Question

Vitamin C has an empirical formula of ${C_3}{H_4}{O_3}$ and a molecular mass of $176amu$ , what is its molecular formula?

Answer 1

Empirical formula is the simplest mole ratio of all the atoms present in the molecular formula of that compound. The empirical mass and molecular mass can be related by the factor $n$ , which is the ratio of the molecular mass and empirical mass. The empirical mass can be determined from the empirical formula.
$n = \dfrac{M}{E}$
$n$ is factor relating the molecular mass and empirical mass
$M$ is molar mass
$E$ is empirical mass.

Complete answer:
Given that Vitamin C has an empirical formula of ${C_3}{H_4}{O_3}$
The mass of molecular formula of Vitamin C is $176amu$
The molar mass of the empirical formula of ${C_3}{H_4}{O_3}$ is determined from the molar mass of carbon, hydrogen, and oxygen atoms.
The molar mass of carbon atoms in empirical formula of ${C_3}{H_4}{O_3}$ is $3 \times 12 = 36amu$
The molar mass of hydrogen atoms in empirical formula of ${C_3}{H_4}{O_3}$ is $4 \times 1 = 4amu$
The molar mass of oxygen atoms in empirical formula of ${C_3}{H_4}{O_3}$ is $3 \times 16 = 48amu$
Thus, the empirical mass of Vitamin C is $36 + 4 + 48 = 88amu$
Substitute the molar mass and empirical mass in the above formula, to get the factor that relates the empirical formula and molecular formula.
$n = \dfrac{{176}}{{88}} = 2$
Thus, molecular formula is equal to the $2 \times$ empirical formula
Molecular formula is equal to ${C_6}{H_8}{O_6}$ .

Note:
Vitamin C is also known as ascorbic acid and is obtained from natural sources like citrus fruits. Vitamin C is an important antioxidant which can be useful in balancing the free radicals of oxygen in the body, free radicals are dangerous to mankind and several organisms.