Question

What is the molecular formula of a compound that has a molecular mass of $54$ and the empirical formula ${C_2}{H_3}$ $?$

Answer 1

First divide the molar mass of the compound by the empirical formula molar mass. Then the result should be a whole number or very close to a whole number. Multiply all the subscripts in the empirical formula by the whole number found in above step. The result is the molecular formula. To find the empirical formula we have to convert the mass of each element to moles using the molar mass from the periodic table. Then divide each mole value by the smallest number of moles calculated. Round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula.

Complete answer:
The molecular mass (m) is the mass of a given molecule: it is measured in $daltons$ ($Da$ or $u$). Different molecules of the same compound may have different molecular masses because they contain different isotopes of an element.
You have been given the empirical formula for a hydrocarbon. You need to determine the empirical formula mass. Then divide the molecular mass by the empirical mass. Then multiply the subscripts of the empirical formula by the result.
Empirical mass of ${C_2}{H_3}$can be find by the following way
Multiply the subscript of each element by its atomic mass and add.
i.e., ${C_2}{H_3}:(2 \times 12.011u\;C) + (3 \times 1.008u\;H) = 27.046\;u\;{C_2}{H_3}$
Molecular formula of the compound can be found by dividing the molecular mass by the empirical mass.
$\dfrac{{54}}{{27.046}} = 2.0$ rounded to two sig figs.
Multiply the subscripts in the empirical formula by 2 to get the molecular formula.
${C_{2 \times 2}}{H_{3 \times 2}} = {C_4}{H_6}$. Hence, the molecular formula of a compound that has a molecular mass of $54$ is ${C_4}{H_6}$.

Note:
Note that the empirical formula of a compound is the simplest whole number ratio of atoms of each element in the compound. For example, the molecular formula of glucose is ${C_6}{H_{12}}{O_6}$ but the empirical formula is $C{H_2}O$. This is because we can divide each number in ${C_6}{H_{12}}{O_6}$ by $6$to make a simpler whole number ratio.