# 3.2 Determining empirical and molecular formulas  (Page 3/6)

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${\text{C1}}_{0.150}{\text{O}}_{0.525}={\text{Cl}}_{\phantom{\rule{0.2em}{0ex}}\frac{0.150}{0.150}\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{\phantom{\rule{0.2em}{0ex}}\frac{0.525}{0.150}\phantom{\rule{0.2em}{0ex}}}={\text{ClO}}_{3.5}$

In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, we must multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl 2 O 7 as the final empirical formula.

In summary, empirical formulas are derived from experimentally measured element masses by:

1. Deriving the number of moles of each element from its mass
2. Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

[link] outlines this procedure in flow chart fashion for a substance containing elements A and X.

## Determining a compound’s empirical formula from the masses of its elements

A sample of the black mineral hematite ( [link] ), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?

## Solution

For this problem, we are given the mass in grams of each element. Begin by finding the moles of each:

$\begin{array}{l}\\ 34.97\phantom{\rule{0.2em}{0ex}}\text{g Fe}\left(\frac{\text{mol Fe}}{55.85\phantom{\rule{0.2em}{0ex}}\text{g}}\right)& =& 0.6261\phantom{\rule{0.2em}{0ex}}\text{mol Fe}\hfill \\ 15.03\phantom{\rule{0.2em}{0ex}}\text{g O}\left(\frac{\text{mol O}}{16.00\phantom{\rule{0.2em}{0ex}}\text{g}}\right)& =& 0.9394\phantom{\rule{0.2em}{0ex}}\text{mol O}\hfill \end{array}$

Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:

$\begin{array}{}\\ \\ \phantom{\rule{0.2em}{0ex}}\frac{0.6261}{0.6261}\phantom{\rule{0.2em}{0ex}}=1.000\phantom{\rule{0.2em}{0ex}}\text{mol Fe}\\ \phantom{\rule{0.2em}{0ex}}\frac{0.9394}{0.6261}\phantom{\rule{0.2em}{0ex}}=1.500\phantom{\rule{0.2em}{0ex}}\text{mol O}\end{array}$

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe 1 O 1.5 ). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

$2\left({\text{Fe}}_{1}{\text{O}}_{1.5}\right)\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}{\text{Fe}}_{2}{\text{O}}_{3}$

The empirical formula is Fe 2 O 3 .

What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?

N 2 O 5

## Deriving empirical formulas from percent composition

Finally, with regard to deriving empirical formulas, consider instances in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.

## Determining an empirical formula from percent composition

The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O ( [link] ). What is the empirical formula for this gas?

## Solution

Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. The calculation is “most convenient” because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element’s mass percentage. This numerical equivalence results from the definition of the “percentage” unit, whose name is derived from the Latin phrase per centum meaning “by the hundred.” Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:

$\begin{array}{l}\\ 27.29%\phantom{\rule{0.2em}{0ex}}\text{C}& =\hfill & \phantom{\rule{0.2em}{0ex}}\frac{27.29\phantom{\rule{0.2em}{0ex}}\text{g C}}{100\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}\hfill \\ 72.71%\phantom{\rule{0.2em}{0ex}}\text{O}& =\hfill & \phantom{\rule{0.2em}{0ex}}\frac{72.71\phantom{\rule{0.2em}{0ex}}\text{g O}}{100\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}\hfill \end{array}$

The molar amounts of carbon and hydrogen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

$\begin{array}{l}\\ 27.29\phantom{\rule{0.2em}{0ex}}\text{g C}\left(\frac{\text{mol C}}{12.01\phantom{\rule{0.2em}{0ex}}\text{g}}\right)& =\hfill & 2.272\phantom{\rule{0.2em}{0ex}}\text{mol C}\hfill \\ 72.71\phantom{\rule{0.2em}{0ex}}\text{g O}\left(\frac{\text{mol O}}{16.00\phantom{\rule{0.2em}{0ex}}\text{g}}\right)& =\hfill & 4.544\phantom{\rule{0.2em}{0ex}}\text{mol O}\hfill \end{array}$

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:

$\begin{array}{}\\ \\ \phantom{\rule{0.2em}{0ex}}\frac{2.272\phantom{\rule{0.2em}{0ex}}\text{mol C}}{2.272}\phantom{\rule{0.2em}{0ex}}=1\\ \phantom{\rule{0.2em}{0ex}}\frac{4.544\phantom{\rule{0.2em}{0ex}}\text{mol O}}{2.272}\phantom{\rule{0.2em}{0ex}}=2\end{array}$

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO 2 .

What is the empirical formula of a compound containing 40.0% C, 6.71% H, and 53.28% O?

CH 2 O

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