Stoichiometry Ch. 9

The Arithmetic of Equations 9-1

Stoichiometry Stoichiometry = the calculation of quantities in chemical reactions. 2H 2 + O 2  2H 2 O

Interpreting Chemical Equations N 2 (g) + 3H 2 (g)  2NH 3 (g) N 2 (g)3H 2 (g)2NH 3 (g) Atoms Molecules Moles (coefficient) 1 mol N 2 3 mol H 2 2 mol NH 3 Molar Mass (don’t use coefficient) 28g N 2 2g H 2 17g NH 3

Practice! H 2 (g) + I 2 (g)  2HI(g) H 2 (g)I 2 (g)2HI(g) Atoms Molecules Moles (coefficient) Mass (don’t use coefficient)

Practice! H 2 (g) + I 2 (g)  2HI(g) H 2 (g)I 2 (g)2HI(g) Atoms Molecules Moles (coefficient) 1 mol H 2 1 mol I 2 2 mol HI Mass (don’t use coefficient) 2g H g x 2 = 253.8g I g + 1g = 127.9g HI

Chemical Calculations 9-2

Mole-Mole Calculations N 2 (g) + 3H 2 (g)  2NH 3 (g) 1 mol N mol H 2  2 mol NH 3 Mole Ratios = used to calculate between moles of reactants and products. –6 possible mole ratio’s from equation above: 1 mol N 2 2 mol NH 3 3 mol H 2 3 mol H 2 1 mol N 2 2 mol NH 3 1 mol N 2 2 mol NH 3 3 mol H 2

Mol A  Mol B Calculation: ____mol A x mol B = _____mol B mol A Example: How many moles of NH 3 are produced when 0.60mol of N 2 reacts with H 2 ? N 2 + 3H 2  2NH mol N 2 x 2 mol NH 3 = 1.2 mol NH 3 1 mol N 2

Stoichiometric Calculations Atoms/Molecules of A Mass of A Volume of A Atoms/Molecules of B Mass of B Volume of B x Mol B x Mol A atoms A x 1 mol A 6.02x10 23 a A Mass A x 1 mol A molar mass A Volume A x 1 mol A 22.4L A x 6.02x10 23 a B = atoms B 1 mol B x molar mass B = mass B 1 mol B x 22.4L B = volume B 1 mol B Mole Ratio

Practice! How many molecules of O 2 are produced from 29.2g of water? 2H 2 O(l)  2H 2 (g) + O 2 (g) mass A  mol A/B  molec B 29.2g H 2 O x 1 mol H 2 O x 1 mol O 2 x 6.02x10 23 molec O 2 = 18g H 2 O2 mol H 2 O 1 mol O 2 = 4.88x10 23 molecules O 2

More Practice! How many liters of oxygen are needed to produce 34.2g of SO 3 ? 2SO 2 (g) + O 2 (g)  2SO 3 (g) Mass (g)  mol  volume (liters) 34.2g SO 3 x 1 mol SO 3 x 1 mol O 2 x 22.4L O 2 80g SO 3 2 mol SO 3 1 mol O 2 = 4.79 L O 2

Percent Yield 9-3

Calculating the Percent Yield Theoretical Yield: amount of product a chemical reaction predicts by stochiometry. Actual Yield: is less than the theoretical yield due to lab techniques, equipment, etc. Percent Yield: ratio of the actual yield to the theoretical yield expressed as a percent. –Shows efficiency of reaction –Percent yield = actual yield x 100% theoretical yield

Example A theoretical yield, using stoichiometry, of 4.55g of ammonia is produced. The actual yield from lab is 3.86g. What is the percent yield of this reaction? –Percent yield = 3.86g x 100 = 84.8% 4.55g

Example What is the theoretical yield of CaO if 24.8g CaCO 3 is used? What is the percent yield if 13.1g CaO is produced in lab? CaCO 3 (s)  CaO(s) + CO 2 (g) 24.8g CaCO 3  ? g CaOmass  mol  mass 1) Theoretical yield = 24.8g CaCO 3 x 1 mol CaCO 3 x 1 mol CaO x 56g CaO 100g CaCO 3 1 mol CaCO 3 1 molCaO = 13.9g CaO 2) Percent yield = 13.1g CaO x 100% = 94.2% 13.9g CaO