1. Chemistry Handbook Review
Calculate the mass of 6.89 mol of Antimony (Sb).
6.89 mol Sb x ( g Sb / 1 mol Sb) =
839 g Sb

2. Chemistry Handbook Review
A chemisty needs mol selenium (Se) for a reaction. What mass of selenium should the chemist use?
mol Se x (78.96 g Se / 1 mol Se) =
5.53 g Se

3. Chemistry Handbook Review
A sample of sulfur (S) has a mass of 223 g. How many moles of sulfur are in the sample?
223 g S x (1 mol S / g S) = mol S

4. Chemistry Handbook Review
9. A tank of compressed helium (He) contains 50.0 g helium. How many moles of helium are in the tank?
50.0 g He x (1 mol He / g He) = mol He

5. Chemistry Handbook Review
A nickel coin is 25.0% nickel and 75.0% copper. If a nickel coin has a mass of 5.00 g, how many moles of Ni are in the coin?
5.00 g x 0.25 = 1.25 g Ni
1.25 g Ni x (1 mol Ni / g Ni) = mol Ni

6. Chemistry Handbook Review
Calculate the number of moles in 17.2 g of benzene (C6H6).
Molar mass of C6H6 is:
C: 6 x = g C
H: 6 x = g H
Total = 78.1 g C6H6
17.2 g C6H6 x (1 mol C6H6 / 78.1 g C6H6) = mol C6H6

7. Chemistry Handbook Review
Calculate the number of moles in g of potassium chlorate (KClO3).
Molar mass of KClO3 is:
K: 1 x = g K
Cl: 1 x = g Cl
O: 3 x = g O
Total = g KClO3
350.0 g KClO3 x (1 mol KClO3 / g KClO3) = mol KClO3

8. Chemistry Handbook Review
18. Determine the mass of mol of tin (II) sulfate (SnSO4).
Molar mass of SnSO4 is:
Sn: 1 x = g Sn
S: 1 x = g S
O: 4 x = g O
Total = g SnSO4
0.187 mol SnSO4 x ( g SnSO4 / 1 mol SnSO4 ) = 40.2 g SnSO4

9. Chemistry Handbook Review
19. A chemist needs 1.35 mol of ammonium dichromate for a reaction. The formula for this substance is (NH4)2Cr2O7. What mass of ammonium dichromate should the chemist measure out?
Molar Mass of (NH4)2Cr2O7 :
N: 2 x = g N
H: 8 x = 8.06 g H
Cr: 2 x = g Cr
O: 7 x = g O
Total: g (NH4)2Cr2O7
1.35 mol (NH4)2Cr2O7 x ( g (NH4)2Cr2O7 / 1 mol (NH4)2Cr2O7) = g (NH4)2Cr2O7

10. Chemistry Handbook Review
19.b. Find the number of H atoms found in 15.5 grams of methane (CH4).
Molar mass of CH4 :
1 mole C x g = g C
4 moles H x g = g H
Total mass of g/mol
15.5 g CH4 x (1 mol CH4 /16.04 g CH4) = mol CH4
0.966 mol CH4 x (6.02 x 1023 molecules CH4 / 1 mol CH4) = 5.82 x 1023 molecules CH4
5.82 x 1023 molecules CH4 x (4 atoms H/ 1 molecule CH4) = 2.33 x 1024 atoms H

11. Chemistry Handbook Review
A student needs mol each of zinc metal and copper (II) nitrate (Cu(NO3)2) for an experiment. What mass of each should the student obtain?
0.200 mol Zn x (65.39 g Zn / 1 mol Zn) = 13.1 g Zn

12. Chemistry Handbook Review
Molar Mass of Cu(NO3)2:
Cu: 1 x = g Cu
N: 2 x = g N
O: 6 x = g O
Total = g Cu(NO3)2
0.200 mol Cu(NO3)2 x ( g Cu(NO3)2 / 1 mol Cu(NO3)2) = 37.5 g Cu(NO3)2

13. Chemistry Handbook Review
Calculate the percent composition of aluminum oxide (Al2O3).
Molar Mass of Al2O3 :
Al: 2 x = g Al
O: 3 x = g O
Total = g Al2O3
% Al = / = 52.93%
% O = / = 47.07%

14. Chemistry Handbook Review
22. Determine the percent composition of magnesium nitrate, which has the formula Mg(NO3)2.
Molar Mass of Mg(NO3)2 :
Mg: 1 x = g Mg
N: 2 x = g N
O: 6 x = g O
Total = g Mg(NO3)2
% Mg = / = 16.39%
% N = / = 18.89%
% O = / = 64.72%

15. Chemistry Handbook Review
23. Calculate the percent oxygen in potassium chlorate KClO3.
Molar Mass of KClO3 :
K: 1 x = g K
Cl: 1 x = g Cl
O: 3 x = g O
Total = g KClO3
% O = / = 39.17%

16. Chemistry Handbook Review
24. Calculate the percent nitrogen in ammoinium hexacyanoiron (II), which has the formula (NH4)4Fe(CN)6.
Molar Mass of (NH4)4Fe(CN)6:
N: 10 x = g N
H: 16 x = g H
Fe: 1 x = g Fe
C: 6 x = g C
Total = g (NH4)4Fe(CN)6
% N = / = 49.30%

17. Chemistry Handbook Review
Acetylene gas has the molecular formula C2H2. How does the percent composition of acetylene compare with that of benzene (C6H6)?
The percent compositions will be the same because the ratio of carbon atoms to hydrogen atoms is the same, 1:1

18. Chemistry Handbook Review
The composition of acetic acid is 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen. Calculate the empirical formula for acetic acid.
40.00 g C x (1 mol C / g C) = mol C
6.71 g H x (1 mol H / g H) = 6.66 mol H
53.29 g O x (1 mol O / g O) = mol O
3.330 mol C / = mol C
6.66 mol H / = 2.00 mol H
3.331 mol O / = mol O
Empirical formula = CH2O

19. Chemistry Handbook Review
27. An unknown compound is analyzed and found to be composed of 14.79% nitrogen, 50.68% oxygen, and 34.53% zinc. Calculate the empirical formula for the compound.
14.79 g N x (1 mol N / g N ) = mol N
50.68 g O x (1 mol O / g O ) = mol O
34.53 g Zn x(1 mol Zn/65.39 g Zn) = mol Zn
1.056 mol N / = mol N
3.168 mol O / = mol O
mol Zn / = mol Zn
Empirical formula = N2O6Zn = Zn(NO3) 2

20. Chemistry Handbook Review
28. A lab investigation is done to determine the empirical formula for a compound that combines aluminum with carbon. An empty crucible and lid is measured at grams, and then again measured after aluminum was added at g. After reacting only with available carbon, the final mass was g. Determine the empirical formula of the compound.

21. Chemistry Handbook Review
Mass of Aluminum:
g – g = g Al
Mass of Carbon:
g – g = g C

22. Chemistry Handbook Review
3.442 g Al x (1 mol Al / g Al) = mol Al
1.149 g C x (1 mol C / g C) = mol C
mol Al / = mol Al
mol C / = mol C
Ratio of (1.334 : 1.000) x 3 = (4.002 : 3.000)
Empirical formula = Al4C3

23. Chemistry Handbook Review
29. The composition of ascorbic acid (vitimin C) is 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen. What is the empirical formula for vitimin C?
40.92 g C x (1 mol C / g C ) = mol C
4.58 g H x (1 mol H/ g H) = 4.54 mol H
54.50 g O x (1 mol O / g O ) = mol O
3.407 mol C / = mol C
4.54 mol H / = mol H
3.406 mol O / = mol O
Ratio of (1.000: : 1.000) x 3 = (3.000: : 3.000)
Empirical formula = C3H4O3

24. Chemistry Handbook Review
30. Ricinine is one of the poisonous compunds found in the castor plant. The composition of ricinine is 58.54% carbon, 4.91% hydrogen, 17.06% nitrogen, and % oxygen. Ricinine’s molar mass is g/mol. Determine its molecular formula.
58.54 g C x (1 mol C / g C ) = mol C
4.91 g H x (1 mol H/ g H) = 4.87 mol H
17.06 g N x (1 mol N / g N ) = mol N
19.49 g O x (1 mol O / g O ) = mol O
4.874 mol C / = mol C
4.87 mol H / = mol H
1.218 mol N / = mol N
1.218 mol O / = mol O
Empirical formula = C4H4NO

25. Chemistry Handbook Review
Molar Mass of C4H4NO:
C: 4 x = g C
H: 4 x = g H
N: 1 x = g N
O: 1 x = g O
Total = g C4H4NO
g/mol / g/mol =
Molecular formula = C8H8N2O2

26. Chemistry Handbook Review
31. The compound borazine consists of 40.29% boron (B), 7.51% hydrogen, and 52.20% nitrogen, and its molar mass is g/mol. Calculate the molecular formula for borazine.
40.29 g B x (1 mol C / g B) = mol B
7.51 g H x (1 mol H/ g H) = 7.45 mol H
52.20 g N x (1 mol N / g N) = mol N
3.727 mol B / = mol B
7.45 mol H / = 2.00 mol H
3.727 mol N / = mol N
Empirical formula = BH2N

27. Chemistry Handbook Review
Molar Mass of BH2N :
B: 1 x = g C
H: 2 x = g H
N: 1 x = g N
Total = g BH2N
80.50 g/mol / g/mol = 3.000
Molecular formula = B3H6N3

28. Chemistry Handbook Review
33. Find the percent of water and the percent of anhydrate in the compound Cr2(SO4)3•18H2O
Molar Mass of Cr2(SO4)3:
2 moles Cr x 52.0 g/mol = 104.0g
3 moles S x 32.1 g/mol = 96.3g
12 moles O x 16.0 g/mol = 192.0g
Total of g/mol
Molar Mass of 18 moles H2O:
18 x g/mol = g/mol

29. Chemistry Handbook Review
392.3 g/mol of anhydrate g/mol water = g hydrate
% anhydrate (Cr2(SO4)3): 392.3g / g = 54.8%
% water (H2O): 324.0g / g = 45.2%

30. Chemistry Handbook Review
Cerium (III) iodide (CeI3) occurs as a hydrate. This was evaulated during a lab by measuring an empty crucible and lid at grams, then adding the hydrate and re-measuring the mass at g. After heating over a bunsen burner, the final mass was grams. Calculate the formula for the hydrate.
Mass of water= g – g = g H2O
Mass of anhydrate= g – = g CeI3

31. Molar Mass of CeI3:
Ce: 1 mol Ce x ( g Ce / 1 mol Ce) = g Ce
I: 3 mol I x ( g I / 1 mol I) = g I
Total = g/mol
Molar Mass of H2O :
H: 2 mol H x (1.008 g H / 1 mol H) = g H
O: 1 mol O x ( g O / 1 mol O) = g O
Total = g/mol

32. Chemistry Handbook Review
1.520g CeI3 x (1 mol CeI3 / g CeI3) = mol CeI3
0.474g H2O x (1 mol H2O / g H2O ) = mol H2O
mol H2O / mol CeI3 = / = 9.01
Formula of hydrate = CeI3•9H2O

33. Chemistry Handbook Review
36. Cobalt (II) nitrate (Co(NO3)2) is used in ceramic glazes and produces a blue color when the ceramic is fired. This compound exists as a hydrate whose composition is 37.1% water and 62.9% Co(NO3)2 by mass. Write the molecular formula for this hydrate.
Molar Mass of Co(NO3)2 :
Co: 1 x = g Co
N: 2 x = g N
O: 6 x = g O
Total = g Co(NO3)2

34. Chemistry Handbook Review
62.9 g Co(NO3)2 x (1 mol Co(NO3)2 / g Co(NO3)2) = mol Co(NO3)2
37.1 g H2O x (1 mol H2O / g H2O ) = 2.06 mol H2O
mol H2O / mol Co(NO3)2 = 2.06 / = 5.99
Formula of hydrate = Co(NO3)2 •6H2O