Percent Composition: Quiz

Percent Composition: Quiz

7 questions

1. Question 1EasyWhat is the formula mass of calcium hydroxide, $\ce{Ca(OH)2}$? (Ca = 40.08, O = 16.00, H = 1.008)

   A57.09 uB58.10 uC74.10 uD148.19 u
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   Answer: 74.10 u

   The subscript 2 after the parentheses applies to both O and H: $40.08 + 2(16.00 + 1.008) = 74.10$ u. The answer 58.10 u doubles only the H.

2. Question 2EasyWhat is the mass percent of oxygen in water, $\ce{H2O}$?

   A11.19%B33.3%C50.0%D88.81%
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   Answer: 88.81%

   $\dfrac{16.00}{18.016} \times 100\% = 88.81\%$. The answer 33.3% is the fraction of atoms that are oxygen (1 of 3), not the fraction of the mass.

3. Question 3MediumWhat is the mass percent of carbon in carbon dioxide, $\ce{CO2}$?

   A12.01%B27.29%C33.3%D72.71%
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   Answer: 27.29%

   Formula mass = 12.01 + 2(16.00) = 44.01 u. % C = $\dfrac{12.01}{44.01} \times 100\% = 27.29\%$. The answer 72.71% is the percentage of oxygen.

4. Question 4MediumWhat is the mass percent of nitrogen in ammonia, $\ce{NH3}$?

   A14.01%B17.75%C25.0%D82.25%
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   Answer: 82.25%

   Formula mass = 14.01 + 3(1.008) = 17.034 u. % N = $\dfrac{14.01}{17.034} \times 100\% = 82.25\%$. The answer 25.0% counts atoms (1 of 4) instead of mass.

5. Question 5HardWhat mass of iron is in 50.0 g of iron(III) oxide, $\ce{Fe2O3}$? (Fe = 55.85, O = 16.00)

   A17.5 gB35.0 gC50.0 gD69.9 g
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   Answer: 35.0 g

   Formula mass = 2(55.85) + 3(16.00) = 159.70 u. Mass of Fe = $\dfrac{111.70}{159.70} \times 50.0\ \text{g} = 35.0$ g. The answer 17.5 g forgets that there are two Fe atoms in the formula.

6. Question 6EasySample A is 1.00 g of pure water and sample B is 1.00 kg of pure water. How do their mass percents of hydrogen compare?

   AThey are the sameBSample B has a higher percentageCSample A has a higher percentageDIt depends on the temperature
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   Answer: They are the same

   A pure compound always has the same composition by mass. Both samples are 11.19% hydrogen; only the total mass is different.

7. Question 7MediumWhat is the mass percent of calcium in calcium carbonate, $\ce{CaCO3}$? (Ca = 40.08, C = 12.01, O = 16.00)

   A12.00%B20.0%C40.04%D47.96%
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   Answer: 40.04%

   Formula mass = 40.08 + 12.01 + 3(16.00) = 100.09 u. % Ca = $\dfrac{40.08}{100.09} \times 100\% = 40.04\%$. The answer 20.0% counts atoms (1 of 5).