Composition of Mixtures - AP Chemistry

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?
Answer: 5.0% 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.
A sample of brass, a mixture of copper and zinc, has a mass of 5.00 g. When the sample is placed in excess hydrochloric acid, only the zinc reacts, producing hydrogen gas according to the equation $\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$ . If 0.820 L of dry $\text{H}_2$ gas is collected at standard temperature and pressure (STP), what is the mass percent of copper in the brass sample?
Answer: 52.0% At STP, 1 mole of an ideal gas occupies 22.4 L. Moles of $H_2$ = $0.820 \text{ L} / 22.4 \text{ L/mol} = 0.0366 \text{ mol}$ . From the balanced equation, the mole ratio of Zn to $H_2$ is 1:1. So, moles of Zn = $0.0366 \text{ mol}$ . Mass of Zn = $0.0366 \text{ mol} \times 65.38 \text{ g/mol} = 2.40 \text{ g}$ . Mass of Cu = Total mass - Mass of Zn = $5.00 \text{ g} - 2.40 \text{ g} = 2.60 \text{ g}$ . Mass percent of Cu = $(2.60 \text{ g} / 5.00 \text{ g}) \times 100\% = 52.0\%$ . Choice (A) represents the mass percent of zinc in the sample.
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?
Answer: 40.0% 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.
A nonreacting gas mixture contains $2.0\,\text{mol}$ of He and $8.0\,\text{mol}$ of Ne. What is the mole fraction of He in the mixture?
Answer: 0.20 This question tests the skill of calculating mole fraction in a gas mixture. Mole fraction is calculated as: moles of component / total moles in mixture. The mixture contains 2.0 mol He and 8.0 mol Ne, giving a total of 10.0 mol. Therefore, mole fraction of He = 2.0 mol / 10.0 mol = 0.20. Students who selected 0.80 (choice B) calculated the mole fraction of Ne instead of He, confusing which component was being asked for. When finding mole fractions, always double-check which component the question is asking about and ensure you're dividing by the total moles of all components.
A nonreacting gas mixture contains 2.0 mol of He and 6.0 mol of Ne. What is the mole fraction of He, $X_{\mathrm{He}}$ , in the mixture?
Answer: 0.25 This question tests the skill of calculating the mole fraction of a gas in a nonreacting mixture. The mole fraction is the ratio of the moles of one component to the total moles in the mixture. For He, it is 2.0 mol / (2.0 mol + 6.0 mol) = 2.0 / 8.0 = 0.25. This value indicates the fractional contribution of He to the total gas molecules present. A tempting distractor is 0.75, which results from the misconception of calculating the mole fraction of Ne instead of He. When approaching mole fraction problems, sum the moles of all components first to establish the denominator before dividing.
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?
Answer: 45% 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.
A student needs to determine the mass percent of $Na_2CO_3$ in a solid mixture containing $Na_2CO_3$ and inert, non-reactive NaCl. The student adds the mixture to an excess of HCl solution. Which of the following measurements, in addition to the initial mass of the mixture, is essential to calculate the mass percent of $Na_2CO_3$ ?
Answer: The total mass of $CO_2$ gas produced by the reaction. $Na_2CO_3$ reacts with HCl to produce $CO_2$ gas ( $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$ ), while NaCl does not react. By measuring the mass of $CO_2$ produced, one can use stoichiometry to calculate the mass of $Na_2CO_3$ that reacted, and thus its percentage in the original mixture. (A), (B), and (C) are not directly useful for a stoichiometric calculation of composition.
An impure sample of silver nitrate, $AgNO_3$ , has a mass of 1.50 g. The sample is dissolved in water, and an excess of sodium chloride solution is added. A precipitate of silver chloride, AgCl, forms and is collected, dried, and weighed. If the mass of the AgCl precipitate is 1.25 g, what is the mass percent of $AgNO_3$ in the original impure sample? (Molar mass of $AgNO_3$ is $169.9 \text{ g/mol}$ ; molar mass of AgCl is $143.4 \text{ g/mol}$ ).
Answer: 98.5% First, calculate the moles of AgCl precipitated: $1.25 \text{ g AgCl} / 143.4 \text{ g/mol} = 0.00872 \text{ mol AgCl}$ . According to the reaction $AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$ , the mole ratio of $AgNO_3$ to AgCl is 1:1. Therefore, the moles of $AgNO_3$ in the sample are $0.00872 \text{ mol}$ . Next, calculate the mass of $AgNO_3$ : $0.00872 \text{ mol} \times 169.9 \text{ g/mol} = 1.48 \text{ g} AgNO_3$ . Finally, calculate the mass percent: ( $1.48 \text{ g} / 1.50 \text{ g}$ ) * 100% = 98.7%. Choice C is closest. (A) is incorrect, possibly from a molar mass error. (B) is the mass percent of precipitate in the original sample mass. (D) is incorrect and physically impossible, likely arising from an inverted molar mass ratio.
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?
Answer: 0.20 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.
A nonreacting mixture contains $3.0\,\text{mol}$ of CO $_2$ and $1.0\,\text{mol}$ of CH $_4$ . What is the mole fraction of CO $_2$ in the mixture?
Answer: 0.75 This question tests the skill of calculating mole fraction in a gas mixture. Mole fraction is found by dividing moles of the component by total moles. The mixture has 3.0 mol CO₂ and 1.0 mol CH₄, totaling 4.0 mol. Therefore, mole fraction of CO₂ = 3.0 mol / 4.0 mol = 0.75. Students who selected 0.25 (choice A) calculated the mole fraction of CH₄ instead, mixing up the gases. When working with mole fractions, create a clear table showing each component and the total to organize your calculation.
A 3.25 g sample of hydrated sodium sulfate, $Na_2SO_4 \cdot nH_2O$ , is heated until all the water of hydration is driven off. The remaining anhydrous $Na_2SO_4$ has a mass of 1.43 g. Based on this data, what is the mass percent of water in the original hydrated salt mixture?
Answer: 56.0% The mass of the water driven off is the difference between the initial mass of the hydrate and the final mass of the anhydrous salt. Mass of $H_2O$ = $3.25 \text{ g} - 1.43 \text{ g} = 1.82 \text{ g}$ . The mass percent of water is calculated by dividing the mass of water by the total mass of the hydrate and multiplying by 100: ( $1.82 \text{ g} / 3.25 \text{ g}$ ) * 100% = 56.0%. (A) represents the mass percent of the anhydrous salt in the hydrate. (C) and (D) are distractors based on incorrect calculations.
A student prepares a nonreacting alloy by mixing $60.0\,\text{g}$ of Cu and $40.0\,\text{g}$ of Zn. What is the mass percent of Zn in the alloy?
Answer: 40.0% This question tests the skill of calculating mass percent in an alloy mixture. Mass percent is calculated as: (mass of component / total mass) × 100%. The alloy contains 60.0 g Cu and 40.0 g Zn, with a total mass of 100.0 g. Therefore, mass percent of Zn = (40.0 g / 100.0 g) × 100% = 40.0%. Students who selected 60.0% (choice A) found the mass percent of Cu instead of Zn, confusing the two components. Always identify which component's mass percent is requested and use that component's mass in the numerator.
A nonreacting mixture has a total mass of $200.\,\text{g}$ and is $15.0\%$ by mass NaNO $_3$ . What mass of NaNO $_3$ is in the mixture?
Answer: 30.0 g This question tests the skill of calculating mass of a component from mass percent. When given mass percent and total mass, multiply them to find component mass: mass = (mass percent / 100%) × total mass. With 200 g total and 15.0% NaNO₃, the mass of NaNO₃ = (15.0/100) × 200 g = 30.0 g. Students who selected 15.0 g (choice A) forgot to convert percent to decimal or multiply by total mass, treating the percentage as the actual mass. Always remember that percent means "per hundred" and must be converted when calculating actual quantities.
A nonreacting mixture contains $2.0\text{ mol}$ of CO $_2$ and $8.0\text{ mol}$ of N $_2$ . What is the mole fraction of CO $_2$ in the mixture?
Answer: 0.20 This question tests the skill of calculating mole fraction in a mixture. Mole fraction is calculated as: mole fraction = moles of component / total moles. The total moles = 2.0 mol CO₂ + 8.0 mol N₂ = 10.0 mol. Therefore, mole fraction of CO₂ = 2.0 mol / 10.0 mol = 0.20. A tempting incorrect answer is 0.80 (choice A), which represents the mole fraction of N₂ instead of CO₂, showing confusion about which component to calculate. To find mole fraction, always divide the moles of the specific component by the total moles of all components.
A nonreacting mixture contains $0.50\text{ mol}$ of O $_2$ and $1.50\text{ mol}$ of N $_2$ . What is the mole fraction of O $_2$ in the mixture?
Answer: 0.25 This question tests the skill of calculating mole fraction in a gas mixture. Mole fraction is found by dividing moles of component by total moles. The total moles = 0.50 mol O₂ + 1.50 mol N₂ = 2.00 mol. Therefore, mole fraction of O₂ = 0.50 mol / 2.00 mol = 0.25. Choice B (0.75) represents the mole fraction of N₂ instead of O₂, showing confusion about which gas to calculate. To avoid this error, clearly identify the component of interest before dividing by total moles.
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?
Answer: 30.0% 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.
A student combines $18.0\text{ g}$ of glucose (C $_6$ H $_{12}$ O $_6$ ) with $72.0\text{ g}$ of water to form a nonreacting mixture. What is the mass percent of glucose in the mixture?
Answer: 20.0% This question tests the skill of calculating mass percent composition in a solution. Mass percent equals (mass of solute / total mass) × 100%. The total mass = 18.0 g glucose + 72.0 g water = 90.0 g. Therefore, mass percent of glucose = (18.0 g / 90.0 g) × 100% = 20.0%. Choice A (25.0%) results from incorrectly calculating 18.0/72.0 × 100%, using only water mass as denominator rather than total mass. When finding mass percent, always divide by the sum of all masses in the mixture.
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 $106.0 \text{ g/mol}$ ).
Answer: 42.0% 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%.
A nonreacting gas mixture contains $2.0\,\text{mol}$ He and $6.0\,\text{mol}$ Ne. What is the mole fraction of He in the mixture?
Answer: 0.25 This question tests the skill of calculating mole fraction of a component in a gas mixture. Mole fraction is calculated as: moles of component / total moles in mixture. We have 2.0 mol He and 6.0 mol Ne, giving total moles = 8.0 mol. The mole fraction of He = 2.0 mol / 8.0 mol = 0.25. A tempting incorrect answer is 0.33 (option C), which might come from mistakenly dividing the moles of He by the moles of Ne alone (2.0/6.0) instead of by the total moles. To find mole fraction correctly, always divide the moles of your component by the sum of moles of all components.
A nonreacting mixture contains $1.5\,\text{mol}$ of $\text{N}_2$ and $3.5\,\text{mol}$ of $\text{O}_2$ . What is the mole fraction of $\text{O}_2$ in the mixture?
Answer: 0.70 This question tests the skill of calculating mole fraction in a gas mixture. Mole fraction equals moles of component divided by total moles. The mixture contains 1.5 mol N₂ and 3.5 mol O₂, giving a total of 5.0 mol. Therefore, mole fraction of O₂ = 3.5 mol / 5.0 mol = 0.70. Students who selected 0.30 (choice A) calculated the mole fraction of N₂ instead, mixing up which gas was being asked about. To avoid this error, always underline or highlight the specific component mentioned in the question before calculating.

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