Stoichiometry Notes.ppt.pptx

Classroom presentation for stoichiometry class. Suitable for use by teachers and students.

1. Stoichiometry Notes Chemistry

2. Stoichiometry Stoichiometry is the quantitative relationship between the amount of reactants used and the amount of products formed. It is based on the law of conservation of mass. The law states that matter is neither created nor destroyed in a chemical reaction. Meaning that the mass of the reactants MUST equal the mass of the products.

3. Using Balanced Equations Remember that the coefficients of reactants and products can represent # of atoms as well as the # of moles that reacted. Moles do not directly tell us the mass of the reactants; however, remember that the number of moles can be converted into grams.

4. Relationship from balanced equations IRON + OXYGEN → Iron (III) Oxide 4Fe + 3O 2 → 2Fe 2 O 3 4 atoms Fe + 3 molecules O 2 → 2 formula units of Fe 2 O 3 4 moles Fe + 3 moles O 2 → 2 moles Fe 2 O 3 223.4 g Fe + 96.0 g O 2 → 319.4 g Fe 2 O 3 319.4 g reactants 319.4 g products

5. Mole Ratios You can use the coefficients in a balanced chemical equation to write mole ratios. A mole ratio is a ratio between the number of moles of any 2 substances in a chemical equation. 2H 2 O 2 → O 2 + 2H 2 O Ex: 2 mol H 2 O 2 1 mol O 2 Ex: 1 mol O 2 2 mol H 2 O

6. Stoichiometric Conversions Mole to Mole Mole to Mass Mass to Mass

7. Stoichiometric Calculations Mole-to-Mole conversions When determining the number of moles produced from a reaction, you must first begin with a balanced equation! After writing your balanced equation, you then need to identify the substance you know and the one you need to determine. Use mole ratios to determine the substance that you need to find.

8. Stoichiometric Calculations cont. Mole-to-Mole conversions Example: A vigorous reaction between potassium and water produces hydrogen gas and potassium hydroxide. Steps: Write a balanced equation: 2K + 2H 2 O → 2KOH + H 2 How many moles of hydrogen are produced if 0.0400 moles of potassium are used? Moles of known x (mole ratio) = moles of unknown 0.0400 mol K x (1 mol H 2 /2 mol K) = 0.0200 mol H 2 Hint: the mole ratio should be the moles of the unknown over the moles of the known

9. Stoichiometric Calculations Mole-to-Mass conversions Now suppose you know the number of moles of a reactant or product and want to calculate the mass of another product or reactant. This is done the same way using mole ratios from the equation to get the number of moles of the unknown. Once you have the number of moles, use the molar mass to determine the number of grams that were produced.

10. Stoichiometric Calculations cont. Mole-to-Mass conversions How many grams of glucose (C 6 H 12 O 6 ) are produced when 24.0 mol of carbon dioxide reacts in excess water? 1. Determine number of moles of glucose produced by the given amount of carbon dioxide. 24.0 mol CO 2 X 1 mol C 6 H 12 O 6 = 4.00 mol C 6 H 12 O 6 6 mol CO 2

11. Stoichiometric Calculations cont. Mole-to-Mass conversions 2. Convert the number of moles of glucose to grams of glucose by multiplying # moles by molar mass. 4.00 mol C 6 H 12 O 6 X 180.18 g C 6 H 12 O 6 = 721 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 721 grams of glucose is produced from 24.0 moles of carbon dioxide.

12. Stoichiometric Calculations Mass-to-Mass conversions Using the mole ratios and mole/mass calculations, you can determine the needed masses to react and the masses that will be produced. First – write the balanced chemical equation for the reaction. Then convert the mass of the reactant to moles. Use the mole ratios to determine the moles of the unknown you are looking for. Then use molar mass to convert moles of the unknown to mass of the unknown.

13. Stoichiometric Calculations cont’d Mass-to-Mass conversions How many grams of NaOH are needed to completely react with 50.0 g of H 2 SO 4 to form Na 2 SO 4 and H 2 O? 1. Write the balanced equation. 2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2

14. Stoichiometric Calculations cont’d Mass-to-Mass conversions 2. Convert grams of sulfuric acid to moles of NaOH 50 g H 2 SO 4 X 1 mol H 2 SO 4 = 0.510 mol H 2 SO 4 98.09 g H 2 SO 4 0.510 mol H 2 SO 4 X 2 mol NaOH = 1.02 mol NaOH 1 mol H 2 SO 4

15. Stoichiometric Calculations cont’d Mass-to-Mass conversions 3. Calculate the mass of NaOH needed. 1.02 mol NaOH X 40.00 g NaOH = 40.8 g NaOH 1 mol NaOH 50 g of H 2 SO 4 reacts completely with 40.8 g of NaOH

16. Limiting Reactants and Percent Yield

17. Limiting Reactants A chemical reaction will proceed until all of one reactant is used up. The reactant that is used up is the limiting reactant. The left over reactants are called excess reactants.

18. Identifying the Limiting Reactant You can determine the limiting reactant in a reaction when given the mass of the reactants. Example: 40 g of sodium hydroxide reacts with 60 g of sulfuric acid. 1. Write the balanced chemical equation: 2 NaOH + H 2 SO 4 → Na 2 SO 4 + 2 H 2 O

19. 2. Convert the mass of the reactants into actual moles of reactants available. 40.0 g NaOH X 1 mol NaOH = 1.00 mol NaOH 40.0 g NaOH 60.0 g H 2 SO 4 X 1 mol H 2 SO 4 = .612 mol H 2 SO 4 98.09 g H 2 SO 4

20. 3. Compare the actual mole ratio calculated to the mole ratio from the balanced chemical equation. Actual mole ratio: 1.00 mol NaOH : 0.612 mol H 2 SO 4 Mole ratio from equation: 2 mol NaOH : 1 mol H 2 SO 4 Divide ratio by 2: 1 mol NaOH : 0.5 mol H 2 SO 4 All of the 1 mol of NaOH is used up (limiting reactant) and there is an excess of H2SO4 (0.612 mol – 0.5 mol = .112 mol left over)

21. Calculating mass of product produced when given mass of reactants. Multiply the number of moles of the limiting reactant by the mole ratio of the product to the limiting reactant. Then multiply by the molar mass of the product. 1.00 mol NaOH x 1 mol Na 2 SO 4 x 142.04 g Na 2 SO 4 = 71.0 g Na 2 SO 4 2 mol NaOH 1 mol Na 2 SO 4

22. Calculating Percent Yield Balance the equation Do a mass → mass conversion between the known and unknown Use the given “actual yield” to set up the ratio: Actual yield is the amount of product that was ACTUALLY produced (found in an experiment) Theoretical yield is the MAXIMUM amount of a reactant produced (from your stoichiometric calculations) Percent yield=(actual/theoretical) x 100

23. Calculating Percent Yield Ex: Aspirin (C 9 H 8 O 4 ) can be made from salicylic acid (C 7 H 6 O 3 ) and acetic anhydride (C 4 H 6 O 3 ). What is the percent yield of aspirin when you mix 13.2 g salicylic acid with an excess of acetic anhydride and obtain 5.9 g of aspirin and some water? Actual yield of aspirin = 5.9 g 1. Write the balanced equation. 2 C 7 H 6 O 3 + C 4 H 6 O 3 → 2 C 9 H 8 O 4 + H 2

24. 2. Calculate the theoretical yield . Salicylic acid is the limiting reactant. 13.2 g C 7 H 6 O 3 X 1 mol C 7 H 6 O 3 = 0.0956 mol C 7 H 6 O 3 138.1 g C 7 H 6 O 3 0.0956 mol C 7 H 6 O 3 X 2 mol C 9 H 8 O 4 = 0.0956 mol C 9 H 8 O 4 2 mol C 7 H 6 O 3 0.0956 mol C 9 H 8 O 4 X 180.2 g C 9 H 8 O 4 = 17.2 g C 9 H 8 O 4 1 mol C 9 H 8 O 4

25. Calculate the Percent Yield Percent yield = 5.9 g C 9 H 8 O 4 X 100 = 34% 17.2 g C 9 H 8 O 4