The questions are on the mole concept test. Take it under real conditions before reading the mark scheme.
Quick answer: A mole is 6.022 × 10²³ particles, and its mass in grams is the molar mass M, so n = m/M links mass to amount. This mark scheme gives the model answers and the marks for the 8 questions of the AP Chemistry test on mole concept, from moles of methane to spot the error, with the equivalent answers to accept.
Marks reward method, correct moles and clear units. Accept answers within rounding of the values shown and award follow-through marks where an earlier value is reused correctly.
Level: AP Chemistry · Curriculum: AP Chem Unit 1
Question 1 Moles of methane (2 marks)
- n = 8.00/16.043 = 0.499 mol.
- N = 0.4987 × 6.022 × 10²³ = 3.00 × 10²³ molecules.
Marking: 1 mark each part.
Question 2 Mass spectrum of an element (3 marks)
- 11 % (peak at 26).
- Average = 24(0.79) + 25(0.10) + 26(0.11) = 18.96 + 2.50 + 2.86 = 24.32.
- This is magnesium.
Marking: 1 mark for each part.
Question 3 Empirical formula (3 marks)
K = 26.6/39.098 = 0.680; Cr = 35.4/51.996 = 0.681; O = 38.0/15.999 = 2.375. Dividing by 0.680: 1 : 1 : 3.49. Multiply by 2: 2 : 2 : 7. K₂Cr₂O₇.Marking: 1 mark for the moles; 1 mark for the ratio; 1 mark for the doubling and formula.
Question 4 Molecular formula (2 marks)
Empirical mass = 12.011 + 1.008 = 13.019. 78.1/13.019 = 6.0, so the formula is C₆H₆.Marking: 1 mark for the factor 6; 1 mark for the formula.
Question 5 Iron in an ore (3 marks)
- %Fe = 2(55.845)/159.69 × 100 = 69.94 %.
- m = 0.6994 × 1.00 kg = 0.699 kg (699 g).
Marking: 1 mark for part (a); 1 mark for using the fraction; 1 mark for the answer with unit.
Question 6 Why a non-integer average? (2 marks)
The average is a weighted mean of the isotope masses, weighted by abundance. No single atom has the average mass; most atoms have one of the integer-like isotope masses.Marking: 1 mark for weighted mean; 1 mark for the idea that individual atoms have whole-number masses.
Question 7 Limestone in a mixture (3 marks)
- n(CO₂) = 0.880/44.009 = 0.0200 mol.
- 1 : 1 ratio, so m(CaCO₃) = 0.0200 × 100.09 = 2.00 g; percent = 2.00/5.00 × 100 = 40.0 %.
Marking: 1 mark for the moles; 1 mark for the mole ratio; 1 mark for the percent.
Question 8 Spot the error (2 marks)
The student forgot the subscripts and the number of atoms: Al₂ has two atoms and (SO₄)₃ has three sulfur and twelve oxygen atoms. Correct value: 2(26.982) + 3(32.06) + 12(15.999) = 53.964 + 96.18 + 191.99 = 342.13 g/mol.Marking: 1 mark for the error; 1 mark for the value.
