KEY2CHEM

Using Mass Data to Determine Purity

Pure substances have a constant composition throughout, while mixtures have varied composition. When working with mixtures, mass percent data can provide insight into how much of the pure substance is present in a mixture.

$$\text{mass percent purity} = \frac{\text{mass of pure substance}}{\text{mass of sample}} \times 100\%$$

Example 1.

A $$10.0\text{ g}$$ piece of jewelry contains $$5.8\text{ g}$$ of gold. What is the mass percent of gold in the jewelry?

A. $$5.8\%$$

B. $$58\%$$

C. $$0.58\%$$

Solution

B. $$58\%$$

\begin{align}\text{mass percent purity} &= \frac{\text{mass of pure substance}}{\text{mass of sample}} \times 100\%\\\text{mass percent gold} &= \frac{\text{mass of gold}}{\text{mass of jewelry}} \times 100\%\\\text{mass percent gold} &= \frac{5.8\text{ g}}{{10.0\text{ g}}} \times 100\% = 58\%\end{align}

Example 2.

A tablet of ibuprofen pain reliever weighs $$0.500\text { g}$$. If the active ingredient in the tablet is $$200 \text{ mg}$$, what mass percent of the tablet is the active ingredient ibuprofen?

A. $$100\%$$

B. $$0.25\%$$

C. $$40\%$$

Solution

C. $$40\%$$

\begin{align}\text{mass percent purity} &= \frac{\text{mass of pure substance}}{\text{mass of sample}} \times 100\%\\\text{mass percent ibuprofen} &= \frac{\text{mass of ibuprofen}}{\text{mass of tablet}} \times 100\%\\\text{mass percent ibuprofen} &= \frac{0.200\text{ g}}{{0.500\text{ g}}} \times 100\% = 40\%\end{align}

Example 3.

A $$5.0\text{ kg}$$ sample of impure iron ore is found to be $$21\%$$ iron by mass. How many grams of iron are in the ore sample?

A. $$1050\text{ g}$$

B. $$105\text{ g}$$

C. $$525\text{ g}$$

Solution

A. $$1050\text{ g}$$

\begin{align}\text{mass percent purity} &= \frac{\text{mass of pure substance}}{\text{mass of sample}} \times 100\%\\\text{mass percent iron} &= \frac{\text{mass of iron}}{\text{mass of ore}} \times 100\%\\21\%&= \frac{\text{ mass of iron}}{{5000\text{ g}}} \times 100\% \\1050\text{ g}&= \text{mass of iron}\end{align}