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AP Chemistry

AP Chemistry Help: Composition Of Mixtures

Review real example questions for Composition Of Mixtures in AP Chemistry.

Question 1

A 4.00 g sample of a mixture of NaHCO3NaHCO_3NaHCO3​ and Na2CO3Na_2CO_3Na2​CO3​ is reacted with excess hydrochloric acid. The reaction produces 0.985 L of CO2CO_2CO2​ gas at 298 K and 1.00 atm. What is the mass percent of NaHCO3NaHCO_3NaHCO3​ in the mixture? (Molar mass of NaHCO3NaHCO_3NaHCO3​ is 84.0 g/mol84.0 \text{ g/mol}84.0 g/mol; Na2CO3Na_2CO_3Na2​CO3​ is 106.0 g/mol106.0 \text{ g/mol}106.0 g/mol).

  1. 16.0%
  2. 42.0%
  3. 58.0%
  4. 84.0%
Explanation: First, find the total moles of CO2CO_2CO2​ produced using the ideal gas law: n=PV/RT=(1.00 atm×0.985 L)/(0.08206 L atm/mol K×298 K)=0.0403 molCO2n = 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_2n=PV/RT=(1.00 atm×0.985 L)/(0.08206 L atm/mol K×298 K)=0.0403 molCO2​. Let x be the mass of NaHCO3NaHCO_3NaHCO3​ and y be the mass of Na2CO3Na_2CO_3Na2​CO3​. We have two equations: 1) x+y=4.00x + y = 4.00x+y=4.00 and 2) Moles of CO2CO_2CO2​ from NaHCO3NaHCO_3NaHCO3​ + Moles of CO2CO_2CO2​ from Na2CO3Na_2CO_3Na2​CO3​ = 0.04030.04030.0403. The reactions are NaHCO3→CO2NaHCO_3 \rightarrow CO_2NaHCO3​→CO2​ (1:1) and Na2CO3→CO2Na_2CO_3 \rightarrow CO_2Na2​CO3​→CO2​ (1:1). So, (x/84.0)+(y/106.0)=0.0403(x/84.0) + (y/106.0) = 0.0403(x/84.0)+(y/106.0)=0.0403. Substitute y=4.00−xy = 4.00 - xy=4.00−x into the second equation: (x/84.0)+((4.00−x)/106.0)=0.0403(x/84.0) + ((4.00 - x)/106.0) = 0.0403(x/84.0)+((4.00−x)/106.0)=0.0403. Solving for x gives x=1.68x = 1.68x=1.68 g. Mass percent of NaHCO3NaHCO_3NaHCO3​ = (1.68 g/4.00 g1.68 \text{ g} / 4.00 \text{ g}1.68 g/4.00 g) * 100% = 42.0%.

Question 2

A solution is prepared by mixing 25 g of ethanol (C2H5OHC_2H_5OHC2​H5​OH) with 75 g of water. This resulting solution is best described as a

  1. pure substance, because its components are chemically bonded.
  2. heterogeneous mixture, because it contains two different types of molecules.
  3. homogeneous mixture, because the components are uniformly distributed.
  4. compound, because the mass ratio of components is fixed at 1:3.
Explanation: Ethanol and water are miscible, meaning they mix uniformly at the molecular level to form a solution, which is a homogeneous mixture. (A) is incorrect because the components are not chemically bonded in a mixture. (B) is incorrect because the mixture is uniform, not heterogeneous. (D) is incorrect because the ratio is fixed for this specific mixture, but it can be varied, which is a characteristic of mixtures, not compounds.

Question 3

An impure 1.80 g sample of calcium carbide (CaC2CaC_2CaC2​) reacts with excess water, producing 0.520 L of acetylene gas (C2H2C_2H_2C2​H2​) 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 CaC2CaC_2CaC2​ sample?

  1. 60.5%
  2. 68.2%
  3. 72.4%
  4. 85.1%
Explanation: First, find the partial pressure of C2H2C_2H_2C2​H2​: PC2H2=Ptotal−PH2O=1.00 atm−0.035 atm=0.965 atmP_{C2H2} = P_{total} - P_{H2O} = 1.00 \text{ atm} - 0.035 \text{ atm} = 0.965 \text{ atm}PC2H2​=Ptotal​−PH2O​=1.00 atm−0.035 atm=0.965 atm. Next, find moles of C2H2C_2H_2C2​H2​ using the ideal gas law: n=PV/RT=(0.965 atm×0.520 L)/(0.08206 L atm/mol K×300 K)=0.0204 moln = 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}n=PV/RT=(0.965 atm×0.520 L)/(0.08206 L atm/mol K×300 K)=0.0204 mol. The reaction is CaC2(s)+2H2O(l)→C2H2(g)+Ca(OH)2(aq)CaC_2(s) + 2H_2O(l) \rightarrow C_2H_2(g) + Ca(OH)_2(aq)CaC2​(s)+2H2​O(l)→C2​H2​(g)+Ca(OH)2​(aq). The mole ratio of CaC2CaC_2CaC2​ to C2H2C_2H_2C2​H2​ is 1:1. Mass of CaC2CaC_2CaC2​ = 0.0204 mol×64.1 g/mol=1.307 g0.0204 \text{ mol} \times 64.1 \text{ g/mol} = 1.307 \text{ g}0.0204 mol×64.1 g/mol=1.307 g. Percent purity = (1.307 g/1.80 g1.307 \text{ g} / 1.80 \text{ g}1.307 g/1.80 g) * 100% = 72.6%.

Question 4

A mixture containing only iron(II) oxide (FeO) and iron(III) oxide (Fe2O3Fe_2O_3Fe2​O3​) 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 g/mol55.85 \text{ g/mol}55.85 g/mol; FeO is 71.85 g/mol71.85 \text{ g/mol}71.85 g/mol; Fe2O3Fe_2O_3Fe2​O3​ is 159.70 g/mol159.70 \text{ g/mol}159.70 g/mol).

  1. 27.8%
  2. 50.0%
  3. 66.7%
  4. 72.2%
Explanation: The mass percent of Fe in FeO is (55.85/71.85)×100%=77.73%(55.85/71.85) \times 100\% = 77.73\%(55.85/71.85)×100%=77.73%. The mass percent of Fe in Fe2O3Fe_2O_3Fe2​O3​ is (2×55.85/159.70)×100%=69.94%(2 \times 55.85 / 159.70) \times 100\% = 69.94\%(2×55.85/159.70)×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)=0.720x(0.7773) + (1-x)(0.6994) = 0.720x(0.7773)+(1−x)(0.6994)=0.720. Expanding: 0.7773x+0.6994−0.6994x=0.7200.7773x + 0.6994 - 0.6994x = 0.7200.7773x+0.6994−0.6994x=0.720, so 0.0779x=0.02060.0779x = 0.02060.0779x=0.0206, and x=0.264=26.4%x = 0.264 = 26.4\%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 Fe2O3Fe_2O_3Fe2​O3​. Choice (D) is the overall iron percentage.

Question 5

A 10.0 g mixture of calcium carbonate (CaCO3CaCO_3CaCO3​) and silicon dioxide (SiO2SiO_2SiO2​) 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 CaCO3CaCO_3CaCO3​ in the mixture? (Molar mass of CaCO3CaCO_3CaCO3​ is 100.1 g/mol100.1 \text{ g/mol}100.1 g/mol; CO2CO_2CO2​ is 44.0 g/mol44.0 \text{ g/mol}44.0 g/mol).

  1. 22.0%
  2. 50.0%
  3. 78.0%
  4. 100.0%
Explanation: The loss of mass is due to the evolution of CO2CO_2CO2​ gas from the decomposition of CaCO3CaCO_3CaCO3​. Mass of CO2CO_2CO2​ lost = $10.0 g−7.8 g=2.2 g\$10.0 \text{ g} - 7.8 \text{ g} = 2.2 \text{ g}$10.0 g−7.8 g=2.2 g. Moles of CO2CO_2CO2​ = 2.2 g/44.0 g/mol=0.050 mol2.2 \text{ g} / 44.0 \text{ g/mol} = 0.050 \text{ mol}2.2 g/44.0 g/mol=0.050 mol. The reaction is CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g)CaCO3​(s)→CaO(s)+CO2​(g). The mole ratio of CaCO3CaCO_3CaCO3​ to CO2CO_2CO2​ is 1:1. So, moles of CaCO3CaCO_3CaCO3​ = 0.050 mol0.050 \text{ mol}0.050 mol. Mass of CaCO3CaCO_3CaCO3​ = 0.050 mol×100.1 g/mol=5.0 g0.050 \text{ mol} \times 100.1 \text{ g/mol} = 5.0 \text{ g}0.050 mol×100.1 g/mol=5.0 g. Initial mass percent of CaCO3CaCO_3CaCO3​ = (5.0 g/10.0 g5.0 \text{ g} / 10.0 \text{ g}5.0 g/10.0 g) * 100% = 50.0%. (A) is the mass percent of CO2CO_2CO2​ lost. (C) is the mass percent of the final residue. (D) incorrectly assumes the entire sample decomposed.

Question 6

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

  1. the original sample was a homogeneous mixture.
  2. a physical change occurred, forming an alloy.
  3. the original sample was a compound that decomposed.
  4. a chemical change occurred, forming a new substance from a mixture.
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 7

A mixture of gases contains equal masses of helium (He) and neon (Ne). What is the mole fraction of helium in the mixture? (Molar mass of He is 4.00 g/mol4.00 \text{ g/mol}4.00 g/mol; Ne is 20.2 g/mol20.2 \text{ g/mol}20.2 g/mol).

  1. 0.165
  2. 0.500
  3. 0.835
  4. 0.899
Explanation: Assume a total mass, for example, 100 g. Then there are 50 g of He and 50 g of Ne. Moles of He = $50 g/4.00 g/mol=12.5 mol\$50 \text{ g} / 4.00 \text{ g/mol} = 12.5 \text{ mol}$50 g/4.00 g/mol=12.5 mol. Moles of Ne = $50 g/20.2 g/mol=2.48 mol\$50 \text{ g} / 20.2 \text{ g/mol} = 2.48 \text{ mol}$50 g/20.2 g/mol=2.48 mol. Total moles = $12.5+2.48=14.98 mol\$12.5 + 2.48 = 14.98 \text{ mol}$12.5+2.48=14.98 mol. Mole fraction of He = Moles of He / Total moles = $12.5/14.98=0.834\$12.5 / 14.98 = 0.834$12.5/14.98=0.834. (B) incorrectly assumes equal moles instead of equal masses. (A) is the mole fraction of neon. (D) is based on a calculation error.

Question 8

A nonreacting mixture contains 1.0 mol1.0\text{ mol}1.0 mol of CH4_44​ and 4.0 mol4.0\text{ mol}4.0 mol of H2_22​. What is the mole fraction of CH4_44​ in the mixture?

  1. 0.10
  2. 0.80
  3. 0.20
  4. 0.50
  5. 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 9

A student makes a nonreacting solution by mixing 15.0 g15.0\text{ g}15.0 g of ethanol (C2_22​H5_55​OH) with 35.0 g35.0\text{ g}35.0 g of water. What is the mass percent of ethanol in the solution?

  1. 70.0%
  2. 42.9%
  3. 30.0%
  4. 50.0%
  5. 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 10

A nonreacting liquid mixture contains 30.0 g30.0\text{ g}30.0 g of acetone and 20.0 g20.0\text{ g}20.0 g of water. What is the mass percent of water in the mixture?

  1. 60.0%
  2. 40.0%
  3. 50.0%
  4. 66.7%
  5. 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.