Using Mass Data to Determine Composition

Pure substances have a constant composition throughout. Pure compounds have a specific mass percent of each element, since the ratio of elements in a compound is constant. The lowest whole number ratio of elements in a compound is called an empirical formula and can be determined from these mass percent values of the elements in a compound.

\(\text{mass percent} = \frac{\text{mass of element}}{\text{mass of compound}}\times 100\%\)

Example 1.

What is the mass percent of carbon in methane (\(\require{mhchem}\ce{CH4}\))?

A. \(25\%\)

B. \(50\%\)

C. \(75\%\)



C. \(75\%\)

\(\begin{align}\text{mass percent} &= \frac{\text{mass of carbon}}{\text{mass of methane}}\times 100\%\\\text{mass percent} &= \frac{12.01\text{ g}}{16.042\text{ g}}\times 100\% = 75\%\\\end{align}\)

Example 2.

Ibuprofen is an analgesic (a pain-reliever) that is composed of \(75.6\% \;\require{mhchem}\ce{C}\), \(8.73\% \;\require{mhchem}\ce{H}\), and \(15.7\% \;\require{mhchem}\ce{O}\) by mass. What is the empirical formula of ibuprofen?

A. \(\require{mhchem}\ce{C13H18O2}\)

B. \(\require{mhchem}\ce{C6H8O}\)

C. \(\require{mhchem}\ce{C11H22O}\)



A. \(\require{mhchem}\ce{C13H18O2}\)

To simplify the arithmetic, assume a sample mass of \(100\text{ g}\); this way, the percent of each element becomes the grams of each element. The masses can then be converted to moles using molar mass.

\(\text{moles carbon}\\75.6\text{ g C}\times\frac{1\text{ mol C}}{12.01\text{ g C}} = 6.29 \text{ mol C}\)

\(\text{moles hydrogen}\\8.73\text{ g H}\times\frac{1\text{ mol H}}{1.008\text{ g H}} = 8.66 \text{ mol H}\)

\(\text{moles oxygen}\\15.7\text{ g O}\times\frac{1\text{ mol O}}{16.00\text{ g O}} = 0.981 \text{ mol O}\)


Next, divide each value by the lowest number of moles (oxygen in this case). 

\(\text{C: } \frac{6.29\text { mol}}{0.981\text{ mol}} = 6.41\)

\(\text{H: } \frac{8.66\text { mol}}{0.981\text{ mol}} = 8.83\)

\(\text{O: } \frac{0.981\text { mol}}{0.981\text{ mol}} = 1.00\)

This would give us the formula \(\text{C}_{6.41}\text{H}_{8.83}\text{O}\). Since the empirical formula is the lowest whole number ratio of elements in the formula, we must multiply all the values by a value that will convert them to integers (with rounding). In this case, that value is \(2\)

\(\text{C: } 6.41\times 2 = 12.82 = 13\\ \text{H: } 8.83\times 2 = 17.66 = 18\\ \text{O:} 1.00\times 2 = 2\)

This gives the empirical formula \(\require{mhchem}\ce{C13H18O2}\)

Example 3.

Ethylene (\(\require{mhchem}\ce{C2H4}\)) is an important compound for fruit ripening and is also used industrial to generate plastics. What are the mass percentages of carbon and hydrogen in ethylene?

A. \(20\%\text{ C}, 40\%\text{ H}\)

B. \(86\%\text{ C}, 14\%\text{ H}\)

C. \(43\%\text{ C}, 57\%\text{ H}\)



B. \(86\%\text{ C}, 14\%\text{ H}\)

Since the compound contains only carbon and hydrogen, the mass percentages of these two elements must add up to the total (\(100\%\)). 

\(\begin{align}\text{mass percent carbon} &= \frac{\text{mass of carbon}}{\text{mass of ethylene}}\times 100\%\\\text{mass percent} &= \frac{2\times12.01\text{ g}}{28.052\text{ g}}\times 100\% = 86\%\\\end{align}\)

\(\begin{align}\text{mass percent hydrogen} &= \frac{\text{mass of hydrogen}}{\text{mass of ethylene}}\times 100\%\\\text{mass percent} &= \frac{4\times1.008\text{ g}}{28.052\text{ g}}\times 100\% = 14\%\\\end{align}\)

Note that the mass percentage of one element could be found by subtracting the other from \(100\%\).