2. Ch. 7 Review Molar Mass Calculations

3. Question #1 (moles  mass) What is the mass of 7.50 moles of sulfur dioxide, SO 2 ? S = 32.07 g/mol O = 16.00 g/mol 7.50 mol SO 2 = g SO 2 1 mol SO 2 g SO 2 X

4. 7.50 mol SO 2 = g SO 2 1 mol SO 2 g SO 2 X 64.07 g SO 2 64.07 g SO 2 SO 2 32.07 g 16.00 g x 2 32.07 g 32.07 g + 32.00 g = x 1

5. 7.50 mol SO 2 = g SO 2 1 mol SO 2 g SO 2 480.525 64.07 X g SO 2 64.07 g SO 2 SO 2 32.07 g 16.00 g x 2 32.07 g 32.07 g + 32.00 g = x 1 481

6. Question 2: (Mass  Moles) How many moles are there in 993.6 g of potassium sulfate K 2 SO 4 ? K=39.10 g/mol; S=32.07 g/mol; O=16.00 g/mol 993.6 g K 2 SO 4 = mol K 2 SO 4 g K 2 SO 4 1 mol K 2 SO 4 X

7. 993.6 g K 2 SO 4 = mol K 2 SO 4 g K 2 SO 4 1 mol K 2 SO 4 X + 64.00 g = 174.27 g K 2 SO 4 39.10 g x 2 32.07 g x 1 78.20 g 78.20 g + 32.07 g 16.00 g x 4 174.27 5.701

8. Question 3: (Moles  Molecules) How many molecules are there in 4.55 moles of nitrogen (N 2 )? 4.55 mol N 2 = molecules N 2 1 mol N 2 molecules N 2 X 1 mole = 6.02 x 10 23 molecules 6.02 x 10 23 2.74 x 10 24

9. Question 4: (Moles  molecules  atoms) How many atoms of nitrogen, N, are there in 2.18 moles of nitrogen (N 2 )? 2.18 mol N 2 = atoms N 1 mol N 2 molecules N 2 X 1 mole = 6.02 x 10 23 molecules 6.02 x 10 23 2.62 x 10 24 1 molecules N 2 atoms N X 2

10. Question 5: (molecules  moles) How many moles are there in 1.326 x 10 12 molecules of carbon tetrachloride (CCl 4 )? 1.326 x 10 12 molecules CCl 4 = mol CCl 4 molecules CCl 4 1 mol CCl 4 X 1 mole = 6.02 x 10 23 molecules 6.02 x 10 23 2.202 x 10 -12

11. Question 6: (moles  volume) What is the volume occupied by 4.20 moles of oxygen gas (O 2 ) at STP? 4.20 mol O 2 = L O 2 1 mol O 2 L O 2 X 1 mole = 22.4 L for gas @STP 22.4 94.08

12. Question 7: (volume  moles) How many moles are there in 0.335 dm 3 of argon gas (Ar) at STP? 0.335 dm 3 Ar = mol Ar dm 3 Ar 1 mol Ar X & 1 mole = 22.4 dm 3 for gas @STP 22.4 0.0150 Note: 1 L = 1 dm 3

13. %NX 100 = 92.02 g 28.02 g = Question 8: (Percent composition) 92.02 g N 2 O 4 92.02 g N 2 O 4 N2O4N2O4N2O4N2O4 14.01 g 16.00 g x 4 28.02 g 28.02 g + 64.00 g = x 2 30.4 % N %OX 100 = 92.02 g 64.00 g = 69.6 % O

14. %OX 100 = 0.432 g 0.128 g = Question 9: (Percent composition) 29.6 % O %F=- 29.6%100% = 70.4 % F A sample of a compound that has a mass of 0.432 g is analyzed. The sample is found to be made up of oxygen and fluorine only. Given that the sample contains 0.128 g of oxygen, calculate the percentage composition of the compound. Alternative method: 0.432 g – 0.128 g = 0.304 g F %FX 100 = 0.432 g 0.304 g = 70.4 % F

15. 47.9 g Zn= 65.39 g Zn 1 mol Zn x Question 10: (Empirical formula) 0.733 mol Zn 52.1 g Cl= 35.45 g Cl 1 mol Cl x 1.47 mol Cl 0.733 = 1 = 2 ZnCl 2 Assume 100g sample. 30 Zn 65.39 Find the empirical formula of a compound, given that the compound is found to be: 47.9% zinc (Zn) and 52.1% chlorine (Cl) by mass. 17 Cl 35.45