1. Stoichiometry Chapter 9

2. Step 1 Balance equations and calculate Formula Mass (FM) for each reactant and product. Example: Tin (II) fluoride, SnF 2, is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride, HF, according to the following equation. Sn (s) + HF(g) SnF 2 (s) + H 2 (g) How many grams of SnF 2 are produced from the reaction of 30.00g of HF with Sn?

3. Step 1 Balance the equation Sn (s) + HF(g) SnF 2 (s) + H 2 (g) Sn (s) + 2HF(g) SnF 2 (s) + H 2 (g) Calculate Formula Mass for all reactants and products. Sn = 119 g/mol HF=1+ 19= 20 g/mol SnF 2 = 119 + (2x19) = 157 g/mol H 2 = 2x1= 2 g/mol

4. Step 2 In the example problem we were given mass. – How many grams of SnF 2 are produced from the reaction of 30 g of HF with Sn? – We must turn the grams of HF into moles of HF. Example: 30gHF 1molHF 20gHF

5. Step 3 Determine the ratio of moles from the balanced equation. Sn (s) + 2HF(g) SnF 2 (s) + H 2 (g) The ratio is 1mol SnF 2 : 2 mol HF 30gHF 1molHF 1molSnF 2 2molHF 20gHF

6. Step 3 BEWARE: – If 2 or more givens are listed, you must determine the limiting reagent: this is the reactant that you have the least amount of moles. – ALWAYS use the limiting reagent to calculate the problem. If you do not, your answer will ALWAYS be WRONG.

7. Step 4 After calculating the mole ratios, proceed to calculate the mass for the unknown. If question only asks for moles of unknown, you are finished. Example: – How many grams of SnF 2 are produced from the reaction of 30 g of HF with Sn? 20gHF2molHF 30gHF 1molHF 1molSnF 2 157g SnF 2 = 1molSnF 2 117.75g SnF 2

8. How To Calculate Percent Yield You would follow all of the steps for solving stoichiometry. Using the example from above, let say that the problem read: – When 30.00 grams of HF reacts with an excess of Sn, the actual yield of SnF 2 is 100.5 g. What is the percent yield?

9. How To Calculate Percent Yield You would have followed all of the steps to calculate that the theoretical yield of SnF 2 was 117.75g. Percent yield = actual yield x 100 theoretical yield Percent yield= 100.5 g x 100 = 85.3% 117.75g