# A compound with an empirical formula of C_4H_4O and a molar mass of 136 grams per mole. What is the molecular formula of this compound?

Jan 2, 2016

Well, you have the reduced formula, and you have the molar mass. There's not much to do other than to define what the terms "empirical formula" and "molecular formula" mean. That would help make the connection.

Molecular Formula - The combination of atomic symbols and subscripts in such a way that they exactly represent a molecule's proportions of each element that composes the molecule.

Empirical Formula - The most reduced form of the molecular formula, i.e. the smallest subscript is odd.

To determine what to do from here, just figure out what the molar mass of the given empirical formula is.

${M}_{\text{C" ~~ "12.011 g/mol}}$
${M}_{\text{H" ~~ "1.0079 g/mol}}$
${M}_{\text{O" ~~ "15.999 g/mol}}$

=> M_("C"_4"H"_4"O") ~~ "68.0746 g/mol"

Now we should see that this is about half of the given "true" molar mass, so the molecular formula is:

$\textcolor{b l u e}{{\text{C"_8"H"_8"O}}_{2}}$

Granted, this could be any number of compounds (like methyl benzoate, o-methylbenzoic acid, p-methylbenzoic acid, etc), but you aren't supposed to figure anything out beyond this.

Jan 2, 2016

The molecular formula is $\text{C"_8"H"_8"O"_2}$.

#### Explanation:

The empirical formula mass of $\text{C"_4"H"_4"O}$ is $\left(4 \times 12\right) + \left(4 \times 1\right) + \left(1 \times 16\right) = \text{68 g/mol}$.

To determine the molecular formula, divide the molecular molar mass by the empirical formula mass.

$\frac{136}{68} = 2$

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

$\text{Molecular Formula"="C"_8"H"_8"O"_2}$