1. It doesn’t matter if you are on the right track, you will still get run over if you don’t keep moving.

2. Stoichiometry Lesson 5 The Journey Continues Lesson 5 The Journey Continues

3. Law of Conservation of Mass Mass is neither created nor destroyed.

4. Problem Balance the following equation. Potassium chloride combines with bromine to produce chlorine gas and potassium bromide. Balance the following equation. Potassium chloride combines with bromine to produce chlorine gas and potassium bromide.

5. Equation 2KCl + Br 2 -----> Cl 2 + 2 KBr

6. What is the mole to mole ratio 2KCl + Br 2 -----> Cl 2 + 2 KBr 2 : 1 : 1 : 2 2KCl + Br 2 -----> Cl 2 + 2 KBr 2 : 1 : 1 : 2

7. Give the mass of the reactants and products. 2KCl + Br 2 -----> Cl 2 + 2 KBr Mass of reactants : [(2 x 39.1) + (2 x 35.45)] + (2 x 79.9) = 308.9 g Mass of products: (2 x 35.45) + [(2 x 39.1) + (2 x 79.9)] = 308.9 g 2KCl + Br 2 -----> Cl 2 + 2 KBr Mass of reactants : [(2 x 39.1) + (2 x 35.45)] + (2 x 79.9) = 308.9 g Mass of products: (2 x 35.45) + [(2 x 39.1) + (2 x 79.9)] = 308.9 g

8. Empirical formula Lowest whole number ratio of the elements in a compound.

9. Problem What is the empirical formula of a compound that is 25.9% N and 74.1 % O? Known : In 100 g, 25.9g are N and 74.1 g are O. What is the empirical formula of a compound that is 25.9% N and 74.1 % O? Known : In 100 g, 25.9g are N and 74.1 g are O.

10. To Solve – Convert to mol 25.9 g N 1 mol N = 1.85 mol N 14.01 g N 74.1 g O 1 mol O = 4.63 mol O 16 g O Remember Empirical formula is lowest whole number. 25.9 g N 1 mol N = 1.85 mol N 14.01 g N 74.1 g O 1 mol O = 4.63 mol O 16 g O Remember Empirical formula is lowest whole number.

11. Now divide by the smallest number. 1.85 = 1 mol N 1.85 4.63 = 2.5 mol O 1.85 Still not a whole number. 1.85 = 1 mol N 1.85 4.63 = 2.5 mol O 1.85 Still not a whole number.

12. Now multiply Multiply the 2.5 by 2. You must get the decimal to be a whole number. What you do to one number you must do to all. Multiply the 2.5 by 2. You must get the decimal to be a whole number. What you do to one number you must do to all.

13. So multiply 2.5 by 2 = 5 Multiply 1 by 2 = 2 So the empirical formula is N 2 O 5 So multiply 2.5 by 2 = 5 Multiply 1 by 2 = 2 So the empirical formula is N 2 O 5

14. Molecular formula Actual amount given.

15. Problem Analysis of a compound reveals this composition: 80% C and 20% H. It's molecular weight is 30.08 g. What is the molecular formula? Analysis of a compound reveals this composition: 80% C and 20% H. It's molecular weight is 30.08 g. What is the molecular formula?

16. Work First find the empirical formula. 80 g C 1 mol C = 6.66 mol C 12.01 g C 20 g H 1 mol H = 19.80 mol H 1.01 g H First find the empirical formula. 80 g C 1 mol C = 6.66 mol C 12.01 g C 20 g H 1 mol H = 19.80 mol H 1.01 g H

17. Divide by the smallest 6.66 mol C = 1 mol C 6.66 mol 20 mol H = 3 mol H 6.66 mol So the empirical formula is CH 3 6.66 mol C = 1 mol C 6.66 mol 20 mol H = 3 mol H 6.66 mol So the empirical formula is CH 3

18. Now find the molar mass (1 x 12.01) + (3 x 1.01) = 15.04 Now divide the original molecular mass given in the problem 30.08 g by 15.04 g. The answer is equal to 2. (1 x 12.01) + (3 x 1.01) = 15.04 Now divide the original molecular mass given in the problem 30.08 g by 15.04 g. The answer is equal to 2.

19. Now Take CH 3 times 2. You have the molecular formula C 2 H 6. Take CH 3 times 2. You have the molecular formula C 2 H 6.