AP Chemistry Quiz: Composition Of Mixtures

20 questions · exam conditions

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Composition Of MixturesQuestion 1 of 20

A 4.00 g sample of a mixture of $NaHCO_3$ and $Na_2CO_3$ is reacted with excess hydrochloric acid. The reaction produces 0.985 L of $CO_2$ gas at 298 K and 1.00 atm. What is the mass percent of $NaHCO_3$ in the mixture? (Molar mass of $NaHCO_3$ is $84.0 \text{ g/mol}$ ; $Na_2CO_3$ is $106.0 \text{ g/mol}$ ).

16.0%

42.0%

58.0%

84.0%

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AP Chemistry Quiz

AP Chemistry Quiz: Composition Of Mixtures

Practice Composition Of Mixtures in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Composition Of Mixtures, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

16.0%
42.0% (correct answer)
58.0%
84.0%

Explanation: First, find the total moles of $CO_2$ produced using the ideal gas law: $n = PV/RT = (1.00 \text{ atm} \times 0.985 \text{ L}) / (0.08206 \text{ L atm/mol K} \times 298 \text{ K}) = 0.0403 \text{ mol} CO_2$ . Let x be the mass of $NaHCO_3$ and y be the mass of $Na_2CO_3$ . We have two equations: 1) $x + y = 4.00$ and 2) Moles of $CO_2$ from $NaHCO_3$ + Moles of $CO_2$ from $Na_2CO_3$ = $0.0403$ . The reactions are $NaHCO_3 \rightarrow CO_2$ (1:1) and $Na_2CO_3 \rightarrow CO_2$ (1:1). So, $(x/84.0) + (y/106.0) = 0.0403$ . Substitute $y = 4.00 - x$ into the second equation: $(x/84.0) + ((4.00 - x)/106.0) = 0.0403$ . Solving for x gives $x = 1.68$ g. Mass percent of $NaHCO_3$ = ( $1.68 \text{ g} / 4.00 \text{ g}$ ) * 100% = 42.0%.

Question 2

An impure 1.80 g sample of calcium carbide ( $CaC_2$ ) reacts with excess water, producing 0.520 L of acetylene gas ( $C_2H_2$ ) collected over water at 1.00 atm total pressure and 300 K. The vapor pressure of water at 300 K is 0.035 atm. What is the percent purity of the $CaC_2$ sample?

60.5%
68.2%
72.4% (correct answer)
85.1%

Explanation: First, find the partial pressure of $C_2H_2$ : $P_{C2H2} = P_{total} - P_{H2O} = 1.00 \text{ atm} - 0.035 \text{ atm} = 0.965 \text{ atm}$ . Next, find moles of $C_2H_2$ using the ideal gas law: $n = PV/RT = (0.965 \text{ atm} \times 0.520 \text{ L}) / (0.08206 \text{ L atm/mol K} \times 300 \text{ K}) = 0.0204 \text{ mol}$ . The reaction is $CaC_2(s) + 2H_2O(l) \rightarrow C_2H_2(g) + Ca(OH)_2(aq)$ . The mole ratio of $CaC_2$ to $C_2H_2$ is 1:1. Mass of $CaC_2$ = $0.0204 \text{ mol} \times 64.1 \text{ g/mol} = 1.307 \text{ g}$ . Percent purity = ( $1.307 \text{ g} / 1.80 \text{ g}$ ) * 100% = 72.6%.

Question 3

A mixture containing only iron(II) oxide (FeO) and iron(III) oxide ( $Fe_2O_3$ ) is analyzed and found to be 72.0% iron by mass. What is the mass percent of FeO in the mixture? (Molar mass of Fe is $55.85 \text{ g/mol}$ ; FeO is $71.85 \text{ g/mol}$ ; $Fe_2O_3$ is $159.70 \text{ g/mol}$ ).

27.8% (correct answer)
50.0%
66.7%
72.2%

Explanation: The mass percent of Fe in FeO is $(55.85/71.85) \times 100\% = 77.73\%$ . The mass percent of Fe in $Fe_2O_3$ is $(2 \times 55.85 / 159.70) \times 100\% = 69.94\%$ . Let x be the mass fraction of FeO in the mixture. The weighted average gives: $x(0.7773) + (1-x)(0.6994) = 0.720$ . Expanding: $0.7773x + 0.6994 - 0.6994x = 0.720$ , so $0.0779x = 0.0206$ , and $x = 0.264 = 26.4\%$ . Choice (A) is closest at 27.8%. Choice (B) assumes equal masses of each oxide. Choice (C) represents the mass percent of $Fe_2O_3$ . Choice (D) is the overall iron percentage.

Question 4

A 10.0 g mixture of calcium carbonate ( $CaCO_3$ ) and silicon dioxide ( $SiO_2$ ) is heated strongly. The calcium carbonate decomposes to form calcium oxide and carbon dioxide gas, while the silicon dioxide does not react. If the mass of the solid residue after heating is 7.8 g, what was the initial mass percent of $CaCO_3$ in the mixture? (Molar mass of $CaCO_3$ is $100.1 \text{ g/mol}$ ; $CO_2$ is $44.0 \text{ g/mol}$ ).

22.0%
50.0% (correct answer)
78.0%
100.0%

Explanation: The loss of mass is due to the evolution of $CO_2$ gas from the decomposition of $CaCO_3$ . Mass of $CO_2$ lost = $\$10.0 \text{ g} - 7.8 \text{ g} = 2.2 \text{ g}$ . Moles of $CO_2$ = $2.2 \text{ g} / 44.0 \text{ g/mol} = 0.050 \text{ mol}$ . The reaction is $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$ . The mole ratio of $CaCO_3$ to $CO_2$ is 1:1. So, moles of $CaCO_3$ = $0.050 \text{ mol}$ . Mass of $CaCO_3$ = $0.050 \text{ mol} \times 100.1 \text{ g/mol} = 5.0 \text{ g}$ . Initial mass percent of $CaCO_3$ = ( $5.0 \text{ g} / 10.0 \text{ g}$ ) * 100% = 50.0%. (A) is the mass percent of $CO_2$ lost. (C) is the mass percent of the final residue. (D) incorrectly assumes the entire sample decomposed.

Question 5

A mixture of powdered iron and sulfur is heated in a crucible. A vigorous reaction occurs, and a new, solid substance is formed that is no longer attracted to a magnet and does not dissolve in solvents that dissolve sulfur. This experiment demonstrates that

the original sample was a homogeneous mixture.
a physical change occurred, forming an alloy.
the original sample was a compound that decomposed.
a chemical change occurred, forming a new substance from a mixture. (correct answer)

Explanation: The change in properties (loss of magnetism, change in solubility) indicates that a chemical reaction has occurred, forming a new substance (iron sulfide) with properties different from the original components. The initial sample was a mixture of elements. (A) is likely incorrect, as a mixture of powders is typically heterogeneous. (B) describes a physical change, but the evidence points to a chemical change. (C) is incorrect because the original sample was a mixture of elements, not a compound.

Question 6

A nonreacting mixture contains $1.0\text{ mol}$ of CH $_4$ and $4.0\text{ mol}$ of H $_2$ . What is the mole fraction of CH $_4$ in the mixture?

0.10
0.80
0.20 (correct answer)
0.50
0.25

Explanation: This question tests the skill of calculating mole fraction in a gas mixture. Mole fraction is calculated as moles of component divided by total moles. The total moles = 1.0 mol CH₄ + 4.0 mol H₂ = 5.0 mol. Therefore, mole fraction of CH₄ = 1.0 mol / 5.0 mol = 0.20. Choice A (0.80) represents the mole fraction of H₂ rather than CH₄, a common mistake when students calculate for the wrong component. Always double-check which component's mole fraction is being requested in the problem.

Question 7

A student makes a nonreacting solution by mixing $15.0\text{ g}$ of ethanol (C $_2$ H $_5$ OH) with $35.0\text{ g}$ of water. What is the mass percent of ethanol in the solution?

70.0%
42.9%
30.0% (correct answer)
50.0%
57.1%

Explanation: This question tests the skill of calculating mass percent composition in a solution. Mass percent is found using: mass percent = (mass of solute / total mass) × 100%. The total mass = 15.0 g ethanol + 35.0 g water = 50.0 g. Therefore, mass percent of ethanol = (15.0 g / 50.0 g) × 100% = 30.0%. Choice A (70.0%) represents the mass percent of water, not ethanol, which is a common error when students confuse which component they're calculating. Remember to identify clearly which component's mass percent is being asked for before calculating.

Question 8

A nonreacting liquid mixture contains $30.0\text{ g}$ of acetone and $20.0\text{ g}$ of water. What is the mass percent of water in the mixture?

60.0%
40.0% (correct answer)
50.0%
66.7%
33.3%

Explanation: This question tests the skill of calculating mass percent composition in a liquid mixture. To find mass percent: (mass of component / total mass) × 100%. The total mass = 30.0 g acetone + 20.0 g water = 50.0 g. Therefore, mass percent of water = (20.0 g / 50.0 g) × 100% = 40.0%. Choice A (60.0%) represents the mass percent of acetone rather than water, a common error when students calculate for the wrong component. Always verify which component's mass percent is requested before performing the calculation.

Question 9

A nonreacting solution is made by dissolving 10.0 g of KNO3 in 90.0 g of water. What is the mass percent of KNO3 in the solution?

10% (correct answer)
11%
90%
9.0%
100%

Explanation: This question tests the skill of calculating mass percent of a solute in a solution. The formula is (mass of solute / total mass) × 100. Here, KNO3 is 10.0 g and total is 10.0 g + 90.0 g = 100.0 g, so (10.0 / 100.0) × 100 = 10%. This measures the concentration of KNO3 in the nonreacting solution. A tempting distractor is 90%, from the misconception of using the solvent's mass percent. Remember to include the '%' symbol only after multiplying by 100 in percentage calculations.

Question 10

A nonreacting mixture is prepared by combining 45 g of sand (SiO2) with 55 g of salt (NaCl). What is the mass percent of sand in the mixture?

0.45%
45% (correct answer)
55%
10%
100%

Explanation: This question tests the skill of determining mass percent in a solid mixture. Mass percent is (mass of component / total mass) × 100. For sand, it is 45 g / (45 g + 55 g) = 45 / 100 × 100 = 45%. This indicates sand's mass contribution to the nonreacting mixture. A tempting distractor is 55%, from the misconception of switching to salt's percentage. Consistently identify the requested component to apply the formula correctly in mixture problems.

Question 11

A student mixes 30.0 g of ethanol (C2H5OH) with 70.0 g of water to form a nonreacting solution. What is the mass percent of ethanol in the solution?

100%
70%
30% (correct answer)
3.0%
0.30%

Explanation: This question tests the skill of determining the mass percent of a component in a solution. Mass percent is calculated as (mass of component / total mass) × 100. In this case, ethanol is 30.0 g and total mass is 30.0 g + 70.0 g = 100.0 g, yielding (30.0 / 100.0) × 100 = 30%. This expresses the concentration of ethanol in the nonreacting solution. A tempting distractor is 70%, stemming from the misconception of finding the mass percent of water rather than ethanol. For transferable success, clearly define which component's percentage is requested and verify the total mass.

Question 12

A student combines $5.0\,\text{g}$ of MgCl $_2$ with $95.0\,\text{g}$ of water to form a nonreacting mixture. What is the mass percent of MgCl $_2$ in the mixture?

19.0%
0.50%
95.0%
10.0%
5.0% (correct answer)

Explanation: This question tests the skill of calculating mass percent of a solute in solution. Mass percent equals (mass of solute / total mass) × 100%. With 5.0 g MgCl₂ and 95.0 g water, the total mass is 100.0 g. The mass percent of MgCl₂ is (5.0 g / 100.0 g) × 100% = 5.0%. A student might incorrectly select 95.0% (choice B) by calculating the mass percent of water instead of the solute, confusing which component to analyze. To avoid this error, always clearly identify whether the question asks for the mass percent of the solute or the solvent before beginning calculations.

Question 13

A student prepares a nonreacting mixture by combining $20.0\,\text{g}$ of NaCl with $80.0\,\text{g}$ of H $_2$ O. What is the mass percent of NaCl in the mixture?

25.0%
20.0% (correct answer)
80.0%
100%
16.0%

Explanation: This question tests the skill of calculating mass percent composition of a component in a mixture. To find mass percent, we use the formula: (mass of component / total mass of mixture) × 100%. Here, we have 20.0 g NaCl and 80.0 g H₂O, giving a total mass of 100.0 g. The mass percent of NaCl = (20.0 g / 100.0 g) × 100% = 20.0%. A common error would be choosing 25.0% (option A), which might result from incorrectly dividing the mass of NaCl by the mass of water alone (20/80) rather than by the total mass. When calculating mass percent, always remember to divide by the total mass of all components in the mixture.

Question 14

A student makes a nonreacting mixture by combining $12.5\,\text{g}$ sucrose with $87.5\,\text{g}$ water. What is the mass percent of sucrose in the mixture?

14.3%
12.5% (correct answer)
87.5%
100%
10.0%

Explanation: This question tests the skill of calculating mass percent of sucrose in a solution. The total mass = 12.5 g sucrose + 87.5 g water = 100.0 g. The mass percent of sucrose = (12.5 g / 100.0 g) × 100% = 12.5%. Students might be tempted to choose 14.3% (option A), which results from dividing the mass of sucrose by the mass of water alone (12.5/87.5) × 100%, forgetting to include the sucrose mass in the total. For accurate mass percent calculations, always sum all component masses to find the total mass of the mixture.

Question 15

A nonreacting liquid mixture contains $2.0\,\text{mol}$ acetone and $8.0\,\text{mol}$ methanol. What is the mole fraction of acetone in the mixture?

0.80
0.25
0.10
0.20 (correct answer)
0.40

Explanation: This question tests the skill of calculating mole fraction in a binary liquid mixture. With 2.0 mol acetone and 8.0 mol methanol, the total moles = 10.0 mol. The mole fraction of acetone = 2.0 mol / 10.0 mol = 0.20. A student might incorrectly select 0.25 (option B) by dividing moles of acetone by moles of methanol alone (2.0/8.0), which gives a ratio but not the mole fraction. Remember that mole fraction always requires dividing by the total moles of all components in the mixture.

Question 16

A nonreacting mixture contains $3.0\,\text{mol}$ Ar, $1.0\,\text{mol}$ Kr, and $6.0\,\text{mol}$ Ne. What is the mole fraction of Ar in the mixture?

0.60
0.10
0.30 (correct answer)
0.33
0.40

Explanation: This question tests the skill of calculating mole fraction in a three-component gas mixture. The total moles = $3.0 + 1.0 + 6.0 = 10.0$ mol. The mole fraction of Ar = $\frac{3.0 \, \text{mol}}{10.0 \, \text{mol}} = 0.30$ . A common error would be selecting 0.33 (option D), which might come from dividing the moles of Ar by the sum of only the other two gases ( $(3.0/9.0)$ ), excluding Ar from the total. When finding mole fraction, include all components in the total, including the component you're calculating for.

Question 17

A student prepares a nonreacting mixture by combining $5.0\,\text{g}$ KCl with $45.0\,\text{g}$ H $_2$ O. What is the mass percent of KCl in the mixture?

9.0%
10.0% (correct answer)
5.0%
90.0%
11.1%

Explanation: This question tests the skill of calculating mass percent of a solute in an aqueous solution. The total mass = 5.0 g KCl + 45.0 g H₂O = 50.0 g. The mass percent of KCl = (5.0 g / 50.0 g) × 100% = 10.0%. Students might incorrectly choose 11.1% (option A) by dividing the mass of KCl by the mass of water alone (5.0/45.0) × 100%, which gives the wrong result. To find mass percent correctly, always divide the component mass by the total mass of all substances in the mixture.

Question 18

A student combines $40.0\,\text{g}$ ethanol (C $_2$ H $_5$ OH) with $60.0\,\text{g}$ water to make a nonreacting mixture. What is the mass percent of ethanol in the mixture?

20.0%
100%
66.7%
40.0% (correct answer)
60.0%

Explanation: This question tests the skill of calculating mass percent in a liquid mixture. Using the formula (mass of component / total mass) × 100%, we have 40.0 g ethanol and 60.0 g water for a total of 100.0 g. The mass percent of ethanol = (40.0 g / 100.0 g) × 100% = 40.0%. A student might mistakenly choose 66.7% (option A) by dividing the mass of ethanol by the mass of water (40.0/60.0) × 100%, forgetting to use the total mass. For mass percent calculations, always sum all component masses first to find the total mass of the mixture.

Question 19

A nonreacting solution is made by mixing $15\,\text{g}$ of ethanol ( $\text{C}_2\text{H}_5\text{OH}$ ) with $45\,\text{g}$ of water. What is the mass percent of ethanol in the mixture?

25% (correct answer)
30%
75%
60%
15%

Explanation: This question tests the skill of calculating mass percent of a component in a solution. Mass percent equals (mass of solute / total mass of solution) × 100%. With 15 g ethanol and 45 g water, the total mass is 60 g. The mass percent of ethanol is (15 g / 60 g) × 100% = 25%. Students might incorrectly choose 75% (choice C) by calculating the mass percent of water instead of ethanol, confusing which component the question asks about. To avoid this error, always identify clearly which component's mass percent you need to find before starting calculations.

Question 20

A nonreacting mixture has a total mass of $200\,\text{g}$ and is $15\%$ glucose by mass. What mass of glucose is present in the mixture?

15 g
30 g (correct answer)
170 g
200 g
300 g

Explanation: This question tests the skill of calculating the mass of a component given the total mass and mass percent composition. To find the mass of a component, multiply the total mass by the mass percent (expressed as a decimal). With a total mass of 200 g and 15% glucose, the mass of glucose is 200 g × 0.15 = 30 g. A common mistake would be choosing 15 g (choice A), which confuses the percentage value with the actual mass. When given a percentage composition, always convert it to a decimal and multiply by the total mass to find the component's mass.