AP Chemistry Quiz: Elemental Composition Of Pure Substances

20 questions · exam conditions

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Elemental Composition Of Pure SubstancesQuestion 1 of 20

A student has a 5.00 g sample of pure iron(III) oxide, $\mathrm{Fe_2O_3}$ . What mass of iron is present? (Atomic masses: Fe = 55.8 g/mol, O = 16.0 g/mol.)

7.00 g

5.00 g

1.50 g

3.50 g

2.00 g

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AP Chemistry Quiz

AP Chemistry Quiz: Elemental Composition Of Pure Substances

Practice Elemental Composition Of Pure Substances 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 Elemental Composition Of Pure Substances, 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

A student has a 5.00 g sample of pure iron(III) oxide, $\mathrm{Fe_2O_3}$ . What mass of iron is present? (Atomic masses: Fe = 55.8 g/mol, O = 16.0 g/mol.)

7.00 g
5.00 g
1.50 g
3.50 g (correct answer)
2.00 g

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula Fe₂O₃ indicates two iron and three oxygen atoms per formula unit. The molar mass is 2×55.8 + 3×16.0 = 159.6 g/mol, with iron contributing 111.6 g/mol. The proportion of iron is 111.6/159.6 ≈ 0.699, so in a 5.00 g sample, the mass of iron is 0.699 × 5.00 ≈ 3.50 g. A tempting distractor is 2.00 g, which could result from incorrectly dividing the sample mass by the number of iron atoms without using molar masses. To solve similar problems, use the formula to count atoms, then convert to mass contribution via the ratio of elemental to total molar mass multiplied by sample mass.

Question 2

A pure compound has the formula $\mathrm{K_2Cr_2O_7}$ . Which statement correctly describes the atom ratio of potassium to oxygen in one formula unit?

K:O = 2:7 (correct answer)
K:O = 2:1
K:O = 1:2
K:O = 7:2
K:O = 1:7

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula K₂Cr₂O₇ shows two potassium, two chromium, and seven oxygen atoms per formula unit. While molar masses would determine mass proportions, here the atom ratio is directly from the formula, giving K:O as 2:7 without needing masses for count-based composition. This ratio reflects the relative number of each atom type in the compound. A tempting distractor is 2:1, which might come from miscounting oxygen atoms or ignoring the subscript on O. To solve similar problems, use the formula to count atoms, then convert to mass contribution if needed, or directly use counts for ratios.

Question 3

A pure sample of $\mathrm{(NH_4)_2SO_4}$ has a mass of $13.2\ \mathrm{g}$ . What mass of nitrogen is present? (Atomic masses: N = 14.0 g/mol, H = 1.0 g/mol, S = 32.1 g/mol, O = 16.0 g/mol)

14.0 g
0.700 g
5.60 g
1.40 g
2.80 g (correct answer)

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula (NH4)2SO4 indicates two nitrogen, eight hydrogen, one sulfur, and four oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 14.0 g/mol for each N, 1.0 g/mol for each H, 32.1 g/mol for S, and 16.0 g/mol for each O, totaling 132.1 g/mol, with nitrogen contributing 28.0 g/mol. For a 13.2 g sample, the mass of nitrogen is (28.0 / 132.1) × 13.2 ≈ 2.80 g. A common error is counting only one NH4 group, leading to 1.40 g, which is incorrect because the formula has two NH4 groups. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the proportion.

Question 4

A 12.5 g sample of pure sodium bicarbonate, $\mathrm{NaHCO_3}$ , is measured. What mass of oxygen is present in the sample? (Atomic masses: Na = 23.0 g/mol, H = 1.0 g/mol, C = 12.0 g/mol, O = 16.0 g/mol.)

2.00 g
4.50 g
7.14 g (correct answer)
9.52 g
10.5 g

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula NaHCO₃ shows one sodium, one hydrogen, one carbon, and three oxygen atoms. The molar mass is 23.0 + 1.0 + 12.0 + 3×16.0 = 84.0 g/mol, with oxygen contributing 48.0 g/mol. The proportion of oxygen is 48.0/84.0 ≈ 0.571, so in a 12.5 g sample, the mass of oxygen is 0.571 × 12.5 ≈ 7.14 g. A tempting distractor is 4.0 g, which might arise from using only one oxygen atom's mass instead of three in the calculation. To solve similar problems, use the formula to count atoms, then convert to mass contribution by determining the elemental mass fraction and applying it to the sample mass.

Question 5

A 18.0 g sample of pure glucose has the formula $\mathrm{C_6H_{12}O_6}$ . What mass of carbon is in the sample? (Atomic masses: C = 12.0 g/mol, H = 1.0 g/mol, O = 16.0 g/mol.)

7.2 g (correct answer)
4.0 g
2.0 g
18.0 g
12.0 g

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula C₆H₁₂O₆ indicates six carbon, twelve hydrogen, and six oxygen atoms per molecule. The molar mass is 6×12.0 + 12×1.0 + 6×16.0 = 180.0 g/mol, with carbon contributing 72.0 g/mol. The proportion of carbon by mass is 72.0/180.0 = 0.40, so in an 18.0 g sample, the mass of carbon is 0.40 × 18.0 = 7.2 g. A tempting distractor is 12.0 g, which could result from mistakenly using only the atomic mass of carbon without accounting for the number of atoms or the total molar mass. To solve similar problems, use the formula to count atoms, then convert to mass contribution by finding the elemental fraction of the molar mass and multiplying by the sample mass.

Question 6

A 30.0 g sample of pure carbon dioxide, $\mathrm{CO_2}$ , is collected. What mass of oxygen is present in the sample? (Atomic masses: C = 12.0 g/mol, O = 16.0 g/mol.)

10.0 g
21.8 g (correct answer)
30.0 g
16.0 g
6.0 g

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula CO₂ indicates one carbon and two oxygen atoms per molecule. The molar mass is 12.0 + 2×16.0 = 44.0 g/mol, with oxygen contributing 32.0 g/mol. The proportion of oxygen is 32.0/44.0 ≈ 0.727, so in a 30.0 g sample, the mass of oxygen is 0.727 × 30.0 ≈ 21.8 g. A tempting distractor is 16.0 g, which might come from using only one oxygen atom's mass without accounting for two or the total molar mass. To solve similar problems, use the formula to count atoms, then convert to mass contribution by calculating the elemental fraction of the molar mass and multiplying by the sample mass.

Question 7

A sample is confirmed to be a pure compound with formula $\mathrm{NH_4Cl}$ . What is the mass percent of nitrogen in $\mathrm{NH_4Cl}$ ? (Atomic masses: N = 14.0 g/mol, H = 1.0 g/mol, Cl = 35.5 g/mol.)

37.3%
31.0%
14.0%
7.5%
26.2% (correct answer)

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula NH₄Cl shows one nitrogen, four hydrogen, and one chlorine atom per formula unit. Using atomic masses, the molar mass is 14.0 + 4×1.0 + 35.5 = 53.5 g/mol, with nitrogen contributing 14.0 g/mol. The mass percent of nitrogen is (14.0/53.5) × 100 ≈ 26.2%, determined by the ratio of nitrogen's mass to the total molar mass. A tempting distractor is 31.0%, which might come from incorrectly using chlorine's mass instead of nitrogen's in the percentage calculation. To solve similar problems, use the formula to count atoms, then convert to mass contribution by computing the percentage based on elemental and total molar masses.

Question 8

A pure sample of $\mathrm{Fe_2O_3}$ has a mass of $16.0\ \mathrm{g}$ . What mass of iron is in the sample? (Atomic masses: Fe = 55.8 g/mol, O = 16.0 g/mol)

11.2 g (correct answer)
5.60 g
8.00 g
2.80 g
16.0 g

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula Fe2O3 indicates two iron atoms and three oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 55.8 g/mol for each Fe and 16.0 g/mol for each O, totaling 159.6 g/mol, with iron contributing 111.6 g/mol. For a 16.0 g sample, the mass of iron is (111.6 / 159.6) × 16.0 ≈ 11.2 g. A common error is calculating for one Fe atom, yielding ≈5.60 g, which is incorrect because the formula has two Fe atoms. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the proportion.

Question 9

A pure compound has formula $\mathrm{H_2O_2}$ . What is the mass percent of hydrogen in $\mathrm{H_2O_2}$ ? (Atomic masses: H = 1.0 g/mol, O = 16.0 g/mol)

11.1%
5.88% (correct answer)
2.94%
50.0%
94.1%

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula H2O2 indicates two hydrogen atoms and two oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 1.0 g/mol for each H and 16.0 g/mol for each O, totaling 34.0 g/mol, with hydrogen contributing 2.0 g/mol. The mass percent of hydrogen is (2.0 / 34.0) × 100 ≈ 5.88%. A common error is calculating percent oxygen instead, yielding 94.1%, which is incorrect because the question asks for hydrogen. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the percentage.

Question 10

A pure sample of $\mathrm{Li_2CO_3}$ has a mass of $7.40 \mathrm{g}$ . What mass of oxygen is present? (Atomic masses: Li = 6.94 g/mol, C = 12.0 g/mol, O = 16.0 g/mol)

7.40 g
2.40 g
3.20 g
1.60 g
4.80 g (correct answer)

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula $\mathrm{Li_2CO_3}$ indicates two lithium, one carbon, and three oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 6.94 g/mol for each Li, 12.0 g/mol for C, and 16.0 g/mol for each O, totaling 73.88 g/mol, with oxygen contributing 48.0 g/mol. For a 7.40 g sample, the mass of oxygen is $(48.0 / 73.88) \times 7.40 \approx 4.80 \mathrm{g}$ . A common error is calculating for two O atoms, yielding $\approx 3.20 \mathrm{g}$ , which is incorrect because the formula has three O atoms. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the proportion.

Question 11

A pure sample of glucose, $\mathrm{C_6H_{12}O_6}$ , has a mass of $18.0 \mathrm{g}$ . What mass of carbon is in the sample? (Atomic masses: C = 12.0 g/mol, H = 1.0 g/mol, O = 16.0 g/mol)

7.20 g (correct answer)
1.00 g
12.0 g
3.00 g
6.00 g

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula $\mathrm{C_6H_{12}O_6}$ indicates six carbon, twelve hydrogen, and six oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: $12.0 \, \mathrm{g/mol}$ for each C, $1.0 \, \mathrm{g/mol}$ for each H, and $16.0 \, \mathrm{g/mol}$ for each O, totaling $180.0 \, \mathrm{g/mol}$ , with carbon contributing $72.0 \, \mathrm{g/mol}$ . For an $18.0 \, \mathrm{g}$ sample, the mass of carbon is $(72.0 / 180.0) \times 18.0 = 7.20 \, \mathrm{g}$ . A common error is miscounting carbon atoms as five, leading to $6.00 \, \mathrm{g}$ , which is incorrect because the formula has six C atoms. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the proportion.

Question 12

Pure ammonia has the formula $\mathrm{NH_3}$ . What is the mass percent of hydrogen in $\mathrm{NH_3}$ ? (Atomic masses: $\mathrm{N}=14.0$ , $\mathrm{H}=1.0\ \mathrm{g\ mol^{-1}}$ .)

$5.9\%$
$17.6\%$ (correct answer)
$21.4\%$
$82.4\%$
$94.1\%$

Explanation: This question requires calculating the elemental composition of pure substances. The formula NH₃ shows that each molecule contains 1 nitrogen atom and 3 hydrogen atoms. The molar mass of NH₃ is: N (14.0) + 3×H (3×1.0) = 17.0 g/mol. The mass contributed by hydrogen is 3.0 g per mole, so the mass percent of hydrogen is (3.0/17.0) × 100% = 17.6%. A tempting error would be to calculate the percent by counting atoms (3 H out of 4 total atoms = 75%), but mass percent requires using actual masses, not atom counts. Always calculate mass percent by dividing the element's total mass contribution by the compound's molar mass, then multiply by 100%.

Question 13

A pure compound has formula $\mathrm{NH_4Cl}$ . Which statement correctly describes the ratio of hydrogen atoms to chlorine atoms in the compound?

4:1 (correct answer)
2:1
4:2
1:4
1:1

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula NH4Cl indicates four hydrogen atoms and one chlorine atom per formula unit. While molar masses determine mass proportions, for atomic ratios, the formula directly gives the counts: four H to one Cl. Thus, the ratio of hydrogen atoms to chlorine atoms is 4:1. A tempting distractor is 1:4, which reverses the ratio, but that's incorrect because the question specifies hydrogen to chlorine. Always use the formula to count atoms of each element, then express the ratio as requested.

Question 14

A 0.500 mol sample of pure aluminum sulfate, $\mathrm{Al_2(SO_4)_3}$ , is used in a laboratory procedure. How many moles of sulfur atoms are present in the sample?

0.167 mol S atoms
1.50 mol S atoms (correct answer)
0.500 mol S atoms
3.00 mol S atoms
6.00 mol S atoms

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula Al₂(SO₄)₃ indicates two aluminum, three sulfur, and twelve oxygen atoms per formula unit. Although molar masses are not directly needed here since the sample is given in moles, they would be used in mass-based problems to find proportions; here, the atomic count directly gives 3 sulfur atoms per formula unit. Thus, in 0.500 mol of the compound, the moles of sulfur atoms are 0.500 × 3 = 1.50 mol. A tempting distractor is 3.00 mol, which could result from confusing the subscript for sulfur with the total number without scaling by sample moles. To solve similar problems, use the formula to count atoms, then convert to mass contribution or moles by multiplying by the number of atoms per unit and the sample quantity.

Question 15

A pure compound has formula $\mathrm{CO_2}$ . What is the mass percent of oxygen in $\mathrm{CO_2}$ ? (Atomic masses: C = 12.0 g/mol, O = 16.0 g/mol)

72.7% (correct answer)
27.3%
50.0%
33.3%
66.7%

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula CO2 indicates one carbon atom and two oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 12.0 g/mol for C and 16.0 g/mol for each O, totaling 44.0 g/mol, with oxygen contributing 32.0 g/mol. The mass percent of oxygen is (32.0 / 44.0) × 100 ≈ 72.7%. A common error is reversing to percent carbon, yielding 27.3%, which is incorrect because the question asks for oxygen. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the percentage.

Question 16

A pure compound has formula $\mathrm{P_2O_5}$ . What is the mass percent of phosphorus in the compound? (Atomic masses: P = 31.0 g/mol, O = 16.0 g/mol)

43.7% (correct answer)
56.3%
28.0%
72.0%
20.0%

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula P2O5 indicates two phosphorus atoms and five oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 31.0 g/mol for each P and 16.0 g/mol for each O, totaling 142.0 g/mol, with phosphorus contributing 62.0 g/mol. The mass percent of phosphorus is (62.0 / 142.0) × 100 ≈ 43.7%. A common error is calculating for one P atom, yielding ≈27.9%, which is incorrect because the formula has two P atoms. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the percentage.

Question 17

A pure sample of $\mathrm{KNO_3}$ has a mass of $10.0\ \mathrm{g}$ . How many moles of oxygen atoms are present? (Atomic masses: K = 39.1 g/mol, N = 14.0 g/mol, O = 16.0 g/mol)

0.025 mol
0.297 mol (correct answer)
0.099 mol
0.792 mol
0.396 mol

Explanation: This question assesses the skill of elemental composition of pure substances. The chemical formula KNO3 indicates one potassium, one nitrogen, and three oxygen atoms per formula unit. To determine the proportion, calculate the molar mass using atomic masses: 39.1 g/mol for K, 14.0 g/mol for N, and 16.0 g/mol for each O, totaling 101.1 g/mol. For a 10.0 g sample, the moles of compound are 10.0 / 101.1 ≈ 0.099 mol, and since there are three oxygen atoms per unit, moles of oxygen = 3 × 0.099 ≈ 0.297 mol. A common error is using only one oxygen atom, leading to ≈0.099 mol, which is incorrect because the formula has three O atoms. Always use the formula to count the number of each type of atom, then convert to mass contribution or moles as needed.

Question 18

A student has a pure sample of aluminum sulfate, $\mathrm{Al_2(SO_4)_3}$ . How many moles of oxygen atoms are present in $0.10\ \mathrm{mol}$ of $\mathrm{Al_2(SO_4)_3}$ ?

$0.70\ \mathrm{mol\ O}$
$0.40\ \mathrm{mol\ O}$
$1.20\ \mathrm{mol\ O}$ (correct answer)
$0.30\ \mathrm{mol\ O}$
$0.80\ \mathrm{mol\ O}$

Explanation: This question tests understanding of elemental composition of pure substances. The formula Al₂(SO₄)₃ shows 2 aluminum atoms, 3 sulfur atoms, and 12 oxygen atoms (since each SO₄ has 4 oxygen atoms and there are 3 SO₄ groups). In 0.10 mol of Al₂(SO₄)₃, we need to find moles of oxygen atoms. A common error would be counting only 4 oxygen atoms instead of 12, or forgetting to multiply by the number of moles of compound. Since each formula unit contains 12 oxygen atoms: 0.10 mol Al₂(SO₄)₃ × 12 mol O/mol Al₂(SO₄)₃ = 1.20 mol O.

Question 19

A student analyzes a pure compound with formula $\mathrm{Fe_2O_3}$ . Which statement correctly describes the mass contribution of oxygen compared with iron in one mole of the compound? (Atomic masses: $\mathrm{Fe}=55.8$ , $\mathrm{O}=16.0\ \mathrm{g\ mol^{-1}}$ .)

Oxygen contributes more mass than iron.
Oxygen and iron contribute equal mass.
Iron contributes more mass than oxygen. (correct answer)
Oxygen contributes $3\times$ the mass of iron.
Iron contributes $2\times$ the mass of oxygen.

Explanation: This question tests understanding of elemental composition of pure substances. The formula Fe₂O₃ contains 2 iron atoms and 3 oxygen atoms per formula unit. To compare mass contributions, calculate the total mass from each element: iron contributes 2 × 55.8 = 111.6 g/mol, while oxygen contributes 3 × 16.0 = 48.0 g/mol. A common error would be to compare atom counts (2 vs 3) rather than actual mass contributions. Since 111.6 > 48.0, iron contributes more mass than oxygen in the compound. The strategy is to multiply the number of atoms by their atomic masses to find actual mass contributions.

Question 20

A pure compound has the formula $\mathrm{Ca_3(PO_4)_2}$ . What is the mole ratio of calcium atoms to oxygen atoms in the compound?

$3:4$
$1:4$
$3:8$ (correct answer)
$3:12$
$3:2$

Explanation: This question tests understanding of elemental composition of pure substances. The formula Ca₃(PO₄)₂ shows 3 calcium atoms and 2 phosphate groups, where each PO₄ contains 4 oxygen atoms. Therefore, the total count is 3 Ca atoms and 2 × 4 = 8 O atoms per formula unit. The mole ratio is simply the ratio of atoms in the formula, not their masses. A common error would be to count only 4 oxygen atoms instead of 8, or to use mass ratios instead of mole ratios. The mole ratio of calcium to oxygen is 3:8.