6.1 Practice Problems
Attempt these problems as if they were real exam questions in an exam environment.
Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem. Having to look up information or the solution should be an indicator that the previous layers (1–5) in the Structured Learning Approach have not been mastered.
What is the mass percent of oxygen in glucose (C6H12O6)?
- 16.0%
- 82.2%
- 53.2%
- 6.7%
- 40.0%
Solution
Answer: C
Concept: Percent composition
One molecule of glucose has a mass of 180.18 amu. Take the ratio of the mass of oxygen to the mass of glucose and multiply by the number of oxygen atoms that are in glucose and convert to a percentage by multiplying by 100%.
\[\begin{align*} \%~\mathrm{O} &= \dfrac{15.99~\mathrm{amu}}{180.18~\mathrm{amu}} \times 6 \times 100\% = 53.25\% \end{align*}\]
An unknown compound was found to contain 79.85% C and 20.15% H with a molar mass of 30.08 g mol–1. What is the empirical formula of the compound?
- C2H6
- CH3
- C4H10
- CH4
- C2H2
Solution
Answer: A
Concept: Percent composition
Find the moles of each element in the unknown compound.
\[\begin{align*} n_{\mathrm{C}} &= \left ( \dfrac{79.85\%}{100\%} \right ) \left ( \dfrac{30.08~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{mol}}{12.01~\mathrm{g}} \right ) = 1.999 \approx 2 \\[2ex] n_{\mathrm{H}} &= \left ( \dfrac{20.15\%}{100\%} \right ) \left ( \dfrac{30.08~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{mol}}{1.01~\mathrm{g}} \right ) = 6.001 \approx 6 \end{align*}\]
Write out the empirical formula and divide each subscript by the smallest number.
\[\mathrm{C}_{2}\mathrm{H}_{6} \longrightarrow \mathrm{C}_{2/2}\mathrm{H}_{6/2} \longrightarrow \mathrm{C}_{1}\mathrm{H}_3 \longrightarrow \mathrm{C}\mathrm{H}_3\]
You measure 50.0 grams of zinc(II) oxide and dissolve it in water to a final volume of 500 mL of solution. What is the molarity of this solution?
- 1.53
- 1.23
- 0.252
- 0.810
- 2.20
Solution
Answer: B
Concept: Concentration
The molar mass of ZnO is 81.39 g mol–1. Find the moles of ZnO.
\[\begin{align*} n_{\mathrm{Zn}} &= \left ( \dfrac{50.0~\mathrm{g~Zn}}{} \right ) \left ( \dfrac{\mathrm{mol}}{81.39~\mathrm{g}} \right )\\ &= 0.6143~\mathrm{mol} \end{align*}\]
Find the concentration of the solution in molarity (g L–1). 500 mL is 0.500 L.
\[\begin{align*} M &= \left ( \dfrac{n_{\mathrm{solu}}}{V_{\mathrm{soln}}} \right ) \left ( \dfrac{0.6143~\mathrm{mol}}{0.500~\mathrm{L}} \right )\\ &= 1.23~M \end{align*}\]
- 1.53
The molecular formula of glucose is C6H12O6. What is the percent composition by mass of carbon in glucose?
- 39.99%
- 6.72%
- 53.28%
- 3.77%
- Impossible to say without knowing the mass of the sample.
Solution
Answer: A
Concept: Percent composition
One molecule of glucose has a mass of 180.18 amu. Take the ratio of the mass of oxygen to the mass of glucose and multiply by the number of oxygen atoms that are in glucose and convert to a percentage by multiplying by 100%.
\[\begin{align*} \%~\mathrm{C} &= \dfrac{12.01~\mathrm{amu}}{180.18~\mathrm{amu}} \times 6 \times 100\% = 39.99\% \end{align*}\]
What is the initial concentration (in M) of a 3.04 L solution that was diluted to 4.43 L with a final concentration of 2.2 M?
- 3.2
- 6.12
- 2.2
- 6.1
- 29.6
Solution
Answer: A
Concept: Dilution
\[\begin{align*} M_1V_1 &= M_2V_2 \\[2ex] M_1 &= \dfrac{M_2V_2}{V_1} \\ &= \dfrac{(2.2~M \times 4.43~\mathrm{L})}{3.04~\mathrm{L}}\\ &= 3.2~M \end{align*}\]