Sample Problems 8

1. Sample Problems 8

2. Sample Problem 1 • What is the formula mass of H2? • The subscript “2” in the formula means that this formula represents two hydrogen atoms. Because the atomic mass of H is 1.008 u . You will get the following: 1.008 + 1.008 = 2.016 u

3. Sample Problem 2 • What is the formula mass of H20? • The formula H2O represents 3 atoms; 2 H and 1 O atom. The formula mass is the sum of the atomic masses of these 3 atoms. 1.0 u + 1.0 u + 16.0 u = 18 u

4. Sample Problem 3 What is the formula mass of calcium hydroxide? Element Atomic Mass Atoms/formula = Ca 40 u 1 40 u O 16 u 2 32 u H 1 u 2 2 u ____ Total 74 u

5. Sample Problem 4 • What is the mass in grams of 1.00 mole of sulfur atoms? • One mole of sulfur atoms is the number of atoms in the gram atomic mass of sulfur. Amu =32.1 u • The gram atomic mass of sulfur is 32.1 g • Therefore the mass of 1.00 mole of sulfur atoms is 32.1 g

6. Sample Problem 5 • What is the mass of 5.0 moles of sulfur? • 5.0 moles S atoms X 32.1 g S 1.00 moles of S atoms = 160 g S or 1.6 X 102 g S

7. Sample Problem 6 • How many moles of carbon atoms are present in a sample with a mass of 24 g? • ? Mol C =24 g C ? Mol C = 24 g C X 1.0 mol C = • 12 g C

8. Sample Problem 6 • (cont)How many carbon atoms would there be in the sample? • ? Atoms C = 2.0 mol C • ?= 2.0 mol C X 6.02 X 1023 atoms C • 1 mol C

9. Sample Problem 7 • What is the mass of 1.0 mole of water? • 1.0 moles of water X 18 grams of water • 1 mole of water • = 18 grams

10. Sample Problem 8 • What is the mass of 1.0 mole of calcium hydroxide, Ca(OH)2 ? • 1.0 mole of Ca(OH)2 X 74 g Ca(OH)2 • 1.0 mole Ca(OH)2

11. Sample Problem 9 • Find the number of Molecules in a sample of oxygen gas with a mass of 64.0 g. • 64 g of O2 X 1 mole/32 g O2 X 6.02 X 1023 molecules / 1 mole of O2 = 1.20 1024 molecules of O2

12. Sample Problem 10 • A sample of a compound containing carbon and oxygen had a mass of 88 g. Experimental procedures showed that 24 g of this sample was carbon and remaining 64 g was oxygen. What is the percentage composition of this compound?

13. 10 (cont) • % Carbon = mass of carbon in sample X 100 • mass of sample • = 24 g X 100% = 27% Carbon • 88 g • % Oxygen = mass of Oxygen in sample X 100% • mass of sample • = 6 • 6 • 64 g / 88 gX 100% = 73%

14. Sample Problem 11 • What is the percentag composition of carbon dioxide? • Solution Find the number of moles to masses by multiplying by the gram atomic masses. • 2.0 moles O X 16 g O/1.0 mol O = 32 g O • 1.0 mole C X 12 g C/1 mol C = 12 g C /44 g CO2

15. 11(cont) • % oxygen in CO2 = 32 g O/44 g CO2 X 100% = 73% oxygen • % Carbon in CO2 = 12 g C/ 44 g CO2 X 100% = 27% carbon

16. Sample Problem 12 • Lab procedures show that a 9.2 g sample of a compound is 2.8 g nitrogen and 6.4 g Oxygen. Find the empirical formula of the compound. • 2.8 g of N X 1.0 mol N atom/ 14 g N = .20 mol N 6.4 g of O X 1 mol of O atoms/16 g O = .40 mol O atoms

17. 12 (cont) • Using the number of moles from Step 1 determine the smallest whole number ratio (divide larest by smallest) • O atoms .40/ .20 = 2 • Therefore it is NO2

18. Sample Problem 13 • What is the Empirical formula of the compound with a percentage composition of 65.2 % arsenic and 34.8% Oxygen by mass? • Change % to grams in 100 g sample. • 65.2 g X 1 mol As atoms/74.9 g As = .870 mol As • 34.8 g O X 1.00 mol O atoms/16.0 g O = 2.18 mol O atoms

19. 13 (cont) • Take smaller divided by larger get 2.51 to 1 ratio then smallest whole number ratio.= • AS2O5.

20. Sample Problem 14 • What is the molecular formula of this compound? Empirical formula is NO2 • The molecular mass is 46 μ.

21. Sample Problem 15