AP Chem 1.4: Composition of Mixtures

How pure is this salt?

Lab reagent bottles state a purity, something like “NaCl ≥ 99.0%”. Where does that number come from? And if someone mixed something else in, how would you find out?

U1.3 showed that a pure substance has a fixed composition. This lesson puts that to work: measure the mass percent of one element in a sample, compare it with the value for the pure substance, and you can tell whether the sample is pure, what might be mixed in, and how much.

Pure substances and mixtures, from particles to data

Lesson 1.2.5 told them apart at the particle level:

A pure substance contains one kind of particle (one kind of molecule or formula unit). Its composition is fixed and matches its formula.
A mixture contains two or more kinds of particles, in proportions that can vary.

In terms of data: different samples of the same pure substance all give the same mass percent of each element, equal to the value calculated from the formula. Different samples of a mixture often disagree with each other and don’t match the value for any single substance.

Here is the mass percent of Cl in three samples each of table salt and lite salt (NaCl mixed with KCl):

Sample	1	2	3	Pure NaCl (calculated)
Table salt	60.6%	60.7%	60.6%	60.66%
Lite salt	56.7%	56.1%	57.4%	60.66%

The table-salt results agree with each other and with the value for NaCl, consistent with a pure substance. The lite-salt results vary and are all below 60.66%, so it’s a mixture, and whatever was added has a lower mass percent of Cl than NaCl does (KCl is 47.55%).

The mass percent in a mixture is a weighted average

Suppose a mixture contains components A and B with mass fractions $x_A$ and $x_B$ (their shares of the mixture’s mass), where $x_A + x_B = 1$ . Then the mass percent of Cl in the mixture is

\text{mass \% Cl}_{\text{mixture}} = x_A \cdot (\text{mass \% Cl})_A + x_B \cdot (\text{mass \% Cl})_B

It’s the same calculation as the average atomic mass in U1.2: a weighted average. The result always lies between the values for the two pure substances, closer to the one there’s more of. That tells you the direction of an error:

If the impurity has a higher mass percent of the element than the main substance, the result comes out high.
If the impurity has a lower mass percent (including none at all), the result comes out low.

Finding purity by elemental analysis

The mass percent of the main substance in a sample is its purity纯度. Measuring how much of each element a sample contains is called elemental analysis元素分析.

If the impurity contains none of the measured element, all of that element comes from the main substance, and purity is the measured value divided by the pure value: $\text{purity} = \frac{\text{measured mass \%}}{\text{mass \% in the pure substance}} \times 100\%$
If the impurity also contains the element, set up the weighted-average equation and solve for $x$ (Example 2).

Pick an impurity and drag the slider to watch the particle view and the scale change. Then try a few samples in “Find the impurity”:

Try it: mix an impurity into NaCl and watch the mass percent of Cl

Impurity:

Example 1: Purity of NaCl by precipitation

A 1.000 g sample of NaCl contaminated with $\ce{NaNO3}$ is dissolved in water, and excess $\ce{AgNO3}$ solution is added. All the chloride precipitates as AgCl, which has a mass of 2.150 g when dry. What is the mass percent of NaCl in the sample?

\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}

Step-by-step solution

Track the chlorine. With excess $\ce{AgNO3}$ , all the Cl in the sample ends up in the AgCl precipitate, so start from the precipitate’s mass. The molar mass of AgCl (g/mol):
$\begin{aligned} M(\text{AgCl}) &= 107.87 + 35.45 \\ &= 143.32 \end{aligned}$
Mass of Cl in the precipitate. Multiply by the fraction of AgCl’s mass that is Cl (result in g):
$\begin{aligned} m(\text{Cl}) &= 2.150 \times \frac{35.45}{143.32} \\ &= 0.5318 \end{aligned}$
Where the chlorine came from. $\ce{NaNO3}$ contains no Cl, so all 0.5318 g came from NaCl. Convert the mass of Cl to a mass of NaCl (g):
$\begin{aligned} m(\text{NaCl}) &= 0.5318 \times \frac{58.44}{35.45} \\ &= 0.8767 \end{aligned}$
Mass percent.
$\begin{aligned} \%\text{NaCl} &= \frac{0.8767}{1.000} \times 100\% \\ &= \boxed{87.67\%} \end{aligned}$
The other 12.33% is $\ce{NaNO3}$ .
Check by another route. The sample is 0.5318 ÷ 1.000 = 53.18% Cl, and pure NaCl is 35.45 ÷ 58.44 = 60.66% Cl:
$\frac{53.18\%}{60.66\%} = 87.67\%$
Both routes agree.

Example 2: When the impurity contains Cl too

A sample of NaCl contaminated with KCl is 58.04% Cl by mass. What is the mass percent of KCl in the sample?

Step-by-step solution

Check the direction first. Pure NaCl is 60.66% Cl, and pure KCl is 35.45 ÷ 74.55 = 47.55% Cl. The sample’s 58.04% lies between them, closer to NaCl, so KCl is the minor component.
Weighted-average equation. Let x be the mass fraction of KCl, so NaCl is 1 − x (percent signs dropped):
$60.66(1 - x) + 47.55x = 58.04$
Solve.
$\begin{aligned} 60.66 - 13.11x &= 58.04 \\ 13.11x &= 2.62 \\ x &= \boxed{0.200} \end{aligned}$
The sample is about 20.0% KCl and 80.0% NaCl.
Substitute back.
$\begin{aligned} &0.800 \times 60.66 + 0.200 \times 47.55 \\ &= 48.53 + 9.51 = 58.04 \end{aligned}$
This matches the measured 58.04%, and the result sits closer to the NaCl value because NaCl is the majority.

Example 3: Identifying an unknown salt by precipitation

A 0.500 g sample of an alkali metal chloride, MCl, is dissolved in water, and excess $\ce{AgNO3}$ solution is added. The AgCl precipitate has a mass of 0.961 g. What is M?

Step-by-step solution

Moles of Cl⁻ from the precipitate. All the Cl ends up in AgCl (in mol):
$\begin{aligned} n(\text{AgCl}) &= \frac{0.961}{143.32} \\ &= 6.705 \times 10^{-3} \end{aligned}$
Link it to MCl. Each formula unit of MCl contains just one Cl, so (in mol):
$\begin{aligned} n(\text{MCl}) &= n(\text{AgCl}) \\ &= 6.705 \times 10^{-3} \end{aligned}$
Molar mass of MCl. In g/mol:
$\begin{aligned} M(\text{MCl}) &= \frac{0.500}{6.705 \times 10^{-3}} \\ &= 74.6 \end{aligned}$
Atomic mass of M. Subtract the chlorine:
$74.6 - 35.45 = 39.1$
Identify M. The alkali metal with atomic mass 39.1 is potassium, so the salt is $\boxed{\text{KCl}}$ .
Check. 0.500 g of KCl is 0.500 ÷ 74.55 = 6.71 × 10⁻³ mol, which gives 6.71 × 10⁻³ × 143.32 = 0.961 g of AgCl, matching the data.

Good gravimetric technique

Examples 1 and 3 both use gravimetric analysis沉淀法（重量分析）: turn the ion you’re measuring completely into a precipitate, then weigh it. How accurate the result is depends on these steps:

Precipitate only the target ion. When you measure $\ce{Cl-}$ with $\ce{AgNO3}$ , the sample can’t contain $\ce{Br-}$ or $\ce{I-}$ , which also precipitate and would make the Cl come out high.
Add excess precipitating reagent so the target ion precipitates completely. The leftover reagent stays in solution and washes away, so it doesn’t affect the result.
Tare the container (or weigh it separately) so you record only the mass of the sample or precipitate.
Wash the precipitate to remove soluble substances stuck to it (such as $\ce{NaNO3}$ ); otherwise it weighs too much.
Dry to constant mass: heat, cool, and weigh repeatedly until the mass stops changing, which shows all the water is gone. For a hydrate, cool it in a desiccator, or it will absorb water again.

Some methods measure a signal proportional to the amount instead of weighing a precipitate (such as how intensely a solution absorbs light, U3.13). Measure a few standards of known mass percent, draw a calibration curve标准曲线, and read the unknown’s mass percent off the line.

Practice: with each of these mixed in, how does the measured mass percent of Cl compare with pure NaCl (60.66%)?

KCl mixed in
$\ce{MgCl2}$ mixed in
$\ce{CaCl2}$ mixed in
$\ce{NaNO3}$ mixed in
The sample got damp and absorbed some water
LiCl mixed in
Pure NaCl, but 5 g weighed out instead of 1 g
Pure NaCl from a different salt mine

Practice: does each error make the measured result too high, too low, or leave it unchanged?

Measuring % Cl by precipitation: the AgCl isn't washed and has $\ce{NaNO3}$ stuck to it
Measuring % Cl by precipitation: the AgCl is weighed before it's dry
Measuring % Cl by precipitation: some AgCl passes through the filter paper
Measuring % Cl by precipitation: the sample also contains a little NaBr
Measuring % Cl by precipitation: more $\ce{AgNO3}$ is added than needed
Measuring % water in a hydrate: after heating, the sample cools in open air and absorbs some water
Measuring % water in a hydrate: the sample isn't heated to constant mass
Measuring % water in a hydrate: some solid spatters out of the crucible during heating

Common mistakes

Mistake 1: “A mixture has no fixed composition, so you can’t calculate anything.” A mixture’s composition can vary, but every individual sample has a definite composition. Knowing the components’ shares, you can calculate an element’s mass percent with a weighted average, and from an element’s mass percent you can work back to the components.

Mistake 2: “The sample is 53.18% Cl, so it’s 53.18% NaCl.” 53.18% is the mass percent of the element Cl, not of NaCl. Even pure NaCl is only 60.66% Cl, so the purity is 53.18 ÷ 60.66 = 87.67% (when the impurity has no Cl).

Mistake 3: “If one element’s mass percent matches the calculated value, the sample must be pure.” Not necessarily. A mixture of about 51% KCl and 49% $\ce{MgCl2}$ is also 60.66% Cl, exactly like pure NaCl. One matching value is only consistent with a pure substance; to confirm it, measure another element (such as Na) or check that several samples agree.

Quick check

A sample labeled NaCl contains a single impurity. It is 62% Cl by mass. The impurity could be

A 2.00 g mixture of $\ce{CaCO3}$ and sand ( $\ce{SiO2}$ ) contains 0.400 g of Ca. The mass percent of $\ce{CaCO3}$ in the mixture is closest to

A sample of glucose ( $\ce{C6H12O6}$ ) has absorbed some water. How does this affect the measured mass percent of carbon?

Which data are enough to find the mole ratio of NaCl to KCl in a mixture of the two?

A method for measuring iron gives a reading proportional to the mass percent of Fe. Standards with 1.0%, 2.0%, and 3.0% Fe read 0.20, 0.40, and 0.60. An unknown reads 0.50. Its mass percent of Fe is closest to

AP-style free response

Three samples labeled “pure KCl” are analyzed. Their mass percents of Cl are 49.8%, 50.6%, and 48.9%.

(a) Calculate the mass percent of Cl in pure KCl.

(b) Are the samples pure KCl? Give two pieces of evidence from the data.

(c) A student suggests that the samples are contaminated with NaCl. Is this hypothesis consistent with the data? Justify your answer.

(d) Assuming NaCl is the only impurity, calculate the mass percent of NaCl in the first sample (49.8% Cl).

Model answer and scoring

(a) $M(\text{KCl}) = 39.10 + 35.45 = 74.55$ g/mol, so mass % Cl $= 35.45 \div 74.55 \times 100\% = 47.55\%$ .

(b) No. First, all three results are higher than the 47.55% for pure KCl. Second, the three samples disagree with each other, while a pure substance has a fixed composition (law of definite proportions). Both point to a mixture.

(c) Yes. NaCl has a higher mass percent of Cl (60.66%) than KCl, so mixing in NaCl raises the mass percent of Cl, matching the data. Different amounts of NaCl in different samples would explain why the values differ.

(d) Let $x$ be the mass fraction of NaCl:

(1 - x) \times 47.55\% + x \times 60.66\% = 49.8\%

x = \frac{49.8 - 47.55}{60.66 - 47.55} = \frac{2.2}{13.11} \approx 0.17

The sample is about 17% NaCl. Since 49.8% is known only to the tenths place, the difference 2.2 has just 2 significant figures, and so does the answer.

Scoring: (a) correct setup; (b) both pieces of evidence: disagreement with the calculated value and disagreement between samples; (c) compares the mass percent of Cl in the two compounds and states the direction (higher); (d) sets up the weighted-average equation and gets about 17%.

Summary

A pure substance contains one kind of particle and has a fixed composition that matches its formula; a mixture contains two or more kinds, and its composition can vary.
The mass percent of an element in a mixture is a weighted average of its components, so it lies between their values.
If the impurity lacks the measured element, purity = measured value ÷ pure value; if it contains the element, solve the weighted-average equation.
An impurity with a higher mass percent of the element pushes the result up; one with a lower mass percent (or none) pushes it down.
Gravimetric analysis: precipitate only the target ion, use excess reagent, tare, wash, and dry to constant mass. The moles of precipitate can also give the molar mass of an unknown salt.

AP tip: Expect a set of elemental-analysis data with questions such as: is the sample pure or a mixture, what impurity could explain the result, and how pure is it? Also expect a gravimetric experiment with questions about which data are needed, or whether a mistake makes the result too high or too low. When judging purity, use two pieces of evidence if you can: whether the value matches the calculated one, and whether different samples agree. Separating mixtures (chromatography, distillation) comes in U3.9, and the precipitation reactions used for analysis in U4.

Composition of Mixtures混合物的组成

Learning goals

Before you start

How pure is this salt?

Pure substances and mixtures, from particles to data

The mass percent in a mixture is a weighted average

Finding purity by elemental analysis

Example 1: Purity of NaCl by precipitation

Example 2: When the impurity contains Cl too

Example 3: Identifying an unknown salt by precipitation

Good gravimetric technique

Common mistakes

Quick check

AP-style free response

Summary

Feedback / Report a problem