# The given percent composition of an organic compound is 48.76% "C" and 8.04% "H". Its molecular mass is "74 g/mol". What are its empirical and chemical formulas?

Aug 19, 2015

Since the empirical formula mass is the same as the molecular mass, the empirical and molecular formulas are the same, $\text{C"_3"H"_6"O"_2}$.

#### Explanation:

The given percent composition of an organic compound is 48.76% "C" and 8.04% "H". The remaining 43.20% is oxygen.

Since the percentages add up to $\text{100%}$, you can assume that you have $\text{100 g}$ of the substance. This allows you to convert percentages directly to mass in grams. Now you need to calculate the moles of each element by dividing its mass in the compound by its molar mass, which is its atomic weight in $\text{g/mol}$.

$48.76 \cancel{\text{g C"xx(1"mol C")/(12.011cancel"g C")="4.0596 mol C}}$

$8.04 \cancel{\text{g H"xx(1"mol H")/(1.008cancel"g H")="7.9761 mol H}}$

$43.20 \cancel{\text{g O"xx(1"mol O")/(15.999cancel"g O")="2.7002 mol O}}$

Determine the Empirical Formula

An empirical formula represents the lowest whole-number ratio of elements in a compound. To determine the mole ratios, divide the moles of each element by the lowest number of moles, $2.7002 \text{mol O}$. The mole ratios will give us the subscripts.

$\text{C} :$(4.0596cancel"mol")/(2.7002 cancel"mol")="1.5034

$\text{H} :$(7.9761cancel"mol")/(2.7002 cancel"mol")="2.9539

$\text{O} :$(2.7002cancel"mol")/(2.7002cancel"mol")="1.0000

An empirical formula gives the lowest whole-number ratio of elements in a compound, represented by the subscripts.

Because the mole ratio of carbon is $1.5$, we need to multiply the mole ratios by $2$ in order to get a whole-number ratio for all of the elements.

$\text{C} :$$1.5034 \times 2 = 3.068 \approx 3$

$\text{H} :$$2.9539 \times 2 = 5.9078 \approx 6$

$\text{O} :$$1.0000 \times 2 = 2$

The empirical formula is $\text{C"_3"H"_6"O"_2}$.

The formula mass for the empirical formula is the sum of the subscripts of each element times its molar mass.

$\left(3 \times 12.011\right) + \left(6 \times 1.008\right) + \left(2 \times 15.999\right) = 74.079 \approx 74 \text{g/mol}$

Since the empirical formula mass is the same as the molecular mass, the empirical and molecular formulas are the same, $\text{C"_3"H"_6"O"_2}$.