1. Mole Calculations

2. The Mole Mole – measurement of the amount of a substance. –We know the amount of different substances in one mole of that substance.

3. Atomic Mass Unit Mass of 1 mole of compound Found by adding the atomic weights of each atom of each element that makes up a compound. H 2 O There are 2 atoms of hydrogen and 1 atom of oxygen.

4. AMU H = 2 x 1.01 = 2.02 O = 1 x 16 = 16 _____ 18.02 = amu 1 mole of any substance = the amu of that substance

5. AMU Molar mass = the sum of the molar masses of atoms of the elements in the formula Calculated the same way

6. The Mole One mole of any substance has Avogadro’s number. 6.022 x 10 23 atoms, molecules, ions

7. The Mole For gases only – 1 mole of a gas occupies 22.4 L

8. Converting Between Units amu  1 mole  6.022 x 10 23 atoms, mlc, ions (expressed in grams)

9. Conversions Change 5.0 grams of sodium chloride to moles of sodium chloride.

10. Change 0.45 moles of barium chlorate to grams of barium chlorate.

11. Determine the number of atoms in 15 grams of water.

12. Empirical Formulas

13. The simplest whole number ratio of moles of each element in the compound. H 2 O NaCl

14. Empirical Formulas Usually given in percentages of each element in the compound. Based on 100% of the compound. Can be compared to a 100 gram sample of the substance. 11.2% Hydrogen 88.8% Oxygen

15. 1. Change percents to grams 11.2% H = 11.2 g H 88.8% O = 88.8 g O

16. 2. Convert grams to moles. (We know an empirical formula is the lowest whole number ratio of moles) 11.2 g H | 1 mol = 11.089 mol | 1.01 g 88.8g O | 1 mol = 5.550 mol | 16 g

17. Cannot have decimals, the numbers must be whole numbers. Divide by the lowest. 11.2 g H | 1 mol = 11.089 mol / 5.550 = 2 | 1.01 g 88.8g O | 1 mol = 5.550 mol / 5.550 = 1 | 16 g H 2 O = empirical formula

18. 36.84% N 63.16% O

19. 35.98% Al 64.02% S

20. Molecular Formula

21. Specifies the actual number of atoms of each element in one molecule or formula unit of the substance.

22. The molar mass of acetylene is 26.04 g/mol and the mass of the empirical formula, CH, is 13.02 g/mol

23. 1. The problem will give you a molar mass of the compound. 2. Calculate the empirical formula as usual.

24. 40.68% carbon 5.08% hydrogen 54.24% oxygen Molar mass = 118.1 g/mol

25. Change % to grams 40.68 g C | 1 mol= 3.387 mol | 12.01 g 5.08 g H | 1 mol= 5.030 mol | 1.01 g 54.24 g O | 1 mol= 3.390 mol | 16 g

26. 3.387 mol / 3.387 mol = 1 x2 = 2 5.030 mol / 3.387 mol = 1.5 x2 = 3 3.390 mol / 3.387 mol = 1 x2 = 2 Empirical Formula = C 2 H 3 O 2

27. To find the molecular formula: (EF amu )x = MM (molar mass)

28. C 2 H 3 O 2 = 59.05 = amu (59.05)x = 118.1 X = 2 (EF) x2 = (C 2 H 3 O 2 ) x2 = C 4 H 6 O 4

29. 65.45% C 5.45% H 29.09% O MM = 110.0 g/mol

30. 49.98 g C 10.47 g H MM = 58.12 g/mol

31. 46.68% N 53.32% O MM = 60.01 g/mol