Per cent Composition: Have you observed the label on the packaged foods? The label consists of the nutritional information of the food. The label also provides information about the masses of various compounds present in each serving. For example, one serving of peanut butter contains \({\rm{7g}}\) of protein, \({\rm{15g}}\) of fat, and \({\rm{3g}}\) of sugar. By calculating the mass fraction of protein, fat, and sugar in one serving size of peanut butter, we can determine the per cent composition of each of the components present in one serving of peanut butter.

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What is Percentage Composition?

For a given compound, the percentage composition is defined as the ratio of the amount of each element present in the compound to the total amount of individual elements multiplied by \(100.\) In simple words, it is the per cent by mass of each component of a compound.

Percentage Composition Formula

For a given element, the percentage composition is expressed by using the following formula:

\({\rm{\% }}{{\rm{C}}_{\rm{e}}}{\rm{ = }}\frac{{{{\rm{G}}_{\rm{e}}}}}{{{{\rm{G}}_{\rm{i}}}}}{\rm{ \times 100}}\)

Where, \({\rm{\% }}{{\rm{C}}_{\rm{e}}}\) represents the percentage composition of the element \({\rm{E,}}{{\rm{G}}_{\rm{e}}}\) represents the total amount of element \({\rm{E}}\) present in the compound, \({{\rm{G}}_{\rm{t}}}\) indicates the total amount of all the elements present in the compound.

Example-

A newly synthesized compound is known to contain the elements zinc and oxygen. When \({\rm{20}}{\rm{.00g}}\) of the sample is decomposed, \({\rm{16}}{\rm{.07g}}\) of zinc remains. Determine the per cent composition of the compound.

Solution:

Given,

1. Mass of compound\( = 20.00\;{\rm{g}}\)
2. Mass of \({\rm{Zn = 16}}{\rm{.07\;g}}\)

Mass of oxygen\( = 20.00\;{\rm{g}} – 16.07\;{\rm{g}} = 3.93\;{\rm{g}}\)

\(\% {\rm{Zn}} = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{zinc}}}}{{{\rm{Total}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{compound}}}} \times 100\)

\({\rm{\% Zn}} = \frac{{16.07}}{{20.00}} \times 100 = 80.35\% \)

\({\rm{\% O = }}\frac{{{\rm{3}}{\rm{.93}}}}{{{\rm{20}}{\rm{.00}}}}{\rm{ \times 100 = 19}}{\rm{.65\% }}\)

The calculations make sense because the sum of the two percentages adds up to \({\rm{100\% }}.\) By mass, the compound is mostly zinc.

Percentage concentrations are also often expressed in terms of relative units (e.g. percentages). The commonly used percentage concentrations are-

Mass Per cent

The mass per cent is the ratio of the mass of the solute to the mass of the solution multiplied by \(100.\)

Mathematically, it is expressed as-

\({\rm{Mass\% = }}\frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{Solute }}}}{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{Solution }}}}{\rm{ \times 100\% }}\)

\({\rm{Mass}}\left( {\frac{{\rm{w}}}{{\rm{W}}}} \right)\% = \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{Solute}}}}{{{\rm{Mass}}{\mkern 1mu} \,{\rm{of}}\,{\mkern 1mu} {\rm{Solute}} + {\rm{Mass}}\,{\mkern 1mu} {\rm{of}}\,{\mkern 1mu} {\rm{Solvent}}}} \times 100\% \)

Volume Per cent:

The volume percent is the ratio of the volume of the solute to the volume of the solution multiplied by \(100.\)

Mathematically, it is expressed as-

\({\rm{Volume\% = }}\frac{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solute }}}}{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solution }}}}{\rm{ \times 100\% }}\)

\({\rm{Volume}}\left( {\frac{{\rm{v}}}{{\rm{V}}}} \right){\rm{\% = }}\frac{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solute }}}}{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solute + Volume}}\,{\rm{of}}\,{\rm{Solvent }}}}{\rm{ \times 100\% }}\)

Mass/Volume Per cent

The mass/volume per cent is the ratio of the mass of the solute to the volume of the solution multiplied by \(100.\) Though the numerator consists of mass or weight of solute (grams) and denominator consists of volume \(\left( {{\rm{mL}}} \right)\) are different units, this concentration unit is not a true relative unit (e.g. percentage); however, it is often expressed in percentage.

Mathematically, it is expressed as-

\({\rm{Mass/Volume\% = }}\frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{Solute }}}}{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solution }}}}{\rm{ \times 100\% }}\)

\(\frac{{{\rm{ Mass }}}}{{{\rm{ Volume }}}}\left( {\frac{{\rm{m}}}{{\rm{V}}}} \right){\rm{\% = }}\frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{Solute }}}}{{{\rm{ Volume}}\,{\rm{of}}\,{\rm{Solute + Volume}}\,{\rm{of}}\,{\rm{Solvent }}}}{\rm{ \times 100\% }}\)

Per cent Composition from a Chemical Formula

The per cent composition of a compound can also be determined from the chemical formula of a compound. The subscripts in the chemical formula give the mass of each element in one mole of the given compound. Per cent composition can be calculated by using the following formula-

\({\rm{Mass\% = }}\frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{an}}\,{\rm{element}}\,{\rm{in}}\,{\rm{one}}\,{\rm{mole}}\,{\rm{of}}\,{\rm{the}}\,{\rm{compound }}}}{{{\rm{ Molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{compound }}}}{\rm{ \times 100\% }}\)

Let us consider the molecule of glucose \({{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}.\) The formula shows that glucose has \(6\) carbon atoms, \(12\) hydrogen atoms, and six oxygen atoms from the formula. By multiplying each of the atoms with its atomic mass, we get each element’s mass in glucose. Adding the masses of each element in glucose, we arrive at its molar mass.

Dividing the mass of each element by the molar mass of glucose, we get the mass fraction of each component. Multiplying this value by \(100\) gives the mass percentage of the individual elements in glucose.

As per the periodic table, each mole of carbon atom weighs \(12.01\;{\rm{g}}.\) 

So, \(6\) moles of carbon atom will weigh \({\rm{12}}{\rm{.01g \times 6 = 72}}{\rm{.06g}}{\rm{.}}\)

Similarly, \(1\) mole of hydrogen weighs \(1.008\;{\rm{g}}{\rm{.}}\) 

Therefore, \(12\) moles of hydrogen will weigh \(12 \times 1.008 = 12.096\;{\rm{g}}.\)

The weight of \(1\) mole of oxygen is \(16.00{\rm{g}}.\) 

Therefore, \(6\) moles of oxygen will weigh \(16.00 \times 6 = 96\;{\rm{g}}.\).

Thus, \(1\) mole of glucose \(\left( {{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}} \right)\) has a total mass of \({\rm{72}}{\rm{.06 + 12}}{\rm{.096 + 96 = 180}}{\rm{.16g}}\)

Calculating the Mass Percentage

To find out the mass \(\% \) of carbon, hydrogen and oxygen in glucose, the mass fraction of each element in \(1\) mole of glucose is calculated. 

Mass fraction of Carbon\( = \frac{{72.06}}{{180.16}} = 0.4000\)
\(\therefore \) mass \(\% \) of Carbon\( = 0.4000 \times 100 = 40.00\% \)

Mass fraction of Hydrogen\( = \frac{{12.096}}{{180.16}} = 0.06714\)
\(\therefore \) mass \(\% \) of Hydrogen\( = 0.06714 \times 100 = 6.714\% \)

Mass fraction of oxygen\( = \frac{{96}}{{180.16}} = 0.53286\)
\(\therefore \) Mass \(\% \) of Oxygen\( = 0.53286 \times 100 = 53.286\% \)

Though carbon and oxygen have an equal number of moles in glucose, their mass percentages vary. As the molar mass of oxygen is higher than that of carbon, its mass per cent is higher than that of carbon.

Solved Examples on Per cent Composition

Q.1. Determine the per cent composition of each element in water.
Ans: The chemical formula for water is \({{\rm{H}}_2}{\rm{O}}.\)
Molar mass of Oxygen\( = 1600 \times 1 = 16\;{\rm{g}}\)
Molar mass of Hydrogen\( = 1 = 1.01 \times 2 = 2.02\;{\rm{g}}\)
Now, using the molar masses, we calculate the percentage composition of each element in \({{\rm{H}}_2}{\rm{O}}.\)
\(\% \) composition of Hydrogen\( = \frac{{2.02}}{{18.02}} \times 100\)
Therefore, \(\% {\rm{H}} = 11.21\% \)
\(\% \) composition of Oxygen\( = \frac{{16}}{{18.02}} \times 100 = 88.79\% \)

Summary

The percentage composition of a compound indicates what percentage of the total mass is made up of each element in the compound. However, it does not tell us the relative ratio by which the atoms combine to form the compound. Percentage composition also comes into the picture with the combustion of fuels, air quality, and even popping popcorn. Percentage composition is a part of our daily lives even when we are unaware of it. This article explains the concept of percentage composition and related calculations.

Frequently Asked Questions (FAQs)

Q.1. How do you find the per cent composition?
Ans: For a given element, the percentage composition is determined by using the following formula:
\({\rm{\% }}{{\rm{C}}_{\rm{e}}}{\rm{ = }}\frac{{{{\rm{G}}_{\rm{e}}}}}{{{{\rm{G}}_{\rm{t}}}}}{\rm{ \times 100}}\)
Where, \({\rm{\% }}{{\rm{C}}_{\rm{e}}}\) represents the percentage composition of the element \({\rm{E,}}{{\rm{G}}_{\rm{e}}}\) represents the total amount of element \({\rm{E}}\) present in the compound, \({{\rm{G}}_{\rm{t}}}\) indicates the total amount of all the elements present in the compound.

Q.2. What is meant by per cent composition?
Ans: For a given compound, the percentage composition is defined as the ratio of the amount of each element present in the compound to the total amount of individual elements multiplied by \(100.\) In simple words, it is the per cent by mass of each element in a compound.

Q.3. Why is per cent composition by mass important?
Ans: Per cent composition allows us to determine the percentage of each element that makes up a specific compound.

Q.4. What is the purpose of per cent composition?
Ans: Per cent composition helps us to determine the chemical composition of certain substances. It is also used to calculate the percentage of an element in a mixture. One can also derive an empirical formula from per cent composition.

Q.5. What is the per cent composition of \({\rm{C}}{{\rm{O}}_{\rm{2}}}\)?
Ans: The atomic masses of carbon and oxygen are found to be \(12.01\;{\rm{g}}/{\rm{mol}}16.00\;{\rm{g}}/{\rm{mol}},\) respectively.
Calculating the molar mass of \({\rm{C}}{{\rm{O}}_{\rm{2}}}.\)
One mole of \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) contains \(1\) mole of carbon atoms and \(2\) moles of oxygen atoms.
Hence, the molar mass of \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) is \({\rm{12}}{\rm{.01g + (2 \times 16}}{\rm{.00\;g)32}}{\rm{.00\;g = 44}}{\rm{.01\;g}}\)
Calculating the mass percent of each atom.
\({\rm{Mass\% = }}\frac{{{\rm{ Mass}}\,{\rm{of}}\,{\rm{an}}\,{\rm{element}}\,{\rm{in}}\,{\rm{one}}\,{\rm{mole}}\,{\rm{of}}\,{\rm{the}}\,{\rm{compound }}}}{{{\rm{ Molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{compound }}}}{\rm{ \times 100\% }}\)
\({\rm{mass\% C = }}\left( {\frac{{{\rm{ mass}}\,{\rm{of}}\,{\rm{1}}\,{\rm{mol}}\,{\rm{of}}\,{\rm{carbon}}\,{\rm{atom }}}}{{{\rm{ mass}}\,{\rm{of}}\,{\rm{1}}\,{\rm{mol}}\,{\rm{of}}\,{\rm{C}}{{\rm{O}}_{\rm{2}}}}}} \right){\rm{ \times 100}}\)
\( = \left( {\frac{{12.01}}{{44.01}}} \right) \times 100 = 27.29\% \)
\({\rm{mass\% O = }}\left( {\frac{{{\rm{ mass}}\,{\rm{of}}\,{\rm{2}}\,{\rm{mol}}\,{\rm{of}}\,{\rm{oxygen}}\,{\rm{atom }}}}{{{\rm{ mass}}\,{\rm{of}}\,{\rm{1}}\,{\rm{mol}}\,{\rm{of}}\,{\rm{C}}{{\rm{O}}_{\rm{2}}}}}} \right){\rm{ \times 100}}\)
\( = \left( {\frac{{32.00}}{{44.01}}} \right) \times 100 = 72.71\% \)

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