Answers to Mole Review Worksheet:

1. a. 180.0 amu b. 55.85 amu

2. a. one molecule of glucose contains 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen.

b. 1.5 x 10 24 atoms

3. a. 36.5 g b. 16.0 g c. 48.0 g d. 132.1 g

4. (5.88 moles) 3.54 x 1024 molecules

5. a. 26.35 g x 1mol = 0.313 moles MgCO3

84.3 g

b. 18.0 mol x 102.0 g = 1836 g Al2O3

1 mol

c. 125.52 g x 1 mol = 0.9678 mol NiCl2

129.7 g

d. 23.2 mol x 154.0 g =3573 g CCl₄

1 mol

6. Al₂O₃ contains 52.9% Al. 52.9% of 200. g is 105.8 g of Al

Al₂O₃ = 102. 0 g Al = 2 x 27.0 = 54.0 g/102.0g x 100 = 52.9 %

200 g compound x 52.9 g Al = 105.8 g Al

100 g compound

7. 73.86% Hg

HgCl₂ = 271.6 g ; Hg = 200.6g/271.6g x 100 = 73.86%B

8. C₄H₉

(reduce 8:18 to 4:9)

9. NiC₂N₂

Ni = 52.6 g x 1 mol = 0.896 mol = 1

58.7 g 0.896

C = 21.9 g x 1 mol = 1.825 mol = 2

12.0 g 0.896

N = 25.5 g x 1 mol = 1.821 mol = 2

14.0 g 0.896

10. CH₂O

C = 7.20 g x 1 mol = 0.600 mol = 1

12.0 g 0.600

H = 1.20 g x 1 mol = 1.20 mol = 2

1.0 g 0.600

O = 9.60 g x 1 mol = 0.600 mol = 1

16.0 g 0.600

11. C₁₁H₂₂O₁₁

n = Molecular formula molar mass (given) = 330 g =11….11 x (CH₂O) = C₁₁H₂₂O₁₁

Empirical formula molar mass (calculated) 30g

12. Fe₂O₃ this compound is called iron(III) oxide

Fe = 18.61 g x 1 mol = 0.334 mol = 1 x 2 = 2

55.8 g 0.334

O = 26.61 g – 18.61 g = 8.00 g x 1 mol = 0.500 mol = 1.5 x 2 = 3

16.0 g 0.334