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Percentage Composition

25th Aug 2019 @ 10 min read

Analytical Chemistry

A compound is a pure substance made up of different elements. So, it consists of atoms of different elements. Examples of compounds are water (H2O), carbon dioxide (CO2), hydrogen chloride (HCl), sulphuric acid (H2SO4), potassium hydroxide (KOH), methane (CH4), acetic acid (CH3COOH). All these compounds contain at least two atoms of elements.

Note: In contrast to compounds, elements are a pure substance made up of the same atom, but the atom can occur more than once. The picture below illustrates the same.

Difference between Elements and Compounds
Figure 1: Difference between Elements and Compounds
Note: The figure does not accurately reflect the sizes of the atoms.

Each element in a compound is always in fixed composition (or percentage). When the percentage is expressed by mass, it is called mass percentage composition (or percentage composition). But when the percentage is explicitly expressed by mole, it is called mole percentage composition.

Definition of Percentage Composition

The percentage composition of an element in a compound is the mass percentage of the element present in the compound.

It tells the mass percentage of each element present in a compound.

How to Calculate Percentage Composition

All we need is the molecular formula and the molar mass of each element present in a compound to determine percentage composition. It is the percentage ratio of the total mass of an element to the total mass of the compound.

\% \,\text{composition} = \frac{\splitfrac{\text{Total mass of an element}} {\text{per mole of the compound}}}{\text{Molar mass of the compound}} \times 100

Now, the total mass of each element is the molar mass of that element times the number of atoms of that element. And the molar mass of a compound is the sum of the molar mass times the number of atoms of each element.

Consider an generic molecular formula: AxByCz. So, the total mass of A per mole of AxByCz is x MA. And the molar mass of AxByCz is MAxByCz = x MA + y MB + z MC.

Mass percentage of A is as follows:

\% \,\text{composition of A} &=\frac{x \,M_\text{A}}{M_{\text{A}_x \text{B}_y \text{C}_z}} \times 100\\ &=\frac{x \,M_\text{A}}{x \,M_\text{A} + y \,M_\text{B} + z \,M_\text{C}} \times 100

Here, MA, MB, and MC are the molar mass of A, B, and C.

Similarly, for B and C,

\% \,\text{composition of B} &=\frac{y \,M_\text{B}}{M_{\text{A}_x \text{B}_y \text{C}_z}} \times 100 \\ &=\frac{y \,M_\text{B}}{x \,M_\text{A} + y \,M_\text{B} + z \,M_\text{C}} \times 100 \% \,\text{composition of C} &=\frac{z \,M_\text{C}}{M_{\text{A}_x \text{B}_y \text{C}_z}} \times 100 \\ &=\frac{z \,M_\text{C}}{x \,M_\text{A} + y \,M_\text{B} + z \,M_\text{C}} \times 100

Example 1: Water

Let take a real example. Water, the most abundant compound in our solar system, is composed of hydrogen and oxygen. The molecular formula is H2O. One mole of water has two moles of hydrogen atoms and one mole of oxygen atoms. The molar mass MH2O of water is as follows:

M_{\text{H}_2 \text{O}} &= 2 \,M_\text{H} + M_\text{O}\\ &= 2 \times 1.008 + 15.999\\ &= 18.015 \,\text{g}\,\text{mol}^{-1}

The percentage composition of hydrogen in water is calculated as follows:

\% \,\text{composition of H} &=\frac{2 \,M_\text{H}}{M_{\text{H}_2 \text{O}}} \times 100 \\ &=\frac{2 \times 1.008}{18.015} \times 100\\ &=11.191 \,\%

Similarly, for oxygen,

\% \,\text{composition of O} &=\frac{M_\text{O}}{M_{\text{H}_2 \text{O}}} \times 100 \\ &=\frac{15.999}{18.015} \times 100\\ &=88.809 \,\%

Note: Sum of % composition of each element is always equal to 100 %.

\sum \,\% \,\text{composition} &=\% \,\text{composition of H} + \% \,\text{composition of O} \\ &= 11.191 \,\% + 88.809 \,\% \\ &=100 \,\%

Example 2: Copper (II) Bromide

Consider an another example of copper (II) bromide CuBr2. The molar mass of CuBr2 is as follows:

M_{\text{CuBr}_2} &=M_\text{Cu} + 2 \,M_\text{Br} \\ &= 63.55 + 2 \times 79.90 \\ &= 223.35 \,\text{g} \,\text{mol}^{-1}

The percentage composition of copper in copper bromide is calculated as follows:

\% \,\text{composition of Cu} &=\frac{M_\text{Cu}}{M_{\text{Cu} \text{Br}_2}} \times 100 \\ &=\frac{63.55}{223.35} \times 100\\ &=28.45 \,\%

Similarly, for bromine,

\% \,\text{composition of Br} &=\frac{2 \,M_\text{Br}}{M_{\text{Cu} \text{Br}_2}} \times 100 \\ &=\frac{2 \times 79.90}{223.45} \times 100\\ &=71.55 \,\%

Mole Percentage Composition

Mole percentage composition is a mole percentage of each element in a compound. We only need the molecular formula of a compound to determine it.

\% \,\text{mole} &= \frac{\splitfrac{\text{Total moles of an element}} {\text{per mole of the compound}}}{\splitfrac{\text{Total mole of element}} {\text{in the compound}}} \times 100 \\ &=\frac{n_i}{\sum n_i} \times 100

Example 3: Carbon Dioxide

The molecular formula of carbon dioxide is CO2. So, one mole of CO2 contains one mole of carbon and two moles of oxygen. Thus, the total number of moles is nC + nO = 1 + 2 = 3 mol.

The mole per cent of carbon in carbon dioxide is as follows:

\% \,\text{mole of carbon} &=\frac{n_\text{C}}{n_\text{C} + n_\text{O}} \times 100 \\ &=\frac{1}{3} \times 100\\ &=33.33 \,\%

Similarly for oxygen,

\% \,\text{mole of oxygen} &=\frac{n_\text{O}}{n_\text{C} + n_\text{O}} \times 100 \\ &=\frac{2}{3} \times 100\\ &=66.67 \,\%

Application of Percentage Composition

To Identify Unknown Chemicals

By knowing the mass percentage of each element in a compound, we can determine the molecular formula (or empirical formula if the molar mass of the compound is unknown). Consider an example below.

Example 4: To Determine Molecular Formula from Mass Percentages

It is known that a chemical compound contains 52.14 % carbon, 13.13 % hydrogen, and 34.73 % oxygen. The molar mass of the chemical is also known; it is 46.069 g mol−1.

Step 1: First, we need to convert the mass percentages into moles. Consider 100 g of the compound. So, it has 52.14 g of carbon, 13.13 g of hydrogen, and 34.73 g of oxygen.

To now, the respective moles are nC = 52.14 ÷ 12.011 = 4.341 mol, nH = 13.13 ÷ 1.008 = 13.026 mol, and nO = 34.73 ÷ 15.999 = 2.170 mol.

Step 2: Now, we have to take the mole to mole ratio of each element such that the divisor is the smallest number.

\frac{n_\text{C}}{n_\text{O}} =\frac{4.341}{2.170} \approx 2 \frac{n_\text{H}}{n_\text{O}} =\frac{13.026}{2.170} \approx 6

Thus, we have nC : nH : nO = 2 : 6 : 1.

Step 3: We can determine the empirical formula from the mole ratios. So, the empirical formula is C2H6O. And the molar mass of the empirical mass is as follows:

M_{\text{C}_2 \text{H}_6 \text{O}} &=2 \times 12.011+ 6 \times 1.008 + 15.999\\ &=46.069 \,\text{g} \,\text{mol}^{-1}

Now, the empirical mass and the molar mass are the same; hence, the empirical and molecular formula must be the same, which is C2H6O.

The compound can be ethanol, CH3CH2OH or dimethyl ether, CH3OCH3.

Labelling

The Ingredients of a product are often labelled in mass percentage composition in by manufacturers. It is easier to understand the amount of an ingredient in the mass percentage in comparison to the mole percentage. Products include food, chemicals, medicine etc. Some examples are mentioned below.

A food pack label having ingredients printed in the mass percentage
Figure 2: A food pack label having ingredients printed in the mass percentage.
[Image Source: Public Domain]
A nitric acid bottle consists of 65 % of nitric acid and the rest is water. The percentage is in mass by mass basis
Figure 3: A nitric acid bottle consists of 65 % of nitric acid and the rest is water. The percentage is in mass by mass basis
[Image Source: Fisher Scientific]

Solved Problems

Problem 1: Sodium Chloride

Sodium chloride is a common salt and has vast variety of applications. The molecular formula of sodium chloride is NaCl. The molar mass of NaCl is MNaCl = 22.99 + 35.45 = 58.44 g mol−1.

The percentage of sodium in sodium chloride is as follows:

\% \,\text{composition of Na} &=\frac{M_\text{Na}}{M_\text{NaCl}} \times 100 \\ &=\frac{22.99}{58.44} \times 100\\ &=39.34 \,\%

For chlorine,

\% \,\text{composition of Cl} &=\frac{M_\text{Cl}}{M_\text{NaCl}} \times 100 \\ &=\frac{35.45}{58.44} \times 100\\ &=60.66 \,\%

Problem 2: Sulphuric Acid

sulphuric acid structure
Figure 4: Sulphuric Acid

Sulphuric acid is one of the largest produced chemicals in the world. It consists of hydrogen, sulphuric, and oxygen. The properties of the acid are colourless, odourless, water-soluble, corrosive when in concentrated form. Its molecular formula is H2SO4.

MH2SO4 be the molar mass of H2SO4.

M_{\text{H}_2 \text{SO}_4} &=2 \times 1.008 + 32.065 + 4 \times 15.999 \\ &= 98.077 \,\text{g} \,\text{mol}^{-1}

The percentage of hydrogen, sulphur, and oxygen in sulphuric acid is as follows:

\% \,\text{composition of H} &=\frac{2 \times M_\text{H}}{M_{\text{H}_2 \text{SO}_4}} \times 100 \\ &=\frac{2 \times 1.008}{98.077} \times 100\\ &=2.055\,\% \% \,\text{composition of S} &=\frac{M_\text{S}}{M_{\text{H}_2 \text{SO}_4}} \times 100 \\ &=\frac{32.065}{98.077} \times 100\\ &=32.694\,\% \% \,\text{composition of O} &=\frac{4 \times M_\text{O}}{M_{\text{H}_2 \text{SO}_4}} \times 100 \\ &=\frac{4 \times 15.999}{98.077} \times 100\\ &=65.251\,\%

Problem 3: Acetone

acetone structure
Figure 5: Acetone

Acetone, also known as propanone, is a common solvent used in chemical industries. It is the smallest ketone, which is colourless, flammable liquid. The molecular formula of acetone is C3H6O.

The molar mass of acetone be MC3H6O.

M_{\text{C}_3 \text{H}_6 \text{O}} &= 3 \times 12.011 + 6 \times 1.008 +15.999 \\ &= 58.080 \,\text{g} \,\text{mol}^{-1}

The percentage of carbon, hydrogen, and oxygen in acetone is as follows:

\% \,\text{composition of C} &=\frac{3 \times M_\text{C}}{M_{\text{C}_3 \text{H}_6 \text{O}}} \times 100 \\ &=\frac{3 \times 12.011}{58.080} \times 100 \\ &=62.04 \,\% \% \,\text{composition of H} &=\frac{6 \times M_\text{C}}{M_{\text{C}_3 \text{H}_6 \text{O}}} \times 100 \\ &=\frac{6 \times 1.008}{58.080} \times 100 \\ &=10.41 \,\% \% \,\text{composition of O} &=\frac{M_\text{O}}{M_{\text{C}_3 \text{H}_6 \text{O}}} \times 100 \\ &=\frac{15.999}{58.080} \times 100 \\ &=27.55 \,\%

Problem 4: Capric Acid (or Decanoic Acid)

Figure 6: Capric Acid or Decanoic Acid

The IUPAC of capric acid is decanoic acid. It is a fatty acid having ten carbons attached to form a straight chain. The carboxylic group (‑COOH) is attached to the end. It has applications in soap, perfumes, lubricants, greases, pharmaceuticals industries.

The molecular formula is C10H20O2. MC10H20O2 be the molar mass of capric acid.

M_{\text{C}_{10} \text{H}_{20} \text{O}_2} &= 10 \times 12.011 + 20 \times 1.008 +2 \times 15.999 \\ &= 172.268 \,\text{g} \,\text{mol}^{-1}

The mass percentage of carbon, hydrogen, and oxygen in acetone is as follows:

\% \,\text{composition of C} &=\frac{10 \times M_\text{C}}{ M_{\text{C}_{10} \text{H}_{20} \text{O}_2}} \times 100 \\ &=\frac{10 \times 12.011}{172.268} \times 100 \\ &=69.72 \,\% \% \,\text{composition of H} &=\frac{20 \times M_\text{H}}{ M_{\text{C}_{10} \text{H}_{20} \text{O}_2}} \times 100 \\ &=\frac{20 \times 1.008}{172.268} \times 100 \\ &=11.70 \,\% \% \,\text{composition of O} &=\frac{2 \times M_\text{O}}{ M_{\text{C}_{10} \text{H}_{20} \text{O}_2}} \times 100 \\ &=\frac{2 \times 15.999}{172.268} \times 100 \\ &=18.58 \,\%

Practice Problems

Find the percentage composition of the following compounds: nitric acid (HNO3), ammonium carbonate (NH4CO3), magnesium iron silicate ({Mg, Fe})2SiO4), benzene (C6H6), glucose (C6H12O6), ethylene diamine (C2H8N2), and Paracetamol (C8H9NO2).

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