Stoichiometry: Free-Response Questions for AP Chemistry
Review the following concepts if necessary:
- Moles, Molar Mass, and Molarity for AP Chemistry
- Percent Composition and Empirical Formulas for AP Chemistry
- Reaction Stoichiometry for AP Chemistry
- Limiting Reactants for AP Chemistry
- Percent Yield for AP Chemistry
Answer the following questions. You have 15 minutes, and you may use a calculator and the tables in the back of the book.
A sample of a monoprotic acid was analyzed. The sample contained 40.0% C and 6.71% H. The remainder of the sample was oxygen.
- Determine the empirical formula of the acid.
- A 0.2720-g sample of the acid was titrated with standard NaOH solution. Determine the molecular weight of the acid if the sample required 45.00 mL of 0.1000 M NaOH for the titration.
- A second sample was placed in a flask. The flask was placed in a hot water bath until the sample vaporized. It was found that 1.18 g of vapor occupied 300.0 mL at 100°C and 1.00 atmospheres. Determine the molecular weight of the acid.
- Using your answer from part a, determine the molecular formula for part b and for part c.
- Account for any differences in the molecular formulas determined in part d.
Answers and Explanations
- The percent oxygen (53.3%) is determined by subtracting the carbon and the hydrogen from 100%.
This gives the empirical formula: CH2O.
You get 1 point for correctly determining any of the elements, and 1 point for getting the complete empirical formula correct.
- Using HA to represent the monoprotic acid, the balanced equation for the titration reaction is:
- HA + NaOH → NaA + H2O
The moles of acid may then be calculated:
(45.00 mL NaOH)(0.1000 mol NaOH/1000 mL) (1 mol HA/1 mol NaOH) = 4.500 × 10–3 mol HA
The molecular mass is:
- 0.2720 g/4.500 × 10–3 mol = 60.44 g/mol
You get 1 point for the correct number of moles of HA (or NaOH) and 1 point for the correct final answer.
First find the moles: n = PV/RT Do not forget, you MUST change to kelvin.
- n = (1.00 atm)(300.0 mL)(l L/1000 mL)/(0.0821 L atm/mol K)(373 K)
- n = 9.80 × 10–3 mol
The molecular mass is: 1.18 g/9.80 × 10–3 mol = 120 g/mol
You get 1 point for getting any part of the calculation correct and 1 point for getting the correct final answer.
For part b: (60.44 g/mol)/(30 g/mol) = 2
- Molecular formula = 2 × Empirical Formula = C2H4O2
For part c: (120 g/mol)/(30 g/mol) = 4
- Molecular formula = 4 × Empirical Formula = C4H8O4
You et 1 point for each correct molecular formula. If you got the wrong answer in part a, you can still get credit for one or both of the molecular formulas if you used the part a value correctly.
You get 1 point if you "combined" two of the smaller molecules.
Total your points. There are 9 points possible.