Stoichiometry Tutorial
Using Molar Mass in Calculations

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1. 12.5 g CaCl2 = 0.113 mol CaCl2

mol CaCl2

=

12.5 g CaCl2

x

1 mol CaCl2

=

0.113 mol

111.1 g CaCl2

CaCl2


2. 10.0 g Na2SO4 = 0.0704 mol Na2SO4

mol Na2SO4
=
10.0 g Na2SO4
x
1 mol Na2SO4
=
0.0704 mol
142.0 g Na2SO4
Na2SO4

3. 16 g NaOH = 0.40 mol NaOH

mol NaOH = 16 g NaOH x 1 mol NaOH = 0.40 mol NaOH
40.0 g NaOH

4. 125 mg NaOH = 3.13 x 10-3 mol NaOH

mol NaOH
=
125 mg NaOH
x
1 g
x
1 mol
=
3.13 x 10-3 mol
1000 mg
40.0 g
NaOH
The only "twist" on this problem is that you first must convert mg to grams. The problem is then identical to the previous problem in terms of the approach to solving it. Notice that the answer was given in scientific notation. Your calculator should have read 0.003125. Remember that the answer should be reported to 3 signficant figures so it was rounded to 0.00313 and then converted to scientific notation because it is a small number.

5. 0.125 mol CaCl2 = 13.9 g CaCl2

g CaCl2 = 0.125 mol CaCl2 x 111.1 g CaCl2 = 13.9 g CaCl2
1 mol CaCl2

6. 0.25 mol Na2SO4 = 36 g Na2SO4

g Na2SO4 = 0.25 mol Na2SO4 x 142.0 gNa2SO4 = 36 g
1 mol Na2SO4 Na2SO4
The answer before rounding was 35.5 g. However, the answer must be reported to 2 significant figures. Therefore, it was rounded to 36 g.

7. 1.55 mol NaOH = 62.0 g NaOH

g NaOH = 1.55 mol NaOH x 40.0 g NaOH = 62.0 g NaOH
1 mol NaOH

 

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