Question

The empirical formula of a compound is \[C{{H}_{2}}O\]. If its vapour density is 90; find out the molecular formula of the compound.

Answer 1

Hint: We should know that an empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers. The molecular formula of butene, \[{{C}_{4}}{{H}_{8}}\], shows that each freely existing molecule of butene contains four atoms of carbon and eight atoms of hydrogen. Its empirical formula is\[C{{H}_{2}}\].

Step by step answer:
We know that chemical composition of a compound is given by the chemical formula. This formula can either be an empirical formula or actual molecular formula. Empirical formula is the simplest formula of a compound which gives a ratio of different atoms present in one molecule of the compound. On the other hand, the molecular formula gives the actual number of atoms of each element present in one molecule of the compound.
We should know the following formula to solve this question.
\[Molecular\text{ }formula\text{ }=\text{ }n\times Empirical\text{ }Formula\]
\[Molecular\text{ }mass=2\times Vapour\text{ }density\]
So, we have to calculate the molecular formula of compound \[C{{H}_{2}}O\]. We will solve this question step by step as follows.
First calculate Empirical mass of \[C{{H}_{2}}O\].
Empirical mass of \[C{{H}_{2}}O\]= $(12\times 1+1\times 2+16)=$30 gram
It is given in the question that vapour density of \[C{{H}_{2}}O\] is 90. So from this we will calculate molecular mass by using following formula:
\[Molecular\text{ }mass=2\times Vapour\text{ }density\] \[{{C}_{6}}{{H}_{12}}{{O}_{6}}\]
Molecular mass of \[C{{H}_{2}}O\] = \[2\times Vapour\text{ }density=\,2\times 90=180\,\dfrac{gram}{mole}\]
Now, by using calculated molecular mass of empirical and molecular mass, we will find “n” which we will use to find molecular formula by using following formula:
\[Molecular\text{ }formula\text{ }=\text{ }n\times Empirical\text{ }Formula\]
“n” will be calculated by taking the ratio of molecular mass calculated by vapour density and empirical mass.
\[n=\dfrac{molecular\text{ }mass\text{ }calculated\text{ }by\text{ }vapour\text{ }density}{empirical\text{ }mass}\]= \[\dfrac{180}{30}=6\]
Now, we calculated “n” which is equal to 6. From this we will calculate the molecular formula.
\[Molecular\text{ }formula\text{ }=\text{ }n\times Empirical\text{ }Formula\]
\[Molecular\text{ }formula\text{ }=\text{ 6}\times C{{H}_{2}}O={{C}_{6}}{{H}_{12}}{{O}_{6}}\]
So, from the above calculation we calculated the molecular formula from the empirical formula. Molecular formula of \[C{{H}_{2}}O\,\]is \[{{C}_{6}}{{H}_{12}}{{O}_{6}}\]which we know by the name of glucose or simple sugar.

Note: We should note that the empirical formula of a compound expresses a ratio between the numbers of atoms of different elements present in a molecule of the compound. This ratio is a mole ratio as well as a ratio between numbers of atoms. From the formula it is possible to calculate the percent composition of a compound. Going in the opposite direction from the composition of a compound, it is possible to calculate its empirical formula. Consider the compound chloroform. The percent composition of chloroform is 10.06% carbon, 0.85% hydrogen, and 89.09% chlorine. We know then that 100 g chloroform contains 10.06 g carbon, 0.85 g hydrogen, and 89.09 g chlorine.