### Percent Composition

We will take the term 'percent composition' to be the percent by mass of each element present in a compound.

Example #1: Calculate the percent composition of water, H2O.

Solution:

1) Assume one mole of water is present. Determine the molar mass of water:

one mole of water weighs 18.0152 grams

2) Determine the mass of hydrogen present in one mole of water:

In one mole of water, there are two moles of H atoms

(2 mol) (1.008 g/mol) = 2.016 g.

3) Determine the mass of oxygen present in one mole of water:

In one mole of water, there is one mole of O atoms

(1 mol) (16.00 g/mol) = 16.00 g.

4) Determine the percentages (by weight) of hydrogen and oxygen in water:

percentage of hydrogen:
2.016 / 18.015 g = 0.1119

multiply by 100, giving 11.19%.

percentage of oxygen:

16.00 g / 18.015 g = 0.8881

multiply by 100 to give 88.81%

Notice that you can also subtract hydrogen's percentage from 100%, giving the percent oxygen as the answer. You can always do this with the last part of this type problem. In fact, subtraction to find the last answer is used often.

Example #2: Calculate the percent composition of methane, CH4.

Solution:

1) Determine the molar mass of CH4:

(4) (1.008 g) + (1) (12.011 g) = 16.043 g

2) Determine the mass of the four moles of hydrogen present in one mole of methane:

(4) (1.008 g) = 4.032 g

3) Determine the percent composition of H in CH4:

4.032 / 16.043 = 25.132% (I skipped the multiply by 100 statement.)

4) Determine the percent composition of C in CH4:

100 − 25.132 = 74.868%

Example #3: Glucose, C6H12O6.

Solution:

1) A brief description of the technique:

(a) figure out the molar mass from the formula.
(b) figure out the grams each atom contributes by multiplying the atomic weight by the number of atoms in the formula.
(c) divide the answer for each atom by the molar mass and multiply by 100 to get a percentage.

Remember, you may figure out the last percentage by subtracting the total percent from 100, as will be done in a moment.

2) Step (a):

mass of one mole of glucose = 180.16 g

3) Step (b):

Carbon ---> 6 x 12.011 g = 72.066 g
Hydrogen ---> 12 x 1.008 = 12.096 g
The oxygen percentage will be arrived at by subtraction.

Step (c):

Carbon's percentage: (72.066 g / 180.16 g) x 100 = 40.00%
Hydrogen's percentage: (12.096 g / 180.16 g) x 100 = 6.71%
Oxygen's percentage: 100 − (40.00 + 6.71) = 53.29%

Example #4: ammonium sulfide, (NH4)2S. Its molar mass is 68.1422 g/mol

Solution:

Comment: notice the presence of the parentheses. They are important.

1) Calculate the mass of nitrogen and hydrogen present in one mole of ammonium sulfide:

nitrogen ---> (14.007) (2) = 28.014 g
hydrogen ---> (1.008) (8) = 8.064 g

Note the influence of the parentheses. There are TWO nitrogens and EIGHT hydrogens in the formula, not one N and four H.

2) Determine the percent composition of ammonium sulfide:

N ---> (28.014 g / 68.1422 g) * 100 = 41.111%
H ---> (8.064 g / 68.1422 g) * 100 = 11.834%
S ---> 100 − (41.111 + 11.834) = 47.055%

3) There are plenty of online percent composition calculators. Here is one.

Notice that the answers in the online calculator are slightly different than mine. I use 14.007 for the atomic weight of nitrogen where the author of the linked file used 14.01.

Make sure you check with your teacher as to his/her wishes regarding which atomic weight values to use.

Example #5: Please calculate the percent composition of the following two substances:

(a) CH2O
(b) CH3COOH

Comment: if you do these two before looking at the answer below, you may have noticed something interesting. They both have the same percentage composition and it's also the same as that of glucose (example #3 above).

Solution for CH3COOH:

1) Determine the molar mass from the formula.

60.05 g/mol

2) Determine the grams each atom contribues by multiplying the atomic weight by the number of atoms in the formula.

Carbon = 2 x 12.011 g = 24.022 g
Hydrogen = 4 x 1.008 = 4.032 g
Oxygen = 2 x 16.00 = 32.00 g

3) Divide the answer for each atom by the molar mass and multiply by 100 to get a percentage.

Carbon's percentage: (24.022 g / 60.05 g) x 100 = 40.00 %
Hydrogen's percentage: (4.032 g / 60.05 g) x 100 = 6.71 %
Oxygen's percentage: (32.00 g / 60.05 g) = 53.29 %

You could have done the last of the three by subtraction, if you wished. I simply decided to not use the subtraction technique the last calculation. Some teachers may demand all calculations be of the dividing type and refuse to allow using the subtraction for the last value.

All three substances (C6H12O6, CH2O, CH3COOH) have the same percentage composition because they all have the same empirical formula. "Empirical formula" is a REAL IMPORTANT concept and we will get to it in the next tutorial.

Example #6: Calculate the percent water in CuCl2 2H2O.

Solution:

1) The first thing we need is the molar mass:

170.4816 g/mol (remember to include the two waters of hydration)

2) How much water is present in one mole of the compound?

(18.0152 g/mol) (2 moles) = 36.0304 g

3) We are now ready to calculate the percent water:

36.0304 g / 170.4816 g = 21.134%

Example #7: Benzene has the formula C6H6. Calculate its percentage composition.

Solution:

Comment: notice how the C and the H are in a 1:1 ratio? Suppose we just use CH as the formula (also a 1:1 ratio) and do the calculation.

1) Determine the "molar mass" of CH:

13.019 g

I wrote "molar mass" because there is no actual compound having the formula of CH.

2) Determine the percent composition:

C ---> 12.011 / 13.019 = 92.257%
H ---> 1.008 / 13.019 = 7.743%

3) I used the formula C6H6 with the above online percent composition calculator. Here is the result.

Example #8: Acetylene has the formula C2H2. Calculate its percentage composition.

Solution:

Notice that acetylene also has a 1:1 ratio between C and H.

Does this mean that acetylene and benzene will have the same percent composition?

Yes. In fact, all substances with the same empirical formula will have the same percentage composition. C4H4 (cyclobutadiene and cumulene are two examples) and C8H8 (cubane and other compounds) are two additional examples.

Example #9: There are nine different compounds all of which have the formula C7H16. What do we know about the percentage composition for each of the nine compounds?

Solution:

They will all have the same percent composition.

Example #10: Determine the percent by weight of water in KAl(SO4)2 12H2O

Solution:

1) Determine the molecular weight of the compound:

KAl(SO4)2
K ---> 39.10 x 1 = 39.10
Al ---> 26.98 x 1 = 26.98
S ---> 32.065 x 2 = 64.13
O ---> 16.00 x 8 = 128.0

Sum = 258.21

12H2O

H ---> 1.008 x 24 = 24.19
O ---> 16.00 x 12 = 192.0

Sum = 216.19

You could have also done 18.0152 times 12.

molecular weight of compound

258.21 + 216.19 = 474.4

2) Percent by weight of water

(216.19 / 474.4) * 100 = 45.57%

Example #11: The mineral mimetite has the formula Pb5(AsO4)3Cl. What percentage (by mass) of oxygen does it contain?

Solution:

1) Determine the molecular weight of mimetite:

1488.207 g/mol

2) In one formula unit of mimetite, there are 12 oxygens. Determine the molecular weight of 12 oxygens:

12 x 16.00 = 192.0 g (in 12 moles of oxygen)

3) Determine the mass percent:

(192 / 1488.207) * 100 = 12.90%

Example #12: A 2.823 gram piece of magnesium metal is heated and reacts with oxygen. The resulting oxide weighed 4.680 grams. Determine the percent composition of the compound.

Solution:

(2.823 g / 4.680 g) * 100 = 60.32% <--- mass percent of Mg

100 − 60.32 = 39.68% <--- mass percent of O

Suppose your teacher demanded that you use the mass of oxygen in your calculations.

4.680 − 2.823 = 1.857 g <--- this is the mass of oxygen in the compound

(2.823 g / 4.680 g) * 100 = 60.32% <--- mass percent of Mg

(1.857 g / 4.680 g) * 100 = 39.68% <--- mass percent of O