Elemental Composition of Pure Substances - AP Chemistry

A pure compound has the formula $\text{Fe}_2\text{O}_3$ . What mass of iron is present in $16.0\ \text{g}$ of $\text{Fe}_2\text{O}_3$ ? (Atomic masses: Fe = 55.8 g/mol, O = 16.0 g/mol.)
Answer: 11.2 g This question tests the skill of elemental composition of pure substances. First, calculate the molar mass of Fe₂O₃: 2×Fe (2×55.8 = 111.6 g/mol) + 3×O (3×16.0 = 48.0 g/mol) = 159.6 g/mol total. Iron contributes 111.6 g/mol out of 159.6 g/mol, so iron's mass fraction is 111.6/159.6 = 0.699. In 16.0 g of Fe₂O₃, the mass of iron is 16.0 g × 0.699 = 11.2 g. A common error would be using the mass of only one iron atom (55.8 g/mol), which would give 5.6 g - this happens when students forget the subscript 2 in Fe₂. To find the mass of an element in a given mass of compound, first find the element's mass fraction, then multiply by the given mass.
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)
Answer: 5.88% 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.
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.)
Answer: 7.14 g 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.
A pure sample of hydrogen peroxide has the formula $\text{H}_2\text{O}_2$ . What mass of oxygen is contained in $34.0\ \text{g}$ of $\text{H}_2\text{O}_2$ ? (Atomic masses: H = 1.0 g/mol, O = 16.0 g/mol.)
Answer: 32.0 g This question tests the skill of elemental composition of pure substances. First, we calculate the molar mass of H₂O₂: 2×H (2×1.0 = 2.0 g/mol) + 2×O (2×16.0 = 32.0 g/mol) = 34.0 g/mol total. Since we have exactly 34.0 g of H₂O₂ (which is 1 mole), and oxygen contributes 32.0 g/mol to the compound, the mass of oxygen present is 32.0 g. A common error would be using the mass of one oxygen atom (16.0 g), forgetting that H₂O₂ contains two oxygen atoms per molecule. To find the mass of an element in a compound, use the ratio: (mass of element in formula/molar mass of compound) × given mass.
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.)
Answer: 7.2 g 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.
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?
Answer: K:O = 2:7 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.
A pure sample of sodium carbonate has the formula $\text{Na}_2\text{CO}_3$ . Which statement correctly describes the atom ratio of sodium to oxygen in $\text{Na}_2\text{CO}_3$ ?
Answer: 2:3 This question tests the skill of elemental composition of pure substances. The formula Na₂CO₃ contains 2 sodium atoms, 1 carbon atom, and 3 oxygen atoms per formula unit. The atom ratio of sodium to oxygen is therefore 2:3, which represents the actual count of atoms in the compound. This ratio is independent of the atomic masses and simply reflects the subscripts in the chemical formula. A common error would be reversing the ratio to 3:2, which happens when students confuse which element comes first in the ratio statement. To find atom ratios, simply count the atoms of each element directly from the chemical formula.
Pure potassium chlorate has the formula $\mathrm{KClO_3}$ . What is the mass percent of oxygen in $\mathrm{KClO_3}$ ? (Atomic masses: $\mathrm{K}=39.1$ , $\mathrm{Cl}=35.5$ , $\mathrm{O}=16.0\ \mathrm{g\ mol^{-1}}$ .)
Answer: $39.2\%$ This problem tests understanding of elemental composition of pure substances. The formula KClO₃ shows one potassium atom, one chlorine atom, and three oxygen atoms per formula unit. The molar mass is: K (39.1) + Cl (35.5) + 3×O (48.0) = 122.6 g/mol. Oxygen contributes 48.0 g per mole, so the mass percent of oxygen is: (48.0/122.6) × 100% = 39.2%. A common mistake would be calculating the atomic percent (3 atoms out of 5 = 60%) instead of mass percent, which requires considering the actual masses of atoms. For mass percent calculations, always use the formula: (mass of element/total molar mass) × 100%.
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)
Answer: 72.7% 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.
A sample of pure magnesium chloride, $\mathrm{MgCl_2}$ , has a mass of $9.5\ \mathrm{g}$ . What mass of chlorine is present in the sample? (Atomic masses: $\mathrm{Mg}=24.3$ , $\mathrm{Cl}=35.5\ \mathrm{g\ mol^{-1}}$ .)
Answer: $7.1\ \mathrm{g}$ This problem tests your understanding of elemental composition of pure substances. The chemical formula MgCl₂ tells us that each formula unit contains 1 magnesium atom and 2 chlorine atoms. To find the mass of chlorine in 9.5 g of MgCl₂, we first calculate the molar mass: Mg (24.3) + 2×Cl (2×35.5) = 95.3 g/mol. The mass fraction of chlorine is (2×35.5)/95.3 = 71.0/95.3 = 0.745. A common error would be using only one chlorine atom (35.5/95.3), which would give about half the correct answer. To find the mass of any element in a compound, multiply the total mass by the element's mass fraction: 9.5 g × 0.745 = 7.1 g.
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)
Answer: 0.297 mol 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.
Pure ammonia has the formula $\mathrm{NH_3}$ . What is the mass percent of hydrogen in $\mathrm{NH_3}$ ? (Atomic masses: $\mathrm{N}=14.0$ , $\mathrm{H}=1.0\ \mathrm{g\ mol^{-1}}$ .)
Answer: $17.6\%$ 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%.
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.)
Answer: 21.8 g 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.
A pure compound has formula $\mathrm{NH_4Cl}$ . Which statement correctly describes the ratio of hydrogen atoms to chlorine atoms in the compound?
Answer: 4:1 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.
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)
Answer: 4.80 g 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.
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)
Answer: 2.80 g 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.
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)
Answer: 43.7% 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.
A pure sample of calcium nitrate has the formula $\text{Ca(NO}_3)_2$ . How many moles of oxygen atoms are present in $0.50\ \text{mol}$ of $\text{Ca(NO}_3)_2$ ?
Answer: 3.0 mol This question tests the skill of elemental composition of pure substances. The formula Ca(NO₃)₂ shows one calcium atom and two nitrate groups, where each nitrate contains 3 oxygen atoms, giving 2×3 = 6 oxygen atoms per formula unit. Since we have 0.50 mol of Ca(NO₃)₂, and each formula unit contains 6 oxygen atoms, the total moles of oxygen atoms is 0.50 mol × 6 = 3.0 mol. A common error would be counting only 3 oxygen atoms (from one NO₃ group), which would give 1.5 mol - this happens when students miss the subscript 2 outside the parentheses. When working with polyatomic ions in parentheses, multiply the atoms inside by the subscript outside to get the total count.
A $6.0\ \mathrm{g}$ sample of pure water, $\mathrm{H_2O}$ , is collected. What mass of hydrogen is present? (Atomic masses: $\mathrm{H}=1.0$ , $\mathrm{O}=16.0\ \mathrm{g\ mol^{-1}}$ .)
Answer: $0.67\ \mathrm{g}$ This question requires calculating the elemental composition of pure substances. The formula H₂O shows that each water molecule contains 2 hydrogen atoms and 1 oxygen atom. The molar mass of water is: 2×H (2.0) + O (16.0) = 18.0 g/mol. Hydrogen contributes 2.0 g per mole, so its mass fraction is 2.0/18.0 = 1/9. In 6.0 g of water, the mass of hydrogen is: 6.0 g × (1/9) = 0.67 g. A common error would be assuming hydrogen makes up half the mass because there are 2 H atoms and 1 O atom, giving 3.0 g, but this ignores that oxygen is much heavier per atom. Always use actual atomic masses to calculate mass fractions, not just atom counts.
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 $\mathrm{Al_2(SO_4)_3}$ ?
Answer: $1.20\ \mathrm{mol\ O}$ 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.

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