Regents Chemistry
Mole Unit Test Review
Formula/Molar mass - Find the formula/molar mass of the following:
1. CO2
1(12) + 2(16) = 44g (44g/mol)
2. Al2O3
2(27) + 3(16) = 102g (102g/mol)
3. C6H10O5 6(12) + 10(1) + 5(16) = 162g (162g/mol)
4. ZnCl2 1(65) + 2(35) = 135g (135g/mol)
5. H2SO4
2(1) + 1(32) + 4(16) = 98g (98g/mol)
6. H2O 2(1) + 1(16) = 18g (18g/mol)
7. KCl 1(39) + 1(35) = 74g (74g/mol)
8. C12H22O11 12(12) + 22(1) + 11(16) = 342g (342g/mol)
9. CaCO3 1(40)+ 1(12)+ 3(16) = 100g (100g/mol)
10. CuSO45H2O 1(64) + 1(32) + 4(16) + 10(1) + 5(16) = 250g (250g/mol)
Draw the mole map in the space below:
Perform the following conversions:
11. 2 moles of NH3 to grams
(2moles)(17g/mol) = 34g
12. 0.4 moles of NaCl to grams
(0.4moles)(58g/mol) = 23.2g
13. 0.8 moles of CO to particles
0.8moles
x 6.02x1023 parts/mole
4.82x1023parts
17. 234.0 grams of NaCl to particles
14. 59.5 grams of NH3 to moles
59.5g = 3.5moles
17g/mol
15. 3 moles of CO to liters
(3moles)(22.4L/mol) = 67.2L
16. 4.5 moles of C2H4 to liters
(4.5moles)(22.4L/mol) = 100.8L
19. 11.2 liters of CO to moles
234.0g = 4.03mol
58g/mol
11.2L = 0.5mol
22.5L/mol
4.03mol
x 6.02x1023parts/mol
2.43 x 1024parts
18. 3.20x1025 particles of CH4 to grams
3.20x1025 particles = 53.16mol
6.02 x 1023parts/mol
20. 89.6 liters of CH4 to moles
89.6 liters = 4moles
22.4L/mol
(53.16mol)(16g/mol) = 850.56g
21. number of atoms of oxygen in 18.0 grams of water
(0.8889)(6.02x1023parts/mol) =
5.35x1023 parts oxygen
16 = 88.89% oxygen in 1 mole H2O
18
22. number of atoms of hydrogen in 18.0g of water
2 = 11.11% hydrogen in 1 mole H2O
18
(0.1111)(6.02x1023) = 6.69x1022parts
Empirical Formulas: elements in a chemical formula are in the lowest possible ratios
Determine the empirical formulas for each of the following molecular formulas:
33. C8H18
C4H9
38. H2O
34. H2O2
HO
39. C4H8
35. Hg2Cl2
HgCl
H2O
CH2
40. C7H12
C7H12
CH2
36. C3H6O3
CH2O
41. C4H8
37. Na2C2O4
NaCO2
42. CH3COOH
CH2O
Steps to find the molecular formula from the empirical formula and the molecular mass:
 Step 1: determine the mass of the empirical formula
 Step 2: divide the molecular mass by the empirical formula mass to find the multiple
 Step 3: multiple the subscripts in the empirical formula by the multiple to find the
molecular formula
Determine the molecular formula for each of the following:
43. Find the molecular formula for a compound with a mass of 78 amu and an
empirical formula of CH. 1(12) + 1(1) = 13g
78 = 6
13
CH  C6H6
44. Find the molecular formula for a compound with a mass of 82 amu and an empirical
formula of C3H5. 3(12) + 5(1) = 41g
82 = 2
41
C3H5  C6H10
45. Find the molecular formula for a compound with a mass of 90 amu and an empirical
formula of HCO2. 1(1) + 1(12) + 2(16) = 45g
90 = 2
45
HCO2  H2C2O4
46. Find the molecular formula for a compound with a mass of 112 amu and an empirical
formula of CH2. 1(12) + 2(1) = 14g
112 = 8
14
CH2  C8H16
47. Find the molecular formula for a compound with a mass of 40 amu and an empirical
formula of C3H4. 3(12) + 4(1) = 40g
40 = 1
40
C 3 H 4  C3H 4
48. What is the definition of percent?
part
whole
= %
Calculate the percent composition of the following compounds:
49. HC2H3O2
1(1) + 2(12) + 3(1) + 2(16) = 60g
% H:
4/60 = 6.67%
% C:
24/60 x 100 = 40.00%
% O:
32/60 = 53.33%
50. NaOH
1(23) + 1(16) + 1(1) = 40g
% Na: 23/40 = 57.50%
% O: 16/40 = 40.00%
% H: 1/40 = 2.50%
51. Calculate the percent of water in compound CaCl210H2O 1(40) + 2(35) + 10(18) = 290g
H 2O
180 x 100 = 62.07% H2O
290
52. Circle the compound which contains the largest percentage of nitrogen?
Ca(NO3)2
1(40) + 2(14) + 6(16) = 164g
28 x 100 = 17.07%
164
(NH4)2SO4
2(14) + 8(1) + 1(32) + 4(16) = 132g
28 x 100 = 21.21%
132
Regents Type Multiple Choice Review Questions:
Q#
1
2
3
4
5
6
answer
C
C
C
B
B
D
Q#
7
8
9
10
11
12
answer
A
A
C
A
C
C
Q#
13
14
15
16
17
33
answer
1.24moles
B
A
B
C
D
Q#
34
35
36
37
38
39
answer
D
B
D
A
B
B