Which is a compound (not an element)?

Answer: H₂O H₂O contains two different elements (compound). He and Au are elements; O₂ is an elemental molecule (just one element).

A separate measurement gives a molar mass of 84 g/mol. What is the molecular formula?

Answer: C6H12 Empirical mass = 14 g/mol. 84/14 = 6 - C6H12.

Mass percent of solute = (mass solute / mass solution) × 100%. A solution made by dissolving 5 g NaCl in 95 g water has mass %

Answer: 5% Mass solution = 5 + 95 = 100 g. Mass % = 5/100 × 100% = 5%.

AP Chemistry · Topic 1.3

Elemental Composition of Pure Substances Practice

Part of Atomic Structure and Properties.(SPQ-2.A)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$Which is a compound (not an element)?$
- A
  $He$
- B
  $Au$
- C
  $O₂$
- D
  $H₂O$
  check_circle
Why
$H₂O contains two different elements (compound). He and Au are elements; O₂ is an elemental molecule (just one element).$
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Sample 2difficulty 1/5
$A 0.250 g sample of a hydrocarbon is burned completely in excess O2. The combustion products are passed through a dehydrating tube (gains 0.306 g) followed by a CO2 absorber (gains 0.747 g). The hydrocarbon contains only C and H.$
$A separate measurement gives a molar mass of 84 g/mol. What is the molecular formula?$
- A
  $C5H10$
- B
  $C7H14$
- C
  $C6H12$
  check_circle
- D
  $CH2$
Why
$Empirical mass = 14 g/mol. 84/14 = 6 -> C6H12.$
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Sample 3difficulty 2/5
$Mass percent of solute = (mass solute / mass solution) \times 100%. A solution made by dissolving 5 g NaCl in 95 g water has mass %$
- A
  $5%$
  check_circle
- B
  $9.5%$
- C
  $5.26%$
- D
  $50%$
Why
$Mass solution = 5 + 95 = 100 g. Mass % = 5/100 \times 100% = 5%.$
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Sample 4difficulty 2/5
$What is the percent composition by mass of carbon in CO₂?$
- A
  $44%$
- B
  $12%$
- C
  $72.7%$
- D
  $27.3%$
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Why
$Mass % C = (12 / 44) \times 100% \approx 27.3%.$
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Sample 5difficulty 2/5
$A 0.250 g sample of a hydrocarbon is burned completely in excess O2. The combustion products are passed through a dehydrating tube (gains 0.306 g) followed by a CO2 absorber (gains 0.747 g). The hydrocarbon contains only C and H.$
$What is the empirical formula?$
- A
  $CH3$
- B
  $CH$
- C
  $C2H3$
- D
  $CH2$
  check_circle
Why
$mol C = 0.747/44.01 = 0.01698; mass C = 0.2039 g. mol H = (0.306/18.02)*2 = 0.03396; mass H = 0.0342 g. Sum = 0.238 g (consistent with 0.250 within error). Ratio C:H = 0.01698:0.03396 = 1:2 -> CH2.$
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