Avogadro's Number
1. Determine the number of atoms in 3 grams of Ca.
1 Mole 
= 
Ca 
= 
40 g 
= 
6.023 X 10^{23} 




____ 

____________ 




3 g 

X 
X 
= 
(6.023 X 10^{23)}(3 g) 


____________ 


(40 g) 
2. Determine the number of atoms in 5 grams of Cu.
1 Mole 
= 
Cu 
= 
63.54 g 
= 
6.023 X 10^{23} 




____ 

____________ 




5 g 

X 
X 
= 
(6.023 X 10^{23)}(5 g) 


____________ 


(63.54 g) 
3. What is the approximate weight of 5 liters of SO_{2}
at S.T.P?
1 Mole 
= 
SO_{2} 
= 
64 g 
= 
6.023 X 10^{23} 
= 
22.4 L 




____________ 



____________ 




X 



5 L 
X 
= 
(64 g)(5 L) 


____________ 


(22.4 L) 
4. What is the weight in grams of 16.8 liters of SO_{3}?
1 Mole 
= 
SO_{3} 
= 
80 g 
= 
6.023 X 10^{23} 
= 
22.4 L 




____________ 



____________ 




X 



16.8 L 
X 
= 
(80 g)(16.8 L) 


____________ 


(22.4 L) 
5. Find the number of molecules in 4 mL of oxygen gas
(O_{2}).
1 Mole 
= 
O_{2} 
= 
32 g 
= 
6.023 X 10^{23} 
= 
22.4 L 
= 
22,400 mL 






____________ 



_________ 






X 



4 mL 
X 
= 
(6.023 X 10^{23)}(4 mL) 


____________ 


(22,400 mL) 
6. What weight of N_{2} contains the same
number of molecules as 48 grams of SO_{2}?
1 Mole 
= 
SO_{2} 
= 
64 g 
= 
6.023 X 10^{23} 
= 
22.4 L 




____________ 

____________ 






48 g 

X 


X 
= 
(48 g)(6.023 X 10^{23}) 


____________ 


(64 g) 
As you can see below, the answer 4.51725 X 10^{23},
is required to complete the second phase of this problem.
1 Mole 
= 
N_{2} 
= 
28 g 
= 
6.023 X 10^{23} 
= 
22.4 L 




____________ 

____________ 






X 

4.51725 X 10^{23} 


X 
= 
(28 g)(4.51725 X 10^{23}) 


____________ 


(6.023 X 10^{23}) 
7. What weight of O_{2} contains the same
number of molecules as 11 grams of CO_{2}?
1 Mole 
= 
CO_{2} 
= 
44 g 
= 
6.023 X 10^{23} 
= 
22.4 L 




____________ 

____________ 






11 g 

X 


X 
= 
(11 g)(6.023 X 10^{23}) 


____________ 


(44 g) 
As you can see below, the answer 1.50575 X 10^{23},
is required to complete the second phase of this problem.
1 Mole 
= 
O_{2} 
= 
32 g 
= 
6.023 X 10^{23} 
= 
22.4 L 




____________ 

____________ 






X 

1.50575 X 10^{23} 


X 
= 
(32 g)(1.50575 X 10^{23}) 


____________ 


(6.023 X 10^{23}) 
8. What volume of N_{2} contains the same number
of molecules as 32 grams of SO_{2}?
1 Mole 
= 
SO_{2} 
= 
64 g 
= 
6.023 X 10^{23} 
= 
22.4 L 




____________ 

____________ 






32 g 

X 


X 
= 
(32 g)(6.023 X 10^{23}) 


____________ 


(64 g) 
As you can see below, the answer 3.0115 X 10^{23},
is required to complete the second phase of this problem.
1 Mole 
= 
N_{2} 
= 
28 g 
= 
6.023 X 10^{23} 
= 
22.4 L 






____________ 

____________ 






3.0115 X 10^{23} 

X 
X 
= 
(22.4 L)(3.0115 X 10^{23}) 


____________ 


(6.023 X 10^{23}) 
9. What is the percentage of carbon in sodium
carbonate (Na_{2}CO_{3})?
Na_{2}CO_{3} 
= 
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 grameqivalent weight of Na_{2}CO_{3}?
See Section 7.13  Normal Solutions
For the purpose of this discussion, we may define the GRAMEQUIVALENT
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 (CaSO_{4}) is 136 grams; its total
positive valence is 2. Therefore, its equivalent weight is onehalf (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 (Al_{2}(SO_{4})_{3})
is 342 grams; its total positive valence is 6. Therefore, it has an equivalent
weight of 57 grams.
Na_{2}CO_{3} 
> 
2Na^{+} 
+ 
CO_{3}^{=} 
Na_{2}CO_{3} 
= 
Na 
= 
2 
X 
23 
= 
46 


C 
= 
1 
X 
12 
= 
12 


O 
= 
3 
X 
16 
= 
48 








___ 








106 