# 3.2 Determining empirical and molecular formulas

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By the end of this section, you will be able to:
• Compute the percent composition of a compound
• Determine the empirical formula of a compound
• Determine the molecular formula of a compound

In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

## Percent composition

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition    , defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

$%\phantom{\rule{0.2em}{0ex}}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass H}}{\text{mass compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%$
$%\phantom{\rule{0.2em}{0ex}}\text{C}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass C}}{\text{mass compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%$

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

$%\phantom{\rule{0.2em}{0ex}}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{2.5\phantom{\rule{0.2em}{0ex}}\text{g H}}{10.0\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=25%$
$%\phantom{\rule{0.2em}{0ex}}\text{C}=\phantom{\rule{0.2em}{0ex}}\frac{7.5\phantom{\rule{0.2em}{0ex}}\text{g C}}{10.0\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=75%$

## Calculation of percent composition

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

## Solution

To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

$\begin{array}{c}\\ %\phantom{\rule{0.2em}{0ex}}\text{C}=\phantom{\rule{0.2em}{0ex}}\frac{7.34\phantom{\rule{0.2em}{0ex}}\text{g C}}{12.04\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=61.0%\\ %\phantom{\rule{0.2em}{0ex}}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{1.85\phantom{\rule{0.2em}{0ex}}\text{g H}}{12.04\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=15.4%\\ %\phantom{\rule{0.2em}{0ex}}\text{N}=\phantom{\rule{0.2em}{0ex}}\frac{2.85\phantom{\rule{0.2em}{0ex}}\text{g N}}{12.04\phantom{\rule{0.2em}{0ex}}\text{g compound}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}100%=23.7%\end{array}$

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

## Check your learning

A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

12.1% C, 16.1% O, 71.8% Cl

## Determining percent composition from formula mass

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH 3 ), ammonium nitrate (NH 4 NO 3 ), and urea (CH 4 N 2 O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH 3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 $×$ 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:

#### Questions & Answers

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