1. Determine the number of atoms in 3 grams of Ca.

 1 Mole = Ca = 40 g = 6.023 X 1023 ____ ____________ 3 g X
 X = (6.023 X 1023)(3 g) ____________ (40 g)
 X = 4.51725 X 1023

2. Determine the number of atoms in 5 grams of Cu.

 1 Mole = Cu = 63.54 g = 6.023 X 1023 ____ ____________ 5 g X
 X = (6.023 X 1023)(5 g) ____________ (63.54 g)
 X = 4.7395 X 1022

3. What is the approximate weight of 5 liters of SO2 at S.T.P?

 1 Mole = SO2 = 64 g = 6.023 X 1023 = 22.4 L ____________ ____________ X 5 L
 X = (64 g)(5 L) ____________ (22.4 L)
 X = 14.28 g

4. What is the weight in grams of 16.8 liters of SO3?

 1 Mole = SO3 = 80 g = 6.023 X 1023 = 22.4 L ____________ ____________ X 16.8 L
 X = (80 g)(16.8 L) ____________ (22.4 L)
 X = 60 g

5. Find the number of molecules in 4 mL of oxygen gas (O2).

 1 Mole = O2 = 32 g = 6.023 X 1023 = 22.4 L = 22,400 mL ____________ _________ X 4 mL
 X = (6.023 X 1023)(4 mL) ____________ (22,400 mL)
 X = 1.075 X 1020

6. What weight of N2 contains the same number of molecules as 48 grams of SO2?

 1 Mole = SO2 = 64 g = 6.023 X 1023 = 22.4 L ____________ ____________ 48 g X
 X = (48 g)(6.023 X 1023) ____________ (64 g)
 X = 4.51725 X 1023

As you can see below, the answer 4.51725 X 1023, is required to complete the second phase of this problem.

 1 Mole = N2 = 28 g = 6.023 X 1023 = 22.4 L ____________ ____________ X 4.51725 X 1023
 X = (28 g)(4.51725 X 1023) ____________ (6.023 X 1023)
 X = 21 g

7. What weight of O2 contains the same number of molecules as 11 grams of CO2?

 1 Mole = CO2 = 44 g = 6.023 X 1023 = 22.4 L ____________ ____________ 11 g X
 X = (11 g)(6.023 X 1023) ____________ (44 g)
 X = 1.50575 X 1023

As you can see below, the answer 1.50575 X 1023, is required to complete the second phase of this problem.

 1 Mole = O2 = 32 g = 6.023 X 1023 = 22.4 L ____________ ____________ X 1.50575 X 1023
 X = (32 g)(1.50575 X 1023) ____________ (6.023 X 1023)
 X = 8 g

8. What volume of N2 contains the same number of molecules as 32 grams of SO2?

 1 Mole = SO2 = 64 g = 6.023 X 1023 = 22.4 L ____________ ____________ 32 g X
 X = (32 g)(6.023 X 1023) ____________ (64 g)
 X = 3.0115 X 1023

As you can see below, the answer 3.0115 X 1023, is required to complete the second phase of this problem.

 1 Mole = N2 = 28 g = 6.023 X 1023 = 22.4 L ____________ ____________ 3.0115 X 1023 X
 X = (22.4 L)(3.0115 X 1023) ____________ (6.023 X 1023)
 X = 11.2 L

9. What is the percentage of carbon in sodium carbonate (Na2CO3)?

 Na2CO3 = Na = 2 X 23 = 46 C = 1 X 12 = 12 O = 3 X 16 = 48 ___ 106
 Partial X 100 = % _______ Total
 12 X 100 = 11.32% _______ 106

10. What is the gram-eqivalent weight of Na2CO3?

See Section 7.13 - Normal Solutions

For the purpose of this discussion, we may define the GRAM-EQUIVALENT WEIGHT OF A SUBSTANCE AS THE WEIGHT OF A MOLE OF THE SUBSTANCE DIVIDED BY THE TOTAL POSITIVE VALENCE OF THE COMPOUND. One mole of NaCl is 58.5 grams; its total positive valence is 1. Therefore, its equivalent weight is also 58.5 grams. One mole of calcium sulfate (CaSO4) is 136 grams; its total positive valence is 2. Therefore, its equivalent weight is one-half (68 grams) its formula weight. One mole of bismuth hydroxide Bi(OH)3 is 396 grams; its total positive valence is 3. Therefore, its equivalent weight is 132 grams. One mole of aluminum sulfate (Al2(SO4)3) is 342 grams; its total positive valence is 6. Therefore, it has an equivalent weight of 57 grams.

 Na2CO3 -----> 2Na+ + CO3=
 Na2CO3 = Na = 2 X 23 = 46 C = 1 X 12 = 12 O = 3 X 16 = 48 ___ 106
 106 = 53 _______ 2