# What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula C2H3?

May 23, 2017

The molecular formula is $\text{C"_4"H"_6}$.

Refer to the explanation for the process.

#### Explanation:

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.

color(blue)("Empirical Mass of C"_2"H"_3"
Multiply the subscript of each element by its atomic mass and add.

${\text{C"_2"H}}_{3} :$(2xx12.011"u C") + (3xx1.008 "u H")="27.046 u C"_2"H"_3"

color(blue)("Determine Molecular Formula"
Divide the molecular mass by the empirical mass.

$\frac{54}{27.046} = 2.0$ rounded to two sig figs.

Multiply the subscripts in the empirical formula by 2 to get the molecular formula.

$\text{C_(2xx2)""H_(3xx2)"=C"_4"H"_6}$

The molecular formula is $\text{C"_4"H"_6}$.