Stoichiometry: Free-Response Questions for AP Chemistry

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By — McGraw-Hill Professional
Updated on Feb 1, 2011

Review the following concepts if necessary:

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.

  1. Determine the empirical formula of the acid.
  2. 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.
  3. 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.
  4. Using your answer from part a, determine the molecular formula for part b and for part c.
  5. Account for any differences in the molecular formulas determined in part d.

Answers and Explanations

  1. 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.

  2. Using HA to represent the monoprotic acid, the balanced equation for the titration reaction is:
      HA + NaOH → NaA + H2O
  3. 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.

  4. This may be done in several ways. One way is to use the ideal gas equation. This will be done here. The equation and the value of R are given in the exam booklet.
  5. 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.

  6. The approximate formula mass from the empirical (CH2O) formula is: 12 + 2(1) + 16 = 30 g/mol
  7. 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.

  8. The one formula is double the formula of the other. Thus, the smaller molecule dimerizes to produce the larger molecule.
  9. You get 1 point if you "combined" two of the smaller molecules.

    Total your points. There are 9 points possible.

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