KEY2CHEM

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. $$75\%$$

Solution

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. $$\require{mhchem}\ce{C11H22O}$$

Solution

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. $$43\%\text{ C}, 57\%\text{ H}$$

Solution

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 $$100\%$$.