s) 1.00 g 7.88g What is the atomic mass of Li relative to H ? 6.88 to 1.00 RELATIVE ATOMIC MASS SCALE • Developed In A Similar Manner For All Elements H C 1 12 16 O Masses relative to that for H. Consequently, a new mass unit (atomic mass unit, amu) is used to express the mass of atoms/molecules/formula units. Define the atomic mass unit (amu): Mass of 1 C-12 atom = 12 amu 1 amu = 1 1.661 661 X 10-24 g Some atomic masses: Hydrogen = 1.008 amu Helium = 4.003 amu Fluorine = 18.998 amu From the periodic table, the mass of C is given as 12.011 amu, not 12.000 amu. Why ??? Carbon consists of two stable isotopes (C-12 & C-13)) Mg 24.3 These very small numbers are awkward to use. use The mass of one C-12 atom is assigned a value of 12 atomic mass units (amu), exactly ! Derived from early experiments involving known substances; e.g. 6.88g Atoms have very, very small masses. The O-16 atom has a mass of only 2.6560 X 10-23 g. 35 Isotope C-12 Mass (amu) 12.00000 % abundance 98.890 C-13 13.00335 1.110 3 Average Atomic Mass of C: Avg. Mass C = (12.00000)(0.98890) + (13.00335)(0.01110) Avg. Mass C = 12.01 amu (Sig. Figures) This number, 6.022 X 1023, is given the special name mole. 6.02214199 X 1023 Average Atomic Mass of chlorine (2 isotopes): Isotope I t Cl-35 Mass (amu) M ( ) 34.968 % abundance b d 75.53 Cl-37 36.956 24.47 Mole SI base unit for the amount of a Mole: substance (mol). 1 mole of anything is 6.022 X 1023 of whatever we a considering. Avg. Mass Cl = (34.968)(0.7553) + (36.956)(0.2447) Avg. Mass Cl = 35.45 amu Average atomic mass is also called atomic weight for an element. Example Problem Copper is C i a metal t l used d in i the th production d ti of f cables, pennies, etc. The atomic masses of its two stable isotopes, Cu-63 (69.09 %) and Cu65 (30.91 %), are 62.93 amu and 64.93 amu, respectively. Calculate the average atomic mass of Cu (63.55 amu). It is difficult to work with individual atoms-too small. Consider a very large number of C-12 atoms. One C-12 atom has a mass of 12 amu exactly. H How many C-12 C 12 atoms are iin exactly l 12 g ?? Mole: The amount of matter that contains as many elementary units (atoms, molecules, etc.) as there are atoms in exactly 12 g of C-12. 1 mole = 6.022 X 1023 (Avogadro’s number) 1 mole C-12 = 6.022 X 1023 atoms = 12.0000 g KNOW !!!!! Molar Mass Molar mass (g) Î mass (g) of one mole of a substance Molar mass in grams is numerically equivalent to atomic mass in atomic mass units (amu). One C-12 atom has a mass of 1.9927 X 10-23 g. 1 C-12 atom = 12.0000 amu; 1 mol C-12 = 12.0000 g ⎛ 1 C - 12 atom ⎞ ⎜⎜ ⎟ (12 g ) = 6.022 X 10 23 C atoms - 23 ⎟ ⎝ 1.9927 X 10 g ⎠ 1 H-1 atom = 1.0078 amu; 1 mol H-1 = 1.0078 g 4 We can use the result (5.81 X 10-23 g/Cl-35 atom) to calculate the mass of the amu is terms of grams: gram ⎛⎜ 5.81 X 10 − 23 g ⎞⎟⎛ 1 Cl - 35 atom ⎞ 1.66 X 10 - 24 g = ⎜ ⎟= amu ⎜⎝ 1 Cl − 35 atom ⎠⎟⎝ 34.968 amu ⎠ amu 1 amu = 1.660538728 X 10-24 g 1 g = 6.02 X 1023 amu Avogadro’s number Figure 3.1 CONVERSION FACTORS: Molar Mass 1 mol C = 12.011 g 1 avg. C atom = 12.011 amu amu/atom is numerically equivalent to grams/mole 1 mol = 6.02 X 1023 ANOTHER CALCULATION How many moles O are in 37.00 g O? First look at given and desired units: gO These concepts apply to all elements!!! Example: What is the mass in grams of one Cl35 atom? Given: 1 mol Cl-35 = 34.968 g conversion factor moles O 1 mol O ⎞ (37.00 g O ) ⎛⎜ ⎟ = 2.312 mol O ⎝ 16.00 g O ⎠ Desired: mass in g/Cl-35 atom Solution to Cl-35 example ANOTHER CALCULATION How many O atoms are in 37.00 g O? ⎞ grams ⎛ 34 .968 g Cl - 35 ⎞⎛ 1 mol Cl - 35 ⎟ = 5.81 X 10 - 23 ⎟⎜ ⎜ atom Cl ⎝ 1 mol Cl - 35 ⎠⎝ 6.02 X 10 23 atoms ⎠ (given) X (conversion factor) = desired Use the correct conversion factor to go from given to desired units. g O Î moles O Î atoms O Avogadro’s # molar mass conversion factors 1 mol O ⎞ ⎛ 6.022 X 10 23 O atoms ⎞ ⎟ (37.00 g O ) ⎛⎜ ⎟ ⎜⎜ ⎟ ⎝ 16.00 g O ⎠ ⎝ 1 mol O ⎠ = 1.393 X 10 24 O atoms 5 Molar Mass: – Mass, in g's, of one mole of a substance. – Numerically equal to the formula mass (in amu) or molecular mass (in amu). – Examples: Molar mass [[Fe(NO ( 3)3] = 241.88 g Molar mass (NH3) = 17.03 g – Another example: – What is the molar mass of C2H5OH ? (46.07 g) Figure 3.2 Formula and Molecular Masses Mass, Moles, Number of Particles molar mass Avogadro' s # Grams ←⎯ ⎯ ⎯⎯→ Moles ←⎯ ⎯ ⎯ ⎯ ⎯→ Molecules Formula Mass (FM): sum of the atomic masses of all atoms in the chemical formula (formula unit) Example: Fe(NO3)3 FM = 1(AM of Fe) + 3(AM of N) + 9(AM of O) FM = 1(55.85 amu) + 3(14.01 amu) + 9(16.00 amu) FM = 241.88 amu Molecular Mass (MM): –The sum of the atomic masses of all atoms of the molecular formula of a molecular substance. –Example: NH3 –MM = 1(AM of N) + 3(AM of H) = 1(14.01 1(14 01 amu) + 3(1 3(1.008 008 amu) = 17.03 amu –Another example: What is the molecular mass of aspartame (nutrasweet), (C14H18N2O5), the artificial sweetener ? (294.32 amu) 1. What is the average mass in grams of one avg. chlorine atom ? (5.89 X 10-23 g) 2. What is the avg. mass in grams of one ethanol (C2H5OH) molecule ? (7.65 X 10-23 g) 3. How many moles of PbCrO4 (Lead Chromate) are in 45.6 grams ? (0.141 mol) 4. How many HCl (hydrogen chloride) molecules are in 46.0 grams ? (7.60 X 1023 molecules) Calculations: Percent Composition of elements by mass: The percent by mass of each element in a compound. Example: ammonia, ammonia NH3 Divide the mass of each element in one mole of the compound by the molar mass of the compound, then multiply by 100 to get a percent. 6 General Approach: %= (# moles of element)(MM element) X 100 MM of compound %H = 3(1.008 g) X 100 = 17.76 % 17.03 g First, calculate mass of each element: • Consider a 100 g sample (any sample size will work). • For a 100 g sample: p = mass of the element: • % composition 85.63 g C 1(14.01 g) X 100 = 82.27 % %N= 17.03 g Percent composition by mass • sum of % composition must equal 100 % • 17.76 % + 82.27 % = 100.03 % • This small deviation from 100 % is due to round-off error. • Another example: What is the percent composition of hydrogen in pentylacetate, CH3CO2C5H11 (responsible for the odor in bananas) ? (10.84 %) Determine Empirical Formulas from a given Percent Composition by Mass (from experiment) • What is the empirical formula of a compound that is 85.63 % C and 14.37 % H by mass ? • How H tto solve l ?? % mass Î mass Î moles Î empirical formula 14.37 g H Second, convert grams of each element into moles of each element: Conversion: molar mass Mass Moles 1 mol C ⎞ (85.63 g C ) ⎛⎜ ⎟ = 7.130 mol C ⎝ 12.01 g C ⎠ 1 mol H ⎞ (14.37 g H ) ⎛⎜ ⎟ = 14.256 mol H ⎝ 1.008 g H ⎠ Third, determine empirical formula: Use the mole amounts as a subscript for each chemical symbol: C7.130 7 130 H14.256 14 256 Convert subscripts to smallest whole-number ratio; divide each by the smallest subscript. 7 C: H: Molecular Formula: 7.130 =1 7.130 14.256 = 1.9994 ≈ 2 7.130 The Empirical Formula is: 28.05 g = 1.9993 empirical formula units ≈ 2 14.03 g The molecular formula is: (CH2)2 Î C2H4 CH2 (smallest whole-number ratio of atoms in the Chemical Formula) Determine Empirical Formulas from a given Percent Composition by Mass (from experiment) Example: What is the empirical formula of a compound that is 48.6% C, 8.2% H, and 43.2% O by mass? Calculate moles of each element in 100 g sample: (48.6 g C)⎛⎜ 1 mol C ⎞ ⎟ = 4.047 mol C 12.01 g C⎠ ⎝ (8.2 g H )⎛⎜ 1 mol H ⎞ ⎟ = 8.135 mol H ⎝ 1.008 g H ⎠ (43.2 g O )⎛⎜ 1 mol O ⎞ ⎟ = 2.700 mol O ⎝ 16.00 g O ⎠ Desire a Molecular Formula ? To determine it we need to know the molar mass. # of empirical formula units in the molecular formula = Molar Mass Empirical Molar Mass Molar mass = 28.05 g (given) Empirical molar mass (CH2) = 14.03 g (determined from the above example) Convert to small mole ratios by dividing each by 2.700: C1.50H3.01O1 Multiply mole ratios (subscripts) by an appropriate multiplier to get smallest whole number ratios. Multiply mole ratios by a multiply of 2 to yield the empirical formula: C3H6O2 8 Required Multipliers for Fractional Subscripts Subscripts Multipliers Whole #’s 1.50 2 3 1.33 .33 3 3.99 Î 4 1.67 3 5.01Î 5 1.25 4 5 1.75 4 7 Calculations Involving Chemical Equations Examples: P1O2.50 X 2 Î P2O5 N1O1.50 X 2 Î N2O3 C1H2.67 X 3 Î C3H8 C1H1.33 X 3 Î C3H4 Mass % elements 100 g sample Grams each element Molar mass Molecular formula Molar atomic mass Empirical formula Calculate mole ratio Moles each element Stoichiometric relationships in chemical equations: CH4 + 2 O2 CO2 + 2 H2O ff in the equation q give the g The coefficients ratios by molecules or moles of reactants and products. For stoichiometric calculations, the coefficients represent mole quantities. “1 mole methane reacts with 2 moles of diatomic oxygen to form 1 mole carbon dioxide and 2 moles water” Stoichiometric Relationships • Also called “stoichiometric equivalents” • Show how reactants and products are related quantitatively (number of molecules or moles). moles) • How ? • Coefficients in the equation gives ratios by molecules or moles of reactants and products. 9 CH4 + 2 O2 CO2 + 2 H2O One “Stoichiometric equivalent” is: 1 mole CH 4 ⇔ 2 moles O 2 “One mole methane reacts with (is stoichiometrically equivalent to) two moles oxygen” CH4 + 2 O2 CO2 + 2 H2O Stoichiometric Equivalents CH4 + 2 O2 conversion factor 1 mole CH 4 ⇔ 2 mole O 2 CO2 + 2 H2O conversion factor conversion factor 1 mole CH 4 ⇔ 1 mole CO 2 1 mole CH 4 ⇔ 2 mole H 2 O 2 mole O 2 ⇔ 1 mole CO 2 2 mole O 2 ⇔ 2 mole H 2 O 1 mole CO 2 ⇔ 2 mole H 2 O Use these relationships to construct conversion factors for stoichiometric (quantitative) calculations. STOICHIOMETRIC CALCULATIONS • Construct conversion factors for calculations from stoichiometric equivalents CH4 + 2 O2 CO2 + 2 H2O • E Example: l How H many grams of f CO2 is i produced from 10.0 g of O2 as a reactant ? given grams O2 X conversion ratios & stoichiometric equivalents given i g O2 desired d i d g CO2 ⎛ 1 mol O 2 ⎞⎛ 1 mol CO 2 ⎞⎛ 44.01 g CO 2 ⎞ ⎟⎜ ⎟⎜ ⎟ ⎝ 32.0 g O 2 ⎠⎝ 2 mol O 2 ⎠⎝ 1 mol CO 2 ⎠ (10.0 g O 2 )⎜ = 6.88 g CO 2 Another example: The smelting (refining) of chalcopyrite ore to produce copper metal. 2 CuFeS2 + 5 O2 2 Cu + 2 FeO + 4 SO2 Chalcopyrite ore = desired grams CO2 10 An analogous part to this problem is to calculate how much of the pollutant, sulfur dioxide (SO2), is produced when 1.00 g of raw ore is refined ? 0.699 g SO2 Chalcopyrite ore (CuFeS2) How many g Copper (Cu) can be produced from 1.00 g of CuFeS2 ore ? Set up the conversion scheme: g CuFeS2 mol CuFeS2 2 CuFeS2 + 5 O2 mol Cu g Cu 2 Cu + 2 FeO + 4 SO2 ⎛ 1 mol CuFeS 2 ⎞⎛ 2 mol Cu ⎞⎛ 63.5 g Cu ⎞ ⎟⎜ ⎟⎜ ⎟ ⎝ 183 g CuFeS 2 ⎠⎝ 2 mol CuFeS 2 ⎠⎝ 1 mol Cu ⎠ (1.00 g CuFeS 2 )⎜ = 0.347 g Cu How many g oxygen (O2) is required to react with 1.00 g of CuFeS2 ore ? mol CuFeS2 2 CuFeS2 + 5 O2 mol O2 g O2 2 Cu + 2 FeO + 4 SO2 ⎛ 1 mol CuFeS 2 ⎞⎛ 5 mol O 2 ⎞⎛ 32.00 g O 2 ⎞ ⎟⎟⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ 183 g CuFeS 2 ⎠⎝ 2 mol CuFeS 2 ⎠⎝ 1 mol O 2 ⎠ (1.00 g CuFeS 2 )⎜⎜ Dictate how much product can be produced. Consider the following reaction: Set up the conversion scheme: g CuFeS2 Limiting Reactants N2 + 3 H2 2 NH3 1 mole N2 reacts with 3 moles H2 to yield 2 moles NH3 Stoichiometric amounts of reactants N2:H2 mole ratio of 1:3 Both reactants are completely consumed in the reaction. = 0.437 g O 2 11 N2 + 3 H2 2 NH3 What would happen if 3 moles N2 and 3 moles H2 are reacted? H2 is completely consumed, however, 2 mol of N2 remain unreacted. H2 Î Limiting reactant N2 Î Excess reactant Limiting reactant dictates (limits) how much product can be produced. N2 + 3 H2 2 NH3 Consider that 3 moles N2 and 3 moles H2 are reacted together. How many moles of N2 react with the H2? moles H2 Î moles N2 ⎛ 1 mol N 2 ⎞ ⎟⎟ = 1 mol N 2 ⎝ 3 mol H 2 ⎠ (3 mol H 2 )⎜⎜ How many moles N2 remain unreacted? 3 mol N 2 − 1 mol N 2 = 2 mol N 2 unreacted Limiting Reactants in Chemical Reactions • How is the limiting reactant determined? • By calculating the amount of product that each reactant would produce if it reacted completely. p y • The reactant yielding the least amount of product is the limiting reactant. Consider our previous reaction: N2 + 3 H2 2 NH3 Consider 3 mol N2 reacting with 3 mol H2 Calculate the amount of product produced from each reactant: ⎛ 2 mol NH 3 ⎞ ⎟ = 6 mol NH 3 ⎝ 1 mol N 2 ⎠ N2 (3 mol N 2 )⎜ H2 (3 mol H 2 )⎜ ⎛ 2 mol NH 3 ⎞ ⎟ = 2 mol NH 3 ⎝ 3 mol H 2 ⎠ H2 is the limiting reactant, N2 remains in excess, and 2 moles of NH3 is produced. Reaction Yield Products TWO TYPES THEORETICAL YIELD PERCENT YIELD Theoretical Yield: product that can be Maximum amount of p obtained from a given amount of reactants based on the limiting reactant. For the N2 + H2 reaction, the theoretical yield of NH3 is 2 mol. Typically, the actual yield of a reaction is somewhat less than the theoretical yield Î Percent Yield % yield = actual yield X 100 theoretical yield Consider that 1.5 mol of NH3 was obtained. % yield = 1.5 mol X 100 = 75 % 2.0 mol % yield < 100 % 12