Chapter 9: Stoichiometry

1. Chapter 9:Stoichiometry Coach Kelsoe Chemistry Pages 298–325

2. Section 9-1:Introduction to Stoichiometry Coach Kelsoe Chemistry Pages 299–301

3. Section 9-1 Objectives • Define stoichiometry • Describe the importance of the mole ratio in stoichiometric calculations. • Write a mole ratio relating two substances in a chemical equation

4. Stoichi-what? • Stoichiometry (pronounced “sto-key-ometry”) comes from two Greek words: • Stoicheion means “element” • Metron means “measure”

5. Introduction to Stoichiometry • A lot of our knowledge of chemistry is based on the careful quantitative analysis of substances involved in chemical reactions. • Composition stoichiometry deals with the mass relationships of elements in compounds. • Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction.

6. Introduction to Stoichiometry • The good news: this is not entirely new material. We’ve already learned about composition stoichiometry in Section 3-2. We just never called it that! • Reaction stoichiometry is based on chemical equations and the law of conservation of matter. • ALL reaction stoichiometry calculations start with a balanced equation.

7. Reaction-Stoichiometry Problems • The problems in this chapter can be classified according to the information given in the problem and the information you are expected to find, called the unknown. • The given and unknown may both be reactants, they may both be products, or one may be a reactant and the other a product. • Masses should be expressed in grams. • Stoichiometric problems are solved by using ratios from the balanced equation to convert the given quantity using the following methods.

8. Reaction-Stoichiometry Problems • Problem Type 1:Given and unknown quantities are in moles. • When you are given the amount of substance in moles and asked to calculate the amount in moles of another substance in the chemical reaction, the general plan is • Amount of given Amount of unknownsubstance (in mol) substance (in mol)

9. Reaction-Stoichiometry Problems • Problem Type 2:Given is an amount in moles and the unknown is a mass that is often expressed in grams. • When you are given the amount in moles of one substance and asked to calculate the mass of another substance in the chemical reaction, the general plan is • Amount of Amount of Mass of unknowngiven substance unknown substance substance(in mol) (in mol) (in grams)

10. Reaction-Stoichiometry Problems • Problem Type 3:Given is a mass in grams and the unknown is an amount in moles. • When you are given the mass of one substance and asked to calculate the amount in moles of another substance in the chemical reaction, the general plan is • Mass of Amount of Amount of unknowngiven substance given substance substance(in grams) (in mol) (in mol)

11. Reaction-Stoichiometry Problems • Problem Type 4:Given is a mass in grams and the unknown is a mass in grams. • When you are given the mass of one substance and asked to calculate the mass of another substance in a chemical reaction, the general plan is • Mass of Amount of Amount of Mass ofgivengivenunknownunknownsubstance subtance substance substance(in grams) (in mol) (in mol) (in grams)

12. Mole Ratio • Solving any reaction-stoichiometry problem requires the use of a mole ratio to convert from moles or grams of one substance in a reaction to moles or grams of another substance. • A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. It is obtained directly from the balanced chemical equation.

13. Mole Ratio • Consider the following equation: 2Al2O3(l)  4Al(s) + 3O2(g) • Remember that the coefficients in a chemical equations satisfy the law of conservation of matter and represent the relative amounts of moles of reactants and products. • Therefore, 2 mol of aluminum oxide decompose to produce 4 mol of aluminum and 3 mol of oxygen gas. See board for the six ratios.

14. Mole Ratio • For the decomposition of aluminum oxide, the appropriate mole ratio would be used as a conversion factor to convert a given amount in moles of one substance to the corresponding amount in moles of another substance. • For example, if given 13 mol of aluminum oxide and need to find out how much oxygen it decomposes into:13.0 mol Al2O3 x 3 mol O2 = 19.5 mol O2 2 mol Al2O3

15. Mole Ratio • Think about this in terms of making a PB&J sandwich:4 slices of bread + 1 cup PB + 1 cup J  2 PB&J sandwiches • Let’s assume this equation is balanced. • If we have 16 slices of bread, how much PB will we need? • 16 bread x 1 cup PB = 4 cups PB • We can do this for any ingredient! 4 Bread

16. Mole Ratio • Mole ratios are exact, so they do not limit the number of significant figures in a calculation. • The number of significant figures in the answer is therefore determined only by the number of significant figures of any measured quantities in a particular problem. • In layman’s terms, you should have the same number of sig figs when you multiply by your mole ratio as you had in your given amount.

17. Molar Mass • Remember that the molar mass is the mass, in grams, of one mole of a substance. • The molar mass is the conversion factor that relates the mass of a substance to the amount in moles of that substance. • To solve reaction-stoichiometry problems, you will need to determine the molar masses using the periodic table.

18. Molar Mass • Returning to the aluminum oxide problem: 2Al2O3(l)  4Al(s) + 3O2(g) • The molar masses can be expressed as so: • Aluminumoxide • Aluminum • Oxygen 101.96 g 1 mol Al2O3 OR 1 mol Al2O3 101.96 g 26.98 g 1 mol Al OR 1 mol Al 26.98 g 32.00 g 1 mol O2 OR 1 mol O2 32.00 g

19. Molar Mass • To find the number of grams of aluminum equivalent to 26.0 mol of aluminum, the calculation would be as follows: • 26.0 mol Al x 26.98 g Al = 701 g Al 1 mol Al

20. Vocabulary • Composition stoichiometry • Mole ratio • Reaction stoichiometry