STOICHIOMETRY Greek: Stoicheon = element metron = element measuring Stoichiometry is the science of measuring the quantitative proportions or mass ratios in which chemical elements stand to one another Molar Ratios: xa + yb ac + zd Stoichiometric relationship: n A = nb = nc = nd x y a z Example 1: Consider the following balanced equation and write the stoichiometric relationship for all reactants and products 2C 2 H 6(g) + 7O 2 (g) 4CO 2 (g) + 6H 2O(l) Solution: nc2h6(g) = no2(g) = nco2(g) = nh2o(g) 2 7 4 6 1. Mole to Mole conversions: mol mol 3H 2 (g) + N 2 (g) 2NH 3 (g) 3 H 2 molecules 1 N 2 molecule 2 NH 3 molecules 3 mols H 2 1 mol N 2 2 mol NH 3 CHEM 110/2014 Page 1
Example 1. If 2.0 mol of N 2 (g) reacts with sufficient H 2 (g), how many mols of NH 3 (g) will be produced? Solution: nn 2 = nnh 3 2.0 mols = nnh 3 1 2 1 2 nnh 3 = 2.0 mols x 2 = 4.0 mols Example 2. 2. Mole to mass conversions: mol grams Example 1: How many grams of oxygen are produced when 1.50 mols of KClO3(s) are decomposed according to the balanced equation? 2 KClO3(s) 2KCl(s) + 3O2(g) nkclo3 = no 2 1.5 mols = no 2 2 3 2 3 Therefore no 2 = 1.50 mols = 2.25 mols no 2 = mass of O2 mass of O2 = 2.25 mols 32.00 g mol -1 Molar mass of O2 = 72.0 g CHEM 110/2014 Page 2
3. Mass to mole conversions 2 KClO3(s) 2 KCl(s) + 3 O2(g) Example: If 80.0 g of O2(g) was produced in the above reaction, calculate the number of moles of KClO3 decomposed. Solution: Stoichiometry from balanced equation is: nkclo3 = no 2 nkclo 3 = no2 2 3 = mass of O2 x Molar mass of O2 = 80.0 g 32.00 g mol -1 = 1.67 mols of KClO 3 decomposed Practice Exercise: Consider the following equation: 2H 2 (g) + O 2 (g) 2H 2 O (a) How many grams of H 2 O are produced when 2.50 moles of O 2 (g) is reacted? (b) If 3.00 moles of H 2 O is produced, calculate the mass of O 2 (g) that was used. (c) How many grams of H 2 (g) must be used, given the data in (b) above? CHEM 110/2014 Page 3
4. Mass to mass conversions Example 1 How many grams of Cl 2 (g) can be liberated from the decomposition of 64.0 g of AuCl 3 in the following reaction: 2AuCl 3 (aq) 2Au(s) + 3Cl 2 Solution: From stoichiometry naucl3 = nau = ncl2 2 2 3 naucl 3 = ncl 2 2 3 ncl 2 = naucl 3 x 3 = 64.0 g x 3 = 0.316 mol 2 303.32 g mol -1 2 mass of Cl 2 (g) = 0.316 mols x 70.9 g mol -1 = 22.40 g Practice Exercises 1. Calculate the mass of AuCl 3 that can be produced from 100 g of Cl 2 in the following reaction: 2Au(s) + 3Cl 2 2 AuCl 3 (aq) 2. Consider the degradation of glucose to carbon dioxide and water: C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O If 856.0g of C 6 H 12 O 6 is consumed by a person over a certain period of time, what is the mass of CO 2 produced? (Ans. Mass of CO 2 = 1.254 x10 3 g) CHEM 110/2014 Page 4
3. Using the following reaction: 2KI(aq) + Pb(NO 3 ) 2 (aq) PbI 2 (s) + 2KNO 3 (aq) Calculate the mass of PbI 2 (s) by reacting 30.0 g of KI with excess Pb(NO 3 ) 2. 4. How many grams of Na(s) are required to react completely with 75.0 g of Cl 2 (g) using the following equation: 2Na(s) + Cl 2 (g) NaCl (unbalanced) 5. A component of acid rain is sulfuric acid which forms when SO 2 (g), a pollutant reacts with oxygen and rain water according to the following reaction: 2SO 2 (g) + O 2 (g) + H 2O(l) H 2 SO 4 (aq) Assuming that there is plenty of O 2 (g) and H 2O(l), how much H 2 SO 4 in kilograms forms from 2.6 x 10 3 kg of SO 2 (g)? 5. Limiting Reagents The limiting reagent (or reactant) is the reactant that is completely consumed in a chemical reaction. The maximum amount of product formed depends on how much of this (limiting) reactant was originally present. Excess reagents are present in quantities greater than necessary to react with the quantity of the limiting reagent. Theoretical yield - the amount of product that can be made in a chemical reaction based on the amount of limiting reagent. Actual or (experimental) yield the amount of product actually produced by a chemical reaction. Percentage Yield = Actual Yield x 100 Theoretical Yield CHEM 110/2014 Page 5
Consider a recipe to bake pancakes: 1 cup flour + 2 eggs + ½ tsp baking powder 5 pancakes Suppose we have: 3 cups flour + 10 eggs + 4 tsp baking powder? pancakes We can make: 3 cups flour 15 pancakes 10 eggs 25 pancakes 4 tsp baking powder 40 pancakes Flour is the limiting reagent as it produces the least amount of pancakes Practice Exercise 1. Consider the following reaction: Ti(s) + 2Cl 2 (g) TiCl4(s) If we begin with 1.8 mol of Ti and 3.2 mol of Cl 2, what is the limiting reagent and calculate the theoretical yield of TiCl 4 in moles? Solution: Given: 1.8 mol Ti Find: limiting reagent 3.2 mol Cl 2 Theoretical yield Stoichiometry: nticl4 = ncl2 = 3.2 mols = 1.6 mols 1 2 2 CHEM 110/2014 Page 6
nticl 4 = nti = 1.8 mols 1 1 Since smaller amount of mols of TiCl 4 is produced from Cl 2, Therefore Cl 2 is the limiting reagent while Ti is the excess reagent. Therefore theoretical yield of TiCl 4 (s) = 1.6 mols Practice Exercise 2. Consider the following reaction: 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) If we begin with 0.552 mol of aluminium and 0.887 mol of chlorine, what is the limiting reagent and the theoretical yield? 6. Limiting Reagent, Theoretical Yield, and Percent Yield from Initial masses of Reactants Consider the following reaction: 2Na(s) + Cl 2 (g) 2NaCl(s) If we begin with 53.2 g of Na and 65.8 g of Cl 2, what is the limiting reactant and theoretical yield? nna = 53.2g = 2.31 mols, ncl 2 = 65.8g = 0.928 mols 22.99g mol -1 70.90g mol -1 nnacl = nna = 2.31 mols 2 2 nnacl = ncl2 nnacl = 0.928 mols x 2 = 1.856 mols 2 1 CHEM 110/2014 Page 7
Therefore the limiting reagent is Cl Theoretical yield (calculated from Cl) Mass of NaCl = 1.856 mols x 58.44 g mol -1 = 108.5 g NaCl Suppose when the synthesis was carried out, the actual yield of NaCl was found to be 86.4 g. What is the percent yield? Percentage Yield = Actual Yield x 100 Theoretical Yield = 86.4 g x 100 108.5 g = 80.0 % Practice Exercise 1. Ammonia can be synthesized by the Haber Process according to the following reaction: 3H 2 (g) + N 2 (g) 2NH 3 (g) (a) What is the maximum amount of ammonia in grams that can be synthesized from 25.2g of N 2 (g) and 8.42g of H 2 (g)? (b) What is the maximum amount of ammonia in grams that can be synthesized from 5.22g of H 2 (g) and 31.5 of N 2 (g)? 2. Consider the following reaction: Cu 2 O(s) + C(s) 2Cu(s) + CO 2 (g) When 11.5 g of C are allowed to react with 114.5 g of Cu 2 O(s), 87.4 g of Cu are obtained. Find the limiting reagent, theoretical yield and percent yield. CHEM 110/2014 Page 8
COMBUSTION REACTIONS A combustion reaction is one in which the elements in a compound react with molecular oxygen to form the oxides of those elements. For example C in a carbon-containing compound will be converted to CO2 and if there is hydrogen it will be converted to H2O Notice that by accurately measuring the mass of CO2 obtained by combustion of the carbon-containing compound, the mass of carbon in the original sample can be calculated. Similarly, by measuring the mass of H2O formed in the reaction, the mass of hydrogen in the original sample can be calculated. These calculations assume that all the carbon in the sample is captured in the CO2 and that all the hydrogen is captured in the H2O EXAMPLE 1: Combustion reaction involving C,H and O only Vitamin C is a compound that contains the elements C. H and O. Complete combustion of a sample of mass 0.2000 g of vitamin C produced 0.2998 g of CO2 and 0.08185 g of H2O. Determine the empirical formula of vitamin C. Molar masses (in g mol -1 ) : CO2 = 44.01 H2O = 18.02 ; C = 12.01 H = 1.008 O = 16.00 C x H y O z + O 2 x CO 2 + H 2 O Solution: All the C is converted to CO2 and all the H is converted to H2O. Step1: Determine the mass of C in CO 2 12.01 g mol -1 x 0.2998 g CO2 = 0.08181 g of C 44.01 g mol 1 CHEM 110/2014 Page 9
Step 2: Calculate the mass of water hydrogen in water 2 x 1.008 g mol -1 x 0.08185 g H2O = 0.009157g of H 18.02 g mol 1 (note: there are 2 mols of hydrogen in water) Step 3: Calculate the mass of O which is obtained by difference Mass of O = Mass of sample (mass of C + mass of H) = 0.2000 g (0.08181 g + 0.00915 g) = 0.1090 g of O Step 4: Convert to moles mass Atomic mass C H O Moles 0.006812 0.009084 0.006814 by smallest number of moles 0.006812 0.009084 0.006814 0.006812 0.006812 0.006812 1 1.33 1 convert to whole numbers by multiplying by 3 1 x 3 1.33 x 3 1 x 3 3 4 3 Therefore the empirical formula of vitamin C is C 3 H 4 O 3 CHEM 110/2014 Page 10
EXAMPLE 2. Combustion reaction involving C,H,O and N only The compound caffeine contains the elements C, H, N and O. Combustion analysis of a 1.500 g sample of caffeine produces 2.737g of CO 2 and 0.6814 g of H 2 O. A separate further analysis of another sample of mass 2.500 g of caffeine produces 0.8677 g of NH. 3 Determine the empirical formula of caffeine. Molar masses (in g mol -1 ): H = 1.008; C = 12.01; N = 14.0 ; O = 16.00 ; CO 2 = 44.01; H 2 O = 18.0 ; NH 3 = 17.04. Important features of this problem are that: 1. You are analysing for 4 elements C, H, N and O. 2. The analysis is performed on two samples (for example sample 1 and 2) which have different masses. NOTE: In the one sample you analyse for carbon and hydrogen. In the other you analyse for nitrogen. Remember that the oxygen is always obtained by difference. CHEM 110/2014 Page 11
Solution: In sample 1-1.500g of caffeine Step1: Determine the mass of C in CO 2 12.01 g mol 1 x 2.737 g CO 2 = 0.7469 g of C 44.01 g mol 1 Step 2: Calculate the mass of water hydrogen in water 2 x 1.008 g mol -1 x 0.06814 g H 2 O = 0.07623 g of H 18.02 g mol 1 In sample 2-2.500 g of caffeine Step 3. Determine the mass of N in NH 3 14.01 g mol 1 x 0.8677 g NH 3 = 0.7134 g of N 17.04 g mol 1 Step 4: Calculate % N in sample 2 = % N in sample 1 Note: The % composition of a pure compound is constant. If you know the percentage of N in sample 1 then the percentage of nitrogen in sample 2 is exactly the same. Therefore % N in sample 1 = 0.7134 x 100 = 28.54 % 2.500 g Step 5. You must determine the masses of all the elements in the compound in a common sample therefore find the mass of N in sample 1 i.e. 28.54 g X 1.500 g = 0.4280 g mass of N 100 g CHEM 110/2014 Page 12
Step 6. The mass of O in sample 1 is obtained by difference Mass of O = Mass of sample 1 (mass of C + mass of H + mass of N) = 1.500 g (0.7469 g + 0.07623 g + 0.4280 g) = 0.2489 g Determine the empirical formula C H N O Mass (g) 0.7469 0.07623 0.4280 0.2489g Atomic mass (g mol -1 ) 12.01 1.008 14.01 16.00 moles: 0.06219 0.07563 0.03057 0.01555 Divide by the smallest number of mol = 0.01555 4 4.9 2 1 ~5 Empirical formula is = C 4 H 5 N 2 O CHEM 110/2014 Page 13
PRACTICE EXERCISES 1. An 0.1888g sample of a hydrocarbon produces 0.6260g of CO2 and 0.1602g of H2O in combustion analysis. Its molecular weight is found to be 106 u. For this hydrocarbon determine (a) its mass percent composition (b) its empirical formula (c) its molecular formula Ans. 90.47% C. 9.50% H Ans. C4H5 Ans. C8H10 2. Dimethylhydrazine is a carbon hydrogen-nitrogen containing compound. Combustion analysis of a 0.312g sample of the compound produces 0.458 g of CO2. From a separate 0.525 g sample the nitrogen content is converted to 0.244 g N2. (a) What is the empirical formula of dimethylhydrazine? Ans. CH4N (b) If the molecular weight was found to be 60.02 what is the molecular formula of the compound? Ans. C2H8N2 3. A 1.35 g sample of a substance containing C. H. N and O is burned in air to produce 0.810 g of H2O and 1.32 g of CO2. In a separate analysis of the same substance all of the N in a sample of mass 0.735 g is converted to 0.284g NH3. Determine the empirical formula of the substance. Ans. CH3NO CHEM 110/2014 Page 14