Chemistry » Quantitative Aspects of Chemical Change » Stoichiometric Calculations Continued

# Percent Purity

## Percent Purity

The final use of stoichiometric calculations that we will look at is to determine the percent purity of a sample. Percent purity is important since when you make a compound you may have a small amount of impurity in the sample and you would need to keep this below a certain level. Or you may need to know how much of a particular ion is dissolved in water to determine if it is below the legally allowed level.

Percent purity can be calculated using: $\% \text{purity} = \frac{\text{mass of compound}}{\text{mass of sample}} \times 100$

## Example: Percent Purity

### Question

Shells contain calcium carbonate ($$\text{CaCO}_{3}$$) as well as other minerals. Faarah wants to know how much calcium carbonate is in a shell. She finds that the shell weighs $$\text{5}$$ $$\text{g}$$. After performing some more experiments she finds that the mass of calcium carbonate and the crucible (a container that is used to heat compounds in) is $$\text{3.2}$$ $$\text{g}$$. The mass of the crucible is $$\text{0.5}$$ $$\text{g}$$. How much calcium carbonate is in the shell?

### Step 1: Write down an equation for percent purity

Percent purity is given by: $\% \text{purity} = \frac{\text{mass of compound}}{\text{mass of sample}} \times 100$

### Step 2: Find the mass of the product

We are given the mass of the crucible and the mass of the crucible with the product. We need to subtract the mass of the crucible from the mass of the crucible with the product to obtain only the mass of the product. \begin{align*} \text{Mass product} & = \text{3.2}\text{ g} – \text{0.5}\text{ g} \\ & = \text{2.7}\text{ g} \end{align*}

### Step 3: Calculate the answer.

Substituting the calculated mass into the equation for percent purity gives: \begin{align*} \% \text{purity} &= \frac{\text{mass of compound}}{\text{mass of sample}} \times 100\\ & = \frac{\text{2.7}}{\text{5}} (\text{100})\\ & = \text{54}\% \end{align*}

## Example: Percent Purity

### Question

Limestone is mostly calcium carbonate ($$\text{CaCO}_{3}$$). Jake wants to know how much calcium carbonate is in a sample of limestone. He finds that the sample weighs $$\text{3.5}$$ $$\text{g}$$. He then adds concentrated hydrochloric acid ($$\text{HCl}$$) to the sample. The equation for this reaction is: \begin{align*} \text{CaCO}_{3}\text{(s)} + 2\text{HCl (aq)} \rightarrow \text{CO}_{2}\text{(g)} + \text{CaCl}_{2}\text{(aq)} + \text{H}_{2}\text{O (l)} \end{align*}

If the mass of calcium chloride produced is $$\text{3.6}$$ $$\text{g}$$, what is the percent purity of the limestone sample?

### Step 1: Calculate the number of moles of calcium chloride

The number of moles of calcium chloride is: \begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{3.6}}{\text{111}} \\ & = \text{0.032}\text{ mol} \end{align*}

### Step 2: Calculate the number of moles of calcium carbonate

The molar ratio of calcium chloride to calcium carbonate is 1:1. Therefore the number of moles of calcium carbonate is $$\text{0.032}$$ $$\text{mol}$$.

### Step 3: Calculate the mass of calcium carbonate

The mass of calcium carbonate is: \begin{align*} m & = nM \\ & = (\text{0.032})(\text{100}) \\ m & = \text{3.24}\text{ g} \end{align*}

### Step 4: Calculate the percent purity

Substituting the calculated mass into the equation for percent purity gives: \begin{align*} \% \text{purity} &= \frac{\text{mass of compound}}{\text{mass of sample}} \times 100\\ & = \frac{\text{3.3}}{\text{3.5}} \times (100)\\ & = \text{94.3}\% \end{align*}

This section includes a recommended experiment for informal assessment. It is recommended that you do this experiment as a demonstration for your class. In this experiment you will show your learners what happens when a small sample (about $$\text{0.5}$$ $$\text{g}$$) of lead(II) nitrate is heated. You must work outside or in a well ventilated space and remember that lead(II) nitrate makes a crackling sound on heating.

A second experiment to determine the percent yield of magnesium carbonate from magnesium sulfate (Epsom salts) and sodium carbonate (washing soda) has also been included. You will need magnesium sulfate, sodium carbonate, mass meter, hot plate or Bunsen burner, small heat resistant beakers, funnel and filter paper. If time and space permits, learners can write their names on their pieces of filter paper before filtering and leave these in an undisturbed place (such as a back room) overnight to dry before weighing them. If this cannot be done remind learners that due to their samples containing some water the sample may not be completely dry when weighing. This makes it slightly harder to take a mass reading and may lead to a percent yield of greater than $$\text{100}\%$$.

When working with a Bunsen burner you must ensure that the room is well ventilated by opening all windows. Also ensure that learners tie long hair up and tuck in any loose clothing.

## Optional Experiment: The thermal decomposition of lead(II) nitrate

### Aim

To observe what happens when lead(II) nitrate is heated.

### Apparatus

• Bunsen burner
• Test tube (glass)
• Test tube holder
• $$\text{0.5}$$ $$\text{g}$$ lead(II) nitrate

### Method

#### Warning:

Lead(II) nitrate produces toxic nitrogen dioxide on heating.

1. Place the lead(II) nitrate sample in the test tube.
2. Light the Bunsen burner and carefully hold the test tube in the flame. Remember to point the mouth of the test tube away from you.
3. Observe what happens.

### Results

A crackling noise is heard on heating the sample and a small amount of a brownish coloured gas is noted. The white powder became yellow.

If a glowing splint is held at the mouth of the test tube the splint reignites.

The balanced chemical for this reaction is:

\begin{align*} 2\text{Pb(NO}_{3}\text{)}_{2}\text{(s)} + \text{ heat } \rightarrow 2\text{PbO (s)} + 4\text{NO}_{2} \text{(g)} + \text{O}_{2}\text{(g)} \end{align*}

## Optional Experiment: Stoichiometry

### Aim

To determine the percentage yield of magnesium carbonate from magnesium sulfate and sodium carbonate.

### Apparatus

• magnesium sulfate (Epsom salts)
• sodium carbonate
• mass meter
• hot plate
• small glass beakers (heat resistant)
• funnel
• filter paper

### Method

1. Weigh out $$\text{5}$$ $$\text{g}$$ of magnesium sulfate.

2. Dissolve the magnesium sulfate in $$\text{20}$$ $$\text{ml}$$ of water. Heat the solution until all the solid dissolves.

3. Dissolve $$\text{5}$$ $$\text{g}$$ of sodium carbonate in $$\text{20}$$ $$\text{ml}$$ of warm water. If necessary heat the solution to dissolve all the sodium carbonate.

4. Carefully pour the sodium carbonate solution into the hot magnesium sulfate solution. You should have a milky looking solution.

5. Weigh the filter paper (if your mass meter is not very accurate then assume the mass of the filter paper is $$\text{0}$$). Carefully filter the final solution.

6. Leave the filter paper to dry slightly (or overnight) and then weigh it. The solid that stays on the filter paper is magnesium carbonate.

### Results

The equation for this reaction is: $$\text{MgSO}_{4}\text{(aq)} + \text{Na}_{2}\text{CO}_{3}\text{(aq)} \rightarrow \text{Na}_{2}\text{SO}_{4}\text{(aq)} + \text{MgCO}_{3}\text{(s)} + \text{H}_{2}\text{O (l)}$$.

Use the above equation to work out the percentage yield of the magnesium sulfate. Remember that you need to determine which of the reactants is limiting. (You may get a percentage yield of greater than $$\text{100}\%$$ if your sample is not completely dry.)

### Conclusion

By performing an experiment we were able to calculate the percentage yield of a reaction.