Question 26 of 34intermediate🔧 ApplyNumerical3 marks

An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.

Correct Answer

The mass of carbon dioxide produced is 0.506 g and the mass of water produced is 0.0864 g.

Exercise: EXERCISES | Q: 8.32
For More Understanding

Explanation

Using the standard formulas for calculating the percentage of carbon and hydrogen from the masses of carbon dioxide and water produced during combustion, we can rearrange the equations to solve for the masses of the products. The percentage of carbon is given by 12×m2×10044×m\frac{12 \times m_2 \times 100}{44 \times m} and the percentage of hydrogen is given by 2×m1×10018×m\frac{2 \times m_1 \times 100}{18 \times m}, where mm is the mass of the organic compound, m1m_1 is the mass of water, and m2m_2 is the mass of carbon dioxide.

Solution Steps

  1. Step 1: Identify the given values Percentage of carbon = 69% Percentage of hydrogen = 4.8% Mass of organic compound, mm = 0.20 g

  2. Step 2: Calculate the mass of carbon dioxide (m2m_2) Using the formula: Percentage of carbon = 12×m2×10044×m\frac{12 \times m_2 \times 100}{44 \times m}

    69=12×m2×10044×0.2069 = \frac{12 \times m_2 \times 100}{44 \times 0.20} 69=1200×m28.869 = \frac{1200 \times m_2}{8.8}

    m2=69×8.81200=607.21200=0.506m_2 = \frac{69 \times 8.8}{1200} = \frac{607.2}{1200} = 0.506 g

  3. Step 3: Calculate the mass of water (m1m_1) Using the formula: Percentage of hydrogen = 2×m1×10018×m\frac{2 \times m_1 \times 100}{18 \times m}

    4.8=2×m1×10018×0.204.8 = \frac{2 \times m_1 \times 100}{18 \times 0.20} 4.8=200×m13.64.8 = \frac{200 \times m_1}{3.6}

    m1=4.8×3.6200=17.28200=0.0864m_1 = \frac{4.8 \times 3.6}{200} = \frac{17.28}{200} = 0.0864 g

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