Stoichiometry and Quantitative Analysis Using Mole Ratios is the study of the relative quantities of reactants and products in a chemical reaction you can use the number of moles for a given reactant or product to find the moles for any other reactant or product

2:7:4:6 Example Consider the following chemical reaction: 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) a) Write the ratio for all components of the reaction. 2:7:4:6 b) What amount, in moles, of CO2(g) is formed if 2.50 mol of C2H6(g) reacts? 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) 4 2 n = 2.50 mol n = 2.50 mol  = 5.00 mol

2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) c) What amount, in moles, of O2(g) is required to react with 10.2 mol of C2H6(g)? 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) 7 2 n = 10.2 mol n = 10.2 mol  = 35.7 mol d) What amount, in moles, of H2O(g) is formed when 100 mmol of CO2(g) is formed? 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) 6 4 n = 100 mmol n = 100 mmol  = 150 mmol = 0.150 mol

B. Gravimetric Stoichiometry mass measurements Steps 1. Write a including the states. Write the information balanced equation given. 2. Find the of the species using moles given n=m M 3. Find the of the species using moles wanted mole ratio wanted given 4. Calculate of the wanted species using mass m=nM

1 Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) Example 1 Iron is produced by the reaction of iron(III) oxide with carbon monoxide to produce iron and carbon dioxide. What mass of iron(III) oxide is required to produce 1000 g of iron? g w 1 Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)   m=? M = 159.70 g/mol m = 1000 g M = 55.85 g/mol 1 2 n = 17.90… x n = m M = 1000 g 55.85g/mol = 17.90… mol = 8.95… mol m = nM = (8.95… mol)(159.70 g/mol) = 1429.7225 g   = 1430 g

 1 Cu2(CO3)(OH)2(s) 2 CuO(s) + 1 CO2(g) + 1 H2O(g) Example 2 The decomposition of the mineral malachite, Cu2(CO3)(OH)2(s), yields copper(II) oxide, carbon dioxide and water vapour. What mass of copper(II) oxide is produced from 1.00 g of malachite? g w  1 Cu2(CO3)(OH)2(s) 2 CuO(s) + 1 CO2(g) + 1 H2O(g) m = 1.00 g M = 221.13 g/mol m=? M = 79.55 g/mol n = 0.00452… x 2 1 n = m M = 1.00 g 221.13 g/mol = 0.00452… mol = 0.00904… mol m = nM = (0.00904… mol)(79.55 g/mol) = 0.7194862 g = 0.719 g