SME
Grade 10 Science · Problem type
Molar mass and percent composition
Use this for a compound’s molar mass and the percent of its mass from each element.
Chart
40.00% + 6.71% + 53.30% = 100% of the whole.
This is an example. Type your own values to start a new problem.
M
g/mol
=
12.01
×
x
+
1.008
×
y
+
16.00
×
z
%C
Percent carbon
calculated · %
%H
Percent hydrogen
calculated · %
%O
Percent oxygen
calculated · %
Formulas
M
=
12.01
×
x
+
1.008
×
y
+
16.00
×
z
180.16 = 12.01 × 6 + 1.008 × 12 + 16.00 × 6
%C
=
12.01
×
x
M
×
100
40.00 = 12.01 × 6/180.16 × 100
%H
=
1.008
×
y
M
×
100
6.71 = 1.008 × 12/180.16 × 100
%O
=
16.00
×
z
M
×
100
53.29 = 16.00 × 6/180.16 × 100
Assumptions
•
Atomic masses: C 12.01, H 1.008, O 16.00 g/mol.
•
Each element’s share of the mass is its atoms’ mass over the molar mass.
•
The percents add to 100.
Step-by-step
WE KNOW
x = 6, y = 12, z = 6
FIND
molar mass (M), percent carbon (%C), percent hydrogen (%H), percent oxygen (%O)
Step 1 · Find molar mass (M)
Use M = 12.01 × x + 1.008 × y + 16.00 × z
Add the mass of every atom in the formula.
M
=
12.01
×
6
+
1.008
×
12
+
16.00
×
6
M
=
72.06
+
12.096
+
96
M
=
180.16
g/mol
Step 2 · Find percent carbon (%C)
Use %C = 12.01 × x/M × 100
The C atoms’ mass as a share of the whole formula’s mass.
%C = 12.01 × 6/180.16 × 100
%C = 72.06/180.16 × 100
%C = 0.4 × 100
%C = 40.00%
Step 3 · Find percent hydrogen (%H)
Use %H = 1.008 × y/M × 100
The H atoms’ mass as a share of the whole formula’s mass.
%H = 1.008 × 12/180.16 × 100
%H = 12.096/180.16 × 100
%H = 0.06714 × 100
%H = 6.71%
Step 4 · Find percent oxygen (%O)
Use %O = 16.00 × z/M × 100
The O atoms’ mass as a share of the whole formula’s mass.
%O = 16.00 × 6/180.16 × 100
%O = 96/180.16 × 100
%O = 0.5329 × 100
%O = 53.29%
Check
Put the numbers back in. Both sides should match.
180.16 ≈ 12.01 × 6 + 1.008 × 12 + 16.00 × 6 ✓
40.00 ≈ 12.01 × 6/180.16 × 100 ✓
6.71 ≈ 1.008 × 12/180.16 × 100 ✓
53.29 ≈ 16.00 × 6/180.16 × 100 ✓
Topic check
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