1.9 Percentage Composition

But many a time, information regarding the percentage of a particular element present in a compound is required. Suppose, an unknown or new compound is given to you, the first question you would ask is: what is its formula or what are its constituents, and in what ratio are they present in the given compound? For known compounds also, such information provides a check whether the given sample contains the same percentage of elements as present in a pure sample. In other words, one can check the purity of a given sample by analysing this data.

Let us understand it by taking the example of water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\). Since water contains hydrogen and oxygen, the percentage composition of both these elements can be calculated as follows:
Mass \(\%\) of an element \(=\frac{\text { mass of that element in the compound } \times 100}{\text { molar mass of the compound }}\)
Molar mass of water \(=18.02 \mathrm{~g}\)
Mass \(\%\) of hydrogen \(=\frac{2 \times 1.008}{18.02} \times 100=11.18\)
Mass \(\%\) of oxygen \(=\frac{16.00}{18.02} \times 100=88.79\)

Example: What is the percentage of carbon, hydrogen, and oxygen in ethanol?

Answer: The molecular formula of ethanol is: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

Molar mass of ethanol is: \((2 \times 12.01+6 \times 1.008+16.00) \mathrm{~g}=46.068 \mathrm{~g}\)
Mass per cent of carbon \( =\frac{24.02 \mathrm{~g}}{46.068 \mathrm{~g}} \times 100=52.14 \%\)
Mass per cent of hydrogen\(=\frac{6.048 \mathrm{~g}}{46.068 \mathrm{~g}} \times 100=13.13 \%\)
Mass per cent of oxygen\(=\frac{16.00 \mathrm{~g}}{46.068 \mathrm{~g}} \times 100=34.73 \%\)
After understanding the calculation of the percent of the mass, let us now see what information can be obtained from the percent composition data.

Chemical Formulae

It is of two types:

Molecular formulae: Chemical formulae that indicate the actual number and type of atoms in a molecule are called molecular formulae.
Example: Molecular formula of benzene is \(\mathrm{C}_6 \mathrm{H}_6\).

Empirical formulae: Chemical formulae that indicate only the relative number of atoms of each type in a molecule are called empirical formulae. Example: Empirical formula of benzene is ” \(\mathrm{CH} “\).

Empirical Formula for Molecular Formula

An empirical formula represents the simplest whole number ratio of various atoms present in a compound, whereas, the molecular formula shows the exact number of different types of atoms present in a molecule of a compound. If the mass percent of various elements present in a compound is known, its empirical formula can be determined. The molecular formula can further be obtained if the molar mass is known. The following example illustrates this sequence.

Determination of Chemical Formulae :

(a) Determination of empirical formulae :
Step (I): Determination of percentage of each element
Step (II): Determination of mole ratio
Step (III): Making it whole number ratio
Step (IV): Simplest whole ratio
(b) Determination of molecular formulae
Step (I): First of all find empirical formulae
Step (II): Calculate the empirical weight
Step (III): Molecular formulae \(=n(\) Empirical formulae \()\)
\(
n=\frac{\text { Molecular weight }}{\text { Empirical weight }}
\)

Example:
A compound contains \(4.07 \%\) hydrogen, \(24.27 \%\) carbon and \(71.65 \%\) chlorine. Its molar mass is \(98.96 \mathrm{~g}\). What are its empirical and molecular formulas?
Solution:
Step 1. Conversion of mass percent to grams

Since we are having mass percent, it is convenient to use \(100 \mathrm{~g}\) of the compound as the starting material. Thus, in the \(100 \mathrm{~g}\) sample of the above compound, 4.07g hydrogen, 24.27g carbon and \(71.65 \mathrm{~g}\) chlorine are present.

Step 2. Convert into number moles of each element

Divide the masses obtained above by the respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound
Moles of hydrogen \(=\frac{4.07 \mathrm{~g}}{1.008 \mathrm{~g}}=4.04\)
Moles of carbon \(=\frac{24.27 \mathrm{~g}}{12.01 \mathrm{~g}}=2.021\)
Moles of chlorine \(=\frac{71.65 \mathrm{~g}}{35.453 \mathrm{~g}}=2.021\)
Step 3. Divide each of the mole values obtained above by the smallest number amongst them

Since \(2.021\) is the smallest value, division by it gives a ratio of \(2: 1: 1\) for \(\mathrm{H}: \mathrm{C}: \mathrm{Cl}\).

In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements \(\mathrm{CH}_{2} \mathrm{Cl}\) is, thus, the empirical formula of the above compound.
Step 5. Writing molecular formula
(a) Determine empirical formula mass by adding the atomic masses of various atoms present in the empirical formula.
For \(\mathrm{CH}_{2} \mathrm{Cl}\), empirical formula mass is
\(
\begin{aligned}
&12.01+(2 \times 1.008)+35.453=49.48 \mathrm{~g} \\
\end{aligned}
\)
(b) Divide Molar mass by empirical formula mass
\(\begin{aligned} \frac{\text { Molar mass }}{\text { Empirical formula mass }} &=\frac{98.96 \mathrm{~g}}{49.48 \mathrm{~g}} \\=& 2=(n) \end{aligned}\)
(c) Multiply empirical formula by \(n\) obtained above to get the molecular formula

Empirical formula \(=\mathrm{CH}_{2} \mathrm{Cl}, n=2\). Hence molecular formula is \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\).