Molecular formulas Transitional Reasoning Practice

Chemistry · The Mole & Stoichiometry · Chemistry · The Mole

5 Chemistry practice questions on Molecular formulas. Questions that build towards applied reasoning.

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Questions

  1. 1.

    An unknown organic compound is found by elemental analysis to consist of 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. In a separate experiment, the complete combustion of a 0.150 mol sample of this compound produces 19.8 g of carbon dioxide (CO2\text{CO}_2). What is the molecular formula of this organic compound? (Atomic masses: C=12.0 g/mol\text{C} = 12.0\text{ g/mol}, H=1.0 g/mol\text{H} = 1.0\text{ g/mol}, O=16.0 g/mol\text{O} = 16.0\text{ g/mol})

    • A) C2H4O2\text{C}_2\text{H}_4\text{O}_2
    • B) C3H6O3\text{C}_3\text{H}_6\text{O}_3
    • C) C4H8O4\text{C}_4\text{H}_8\text{O}_4
    • D) C6H12O6\text{C}_6\text{H}_{12}\text{O}_6
  2. 2.

    Under identical conditions of temperature and pressure, a sealed 2.00 L flask contains 2.40 g of an unknown gaseous hydrocarbon. When the flask is completely evacuated and filled with oxygen gas (O2\text{O}_2) under the same temperature and pressure, the oxygen gas has a mass of 2.56 g. If elemental analysis indicates that the hydrocarbon consists of 80.0% carbon and 20.0% hydrogen by mass, what is the molecular formula of the compound? (Atomic masses: C=12.0 g/mol\text{C} = 12.0\text{ g/mol}, H=1.0 g/mol\text{H} = 1.0\text{ g/mol}, O=16.0 g/mol\text{O} = 16.0\text{ g/mol})

    • A) CH3\text{CH}_3
    • B) C2H4\text{C}_2\text{H}_4
    • C) C2H6\text{C}_2\text{H}_6
    • D) C3H8\text{C}_3\text{H}_8
  3. 3.

    A materials chemist analyzes a volatile oxide of nitrogen utilized as an oxidizer in rocket propulsion systems. A pure gaseous sample containing 3.01×10223.01 \times 10^{22} molecules is found to have a mass of 4.60 g. If nitrogen accounts for 30.4% of the compound's total mass, what is the molecular formula of this substance? (Avogadro's number = 6.02×1023 mol−16.02 \times 10^{23}\text{ mol}^{-1}; atomic masses: N=14.0 g/mol\text{N} = 14.0\text{ g/mol}, O=16.0 g/mol\text{O} = 16.0\text{ g/mol})

    • A) N2O4\text{N}_2\text{O}_4
    • B) NO2\text{NO}_2
    • C) N2O3\text{N}_2\text{O}_3
    • D) N2O5\text{N}_2\text{O}_5
  4. 4.

    An analytical chemist isolates a fragrant liquid containing only carbon, hydrogen, and oxygen, determining its empirical formula to be C2H4O\text{C}_2\text{H}_4\text{O}. To determine its molecular formula, a 0.880 g sample of the compound is completely vaporized, occupying 224 mL at standard temperature and pressure (STP). What is the molecular formula of this compound? (Atomic masses: C=12.0 g/mol\text{C} = 12.0\text{ g/mol}, H=1.0 g/mol\text{H} = 1.0\text{ g/mol}, O=16.0 g/mol\text{O} = 16.0\text{ g/mol}; molar volume of a gas at STP = 22.4 L/mol22.4\text{ L/mol})

    • A) C2H4O\text{C}_2\text{H}_4\text{O}
    • B) C3H6O2\text{C}_3\text{H}_6\text{O}_2
    • C) C4H8O2\text{C}_4\text{H}_8\text{O}_2
    • D) C6H12O3\text{C}_6\text{H}_{12}\text{O}_3
  5. 5.

    During an investigation of an unknown hydrocarbon fuel, elemental analysis reveals that it contains 85.6% carbon and 14.4% hydrogen by mass. In a companion measurement, a 0.250 mol sample of this gaseous compound is found to have a mass of 14.0 g. Based on these experimental results, what is the molecular formula of the fuel? (Atomic masses: C=12.0 g/mol\text{C} = 12.0\text{ g/mol}, H=1.0 g/mol\text{H} = 1.0\text{ g/mol})

    • A) C4H8\text{C}_4\text{H}_8
    • B) CH2\text{CH}_2
    • C) C2H4\text{C}_2\text{H}_4
    • D) C3H6\text{C}_3\text{H}_6

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