Avogadro's Number


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