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Explain why a burning magnesium continues to burn in a gas jar full of carbon (IV) oxide while a burning candle would be extinguished.

Chemistry

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CH2\text{CH}_2

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Part 7

Explain why a burning magnesium continues to burn in a gas jar full of carbon (IV) oxide while a burning candle would be extinguished.

  • A burning candle requires oxygen (O2\text{O}_2) from the air to sustain its combustion. Carbon (IV) oxide (CO2\text{CO}_2) is a stable compound and does not provide oxygen to support the combustion of most substances, including a candle. Therefore, a candle would be extinguished in a gas jar full of CO2\text{CO}_2.
  • Magnesium is a highly reactive metal. When burning, it can react directly with carbon (IV) oxide. In this reaction, magnesium reduces the carbon dioxide to elemental carbon and forms magnesium oxide. This reaction is highly exothermic, releasing enough energy to sustain the burning of magnesium. 2Mg(s)+CO2(g)⟶2MgO(s)+C(s)2\text{Mg}(s) + \text{CO}_2(g) \longrightarrow 2\text{MgO}(s) + \text{C}(s)

Part 8

CH2\text{CH}_2

Determine the empirical formula of the hydrocarbon.

Given:
Mass of CO2=8.4 g\text{CO}_2 = 8.4 \text{ g}
Mass of H2O=3.42 g\text{H}_2\text{O} = 3.42 \text{ g}
Atomic mass of H=1.0\text{H} = 1.0
Atomic mass of C=12.0\text{C} = 12.0
Atomic mass of O=16.0\text{O} = 16.0

  1. Calculate the molar masses of CO2\text{CO}_2 and H2O\text{H}_2\text{O}.

    Molar mass of CO2=12.0+(2×16.0)=12.0+32.0=44.0 g/mol\text{Molar mass of CO}_2 = 12.0 + (2 \times 16.0) = 12.0 + 32.0 = 44.0 \text{ g/mol} Molar mass of H2O=(2×1.0)+16.0=2.0+16.0=18.0 g/mol\text{Molar mass of H}_2\text{O} = (2 \times 1.0) + 16.0 = 2.0 + 16.0 = 18.0 \text{ g/mol}

  2. Calculate the moles of CO2\text{CO}_2 and H2O\text{H}_2\text{O} produced.

    Moles of CO2=Mass of CO2Molar mass of CO2=8.4 g44.0 g/mol≈0.1909 mol\text{Moles of CO}_2 = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2} = \frac{8.4 \text{ g}}{44.0 \text{ g/mol}} \approx 0.1909 \text{ mol} Moles of H2O=Mass of H2OMolar mass of H2O=3.42 g18.0 g/mol=0.19 mol\text{Moles of H}_2\text{O} = \frac{\text{Mass of H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} = \frac{3.42 \text{ g}}{18.0 \text{ g/mol}} = 0.19 \text{ mol}

  3. Determine the moles of carbon and hydrogen atoms in the hydrocarbon.

    Since all carbon in the hydrocarbon forms CO2\text{CO}_2, the moles of carbon atoms are equal to the moles of CO2\text{CO}_2.

    Moles of C atoms=Moles of CO2=0.1909 mol\text{Moles of C atoms} = \text{Moles of CO}_2 = 0.1909 \text{ mol}

    Since all hydrogen in the hydrocarbon forms H2O\text{H}_2\text{O}, and each molecule of H2O\text{H}_2\text{O} contains two hydrogen atoms, the moles of hydrogen atoms are twice the moles of H2O\text{H}_2\text{O}.

    Moles of H atoms=2×Moles of H2O=2×0.19 mol=0.38 mol\text{Moles of H atoms} = 2 \times \text{Moles of H}_2\text{O} = 2 \times 0.19 \text{ mol} = 0.38 \text{ mol}

  4. Determine the simplest whole-number ratio of C to H atoms.

    Divide the moles of each element by the smallest number of moles (0.1909 mol).
    For Carbon:

    0.1909 mol0.1909 mol≈1\frac{0.1909 \text{ mol}}{0.1909 \text{ mol}} \approx 1

    For Hydrogen:

    0.38 mol0.1909 mol≈1.99≈2\frac{0.38 \text{ mol}}{0.1909 \text{ mol}} \approx 1.99 \approx 2

    The ratio of C:H is approximately 1:2.

    The empirical formula of the hydrocarbon is CH2\boxed{\text{CH}_2}.

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