1. Percentage Composition Sec 7.3

2. Percentage Composition Review –What does the formula H 2 O tell us? –What does % mean? –You get 7 out of 10 on a test. What is your %? What if you get 23 out of 27? % means part out of the whole, times 100.

3. Percentage Composition Percent Composition—the mass of each element divided by the total mass of the compound. Ex: In the lab an unknown compound has a mass of 0.2370 g. Decomposition of the sample produces 0.09480 g of C, 0.1264 g of O, and 0.01580 g of H. What is the % comp?

4. Percentage Composition C:0.09480 g x 100% = 40.00% 0.2370 g O:0.1264 g x 100% = 53.33% 0.2370 g H:0.01580 g x 100% = 6.667% 0.2370 g

5. Percentage Composition Ex: A 1.5 g sample is made up of C and H only. If 0.80 g is C, what is the % comp?

6. Percentage Composition C:0.80 g x 100% = 53% 1.5 g H:0.7 g x 100% = 47% 1.5 g

7. Percentage Composition What is the percentage composition of ammonia? What’s the formula? NH 3 Assume: 1 mol of NH 3 So, how many moles of N? Of H?

8. Percentage Composition What’s the mass of N? Of H? Total? N:14.01 g x 100% = 82% 17.04 g H: 3.03g x 100% = 18% 17.04 g

9. Percentage Composition Ex: What is the % of water in magnesium sulfate heptahydrate? MgSO 4 ·7H 2 O Mass of H 2 O is 7 X 18.02 = 126.14 g. Total mass of the compound is 246.50 g.

10. Percentage Composition H 2 O:126.14 g x 100% = 51% 246.50 g So, there are 49% solids. Future lab calculation!

11. Empirical Formula Sec 7.4

12. Empirical Formula Empirical Formula: The simplest whole number ratio of atoms in a compound. H2O2H2O2 HO C2H6C2H6 CH 3 C 6 H 12 O 6 CH 2 O CH 4

13. Empirical Formula Ex: A given compound is analyzed in a mass spectrometer. It contains 89% O and 11% H by mass. What is the empirical formula?

14. Empirical Formula You could assume any starting mass, but what is a good number to assume? Assume: 100 g sample. How many grams of each element are there?

15. Empirical Formula O:89 g O 1 mol O = 5.56 mol O 16.00 g H:11 g H 1 mol H = 11 mol H 1.01 g

16. Empirical Formula So, the formula of your compound is H 11 O 5.56. Does this make sense? Need whole numbers for formulas. So, divide by the smallest number of moles to get a whole number.

17. Empirical Formula O:5.56 mol O = 1 mol O 5.56 mol H:11 mol H = 2 mol H 5.56 mol Formula = H 2 O

18. Empirical Formula Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca.

19. Empirical Formula Cl:2.128 g Cl 1 mol Cl = 0.0600 mol Cl 35.45 g Ca:1.203 g Ca 1 mol O = 0.0300 mol Ca 40.08 g

20. Empirical Formula Cl:0.0600 mol Cl = 2 mol Cl 0.0300 mol Ca:0.0300 mol Ca = 1 mol Ca 0.0300 mol Formula = Cl 2 Ca? NO!!!!! CaCl 2 !

21. Empirical Formula A 16.2 g sample contains only N and O. There are 4.2 g of N. What is the empirical formula?

22. Empirical Formula N: 4.2 g N 1 mol N = 0.3 mol N 14.01 g N O: 12 g O 1 mol O = 0.75 mol O 16.00 g O

23. Empirical Formula N:0.3 mol N = 1 mol N 0.3 mol O:0.75 mol O = 2.5 mol O? 0.3 mol Formula = NO 2.5 ?

24. Empirical Formula Need a whole number ratio. Multiply each by 2. 2.5 x 2 = 5 moles O 1 x 2 = 2 moles N Formula: N 2 O 5

25. Molecular Formula Sec 7.4

26. Molecular Formula A sample contains 4.90 g N and 11.2 g O. The molar mass is 92.0 g/mol. What is the molecular formula? Step #1: find the empirical formula. Step #2: Divide the actual molar mass by the molar mass of the empirical formula to get a ratio. Step #3: Multiply the empirical formula by the ratio you just found.

27. Molecular Formula N: 4.90 g N 1 mol N = 0.35 mol N 14.01 g N O: 11.2 g O 1 mol O = 0.7 mol O 16.00 g O

28. Molecular Formula N:0.35 mol N = 1 mol N 0.35 mol O:0.7 mol O = 2 mol O 0.35 mol Empirical Formula = NO 2

29. Molecular Formula Molar mass of NO 2 = 46 g/mol. But you know the molar mass of the actual compound is 92 g/mol. Take: 92/46 = 2 Multiply NO 2 by 2 to get the molecular formula. Molecular formula = N 2 O 4.

30. Molecular Formula Suppose you determine the empirical formula of a compound is CH 2 O. It’s molar mass is 120 g/mol. What is the molecular formula?

31. Molecular Formula Molar mass of CH 2 O is 30.03 g/mol. 120/30.03 = 4. 4 times CH 2 O = C 4 H 8 O 4.

32. Molecular Formula You do a lab to determine the formula of a hydrated salt. You know it is CuSO 4 ?H 2 O. The original mass of the compound is 1.179 g, and after heating it is.6585 g. What is the formula of the hydrate?

33. Molecular Formula First determine the mass of the dry salt and the water. Mass of dry salt =.6585 g CuSO 4. Mass of H 2 O = 1.179 -.6585 =.5205 g H 2 O Now determine moles of each.

34. Molecular Formula.5205 g H 2 O 1 mol H 2 O = 0.02888 mol H 2 O 18.02 g H 2 O.6585 g CuSO 4 1 mol = 0.004126 mol CuSO 4 159.61 g

35. Molecular Formula 0.02888 mol = 7 mol H 2 O.004126 mol 0.004126 mol = 1 mol CuSO 4 0.004126 mol Formula: CuSO 4 7 H 2 O