Question

Empirical formula of a compound is $\text{C}{{\text{H}}_{\text{2}}}\text{O}$ and its molecular mass is 90, the molecular formula of the compound is :
A ) ${{\text{C}}_{3}}{{\text{H}}_{\text{6}}}{{\text{O}}_{3}}$
B ) ${{\text{C}}_{2}}{{\text{H}}_{\text{4}}}{{\text{O}}_{2}}$
C ) ${{\text{C}}_{6}}{{\text{H}}_{\text{12}}}{{\text{O}}_{6}}$
D ) $\text{C}{{\text{H}}_{\text{2}}}\text{O}$

Answer 1

Multiply the empirical formula with the ratio of the molecular formula weight to the Empirical formula weight to obtain the molecular formula.

Complete step by step answer:
The molecular formula gives the actual number of atoms of each element present in one molecule of a compound. The empirical formula gives the smallest whole number ratio of atoms of various elements present in a molecule of the given compound. In some cases, the empirical formula is same as the molecular formula. However in other compounds, empirical formula and the molecular formula are different. The relationship between the empirical formula and the molecular formula is as given below:

$\text{Molecular formula = n }\times \text{ empirical formula }$
Here, n is the ratio of the molecular formula weight to the empirical formula weight:
$\text{n=}\dfrac{\text{Molecular formula weight}}{\text{Empirical formula weight}}$

The atomic masses of carbon, hydrogen and oxygen are $12\text{ }g/mol,\text{ }1\text{ }g/mol\text{ }and\text{ }16\text{ }g/mol$ respectively.
Empirical formula of a compound is $\text{C}{{\text{H}}_{\text{2}}}\text{O}$.
Calculate the empirical formula mass.

$12\text{ }+2\left( 1 \right)+16=30\text{ }g/mol$
The molecular mass is $90\text{ }g/mol$.
Divide molecular formula weight with empirical formula weight to calculate n.

& \text{n=}\dfrac{90\text{ g/mol}}{30\text{ g/mol}} \\\ & \text{n=3} \\\ \end{aligned}$$ Multiply empirical formula with 3 to obtain the molecular formula. $$\begin{aligned} & \text{Molecular formula = n }\times \text{ empirical formula } \\\ & \text{Molecular formula = 3 }\times \text{ C}{{\text{H}}_{2}}\text{O } \\\ & \text{Molecular formula = }{{\text{C}}_{3}}{{\text{H}}_{6}}{{\text{O}}_{3}} \\\ \end{aligned}$$ Hence, the molecular formula of the compound is $${{\text{C}}_{3}}{{\text{H}}_{6}}{{\text{O}}_{3}}$$: **Hence, the option A) $${{\text{C}}_{3}}{{\text{H}}_{6}}{{\text{O}}_{3}}$$ is the correct option.** **Note:** We can also calculate the mass of molecular given in the four options and match it with the options For $${{\text{C}}_{3}}{{\text{H}}_{\text{6}}}{{\text{O}}_{3}}$$, $$3\left( 12 \right)+6\left( 1 \right)+3\left( 16 \right)=36+6+48=90$$ For $${{\text{C}}_{2}}{{\text{H}}_{\text{4}}}{{\text{O}}_{2}}$$, $$2\left( 12 \right)+4\left( 1 \right)+2\left( 16 \right)=24+4+32=60$$ For $${{\text{C}}_{6}}{{\text{H}}_{\text{12}}}{{\text{O}}_{6}}$$, $$6\left( 12 \right)+12\left( 1 \right)+6\left( 16 \right)=72+12+96=180$$ For $$\text{C}{{\text{H}}_{\text{2}}}\text{O}$$ $$12+2\left( 1 \right)+16=30$$ The molecular mass of $${{\text{C}}_{3}}{{\text{H}}_{\text{6}}}{{\text{O}}_{3}}$$ matches with that of given compound hence it is correct.