Question

The empirical formula of a compound is $C{{H}_{4}}$. One mole of this compound has a mass of 42 g. Its molecular mass is:
A) $C{{H}_{2}}$
B) ${{C}_{2}}{{H}_{2}}$
C)${{C}_{3}}{{H}_{6}}$
D)${{C}_{3}}{{H}_{8}}$

Answer 1

Hint: An empirical formula is the simplest whole number ratio of atoms present in a compound whereas molecular formula shows the exact number of different types of atoms present in a molecule of a compound. Refer to the relation between molecular mass and empirical formula.

Complete answer:
We are given that the empirical formula of a compound is $C{{H}_{2}}$. One mole of this compound has a mass of 42 g. This means, the molecular mass of the compound is given because the mass of one mole of a substance in grams is called its molecular mass.
We are asked to find the molecular formula of the compound. Let us proceed in the steps.
Step 1. Determine empirical formula mass.
Add the atomic masses of various atoms present in the empirical formula to calculate empirical formula mass.
For $C{{H}_{2}}$, Empirical formula mass: $12+(2\times 1)$ = 14 g.
Step 2. Divide the molecular mass by empirical formula mass.
The molecular mass of the compound is 42 g.
Therefore,$\dfrac{Molecular\,mass}{Empirical\,mass}=\dfrac{42}{14}=3=(n)$
Step 3. Multiply empirical formula by $n$ obtained above to get the molecular formula.
 Molecular formula = $Empirical\,formula\times n$
Where, n=3 and empirical formulas is $C{{H}_{2}}$
Therefore, molecular formula is ${{(C{{H}_{2}})}_{3}}$, or ${{C}_{3}}{{H}_{6}}$

Hence, option C is correct.

Note:
The molecular formula obtained i.e., the compound ${{C}_{3}}{{H}_{6}}$is known with the common name ‘propylene’ and IUPAC name as propene. Its structure is $C{{H}_{2}}=CH-C{{H}_{3}}$. It is an unsaturated compound. If the mass percent of various atoms present in a compound is known, its empirical formula can be determined and consequently, molecular formula can also be determined.