Prior to their phaseout in the 1980s, chemicals containing lead were commonly added to gasoline as anti-knocking agents. A 4.083 g sample of one such additive containing only lead, carbon, and hydrogen was burned in an oxygen-rich environment. The products of the combustion were 4.444 g CO, (g) and 2.274 g H,O(g).

Question

Determine the empirical formula of the gasoline additive.

Answer 1

To determine the empirical formula of the gasoline additive, we can follow these steps:

1. Calculate the moles of CO and H2O produced:

   • For CO (carbon monoxide): \[ \text{Molar mass of CO} = 12.01 , \text{g/mol (C)} + 16.00 , \text{g/mol (O)} = 28.01 , \text{g/mol} \] \[ \text{Moles of CO} = \frac{4.444 , \text{g}}{28.01 , \text{g/mol}} \approx 0.1585 , \text{mol} \]

   • For H2O (water): \[ \text{Molar mass of H2O} = 2 \times 1.008 , \text{g/mol (H)} + 16.00 , \text{g/mol (O)} = 18.016 , \text{g/mol} \] \[ \text{Moles of H2O} = \frac{2.274 , \text{g}}{18.016 , \text{g/mol}} \approx 0.1266 , \text{mol} \]

2. Determine the moles of carbon and hydrogen in the products:

   • Each mole of CO contains 1 mole of C, so the moles of C from CO is approximately \(0.1585 \).
   • Each mole of H2O contains 2 moles of H, so the moles of H from H2O is: \[ 2 \times 0.1266 \approx 0.2532 , \text{mol} \]

3. Calculate the mass of carbon and hydrogen:

   • Mass of Carbon: \[ \text{Mass of C} = 0.1585 , \text{mol} \times 12.01 , \text{g/mol} \approx 1.903 , \text{g} \]
   • Mass of Hydrogen: \[ \text{Mass of H} = 0.2532 , \text{mol} \times 1.008 , \text{g/mol} \approx 0.255 , \text{g} \]

4. Calculate the mass of lead:

   • Total mass of the additive = 4.083 g
   • Mass of Lead = Total mass - (Mass of C + Mass of H): \[ \text{Mass of Pb} = 4.083 , \text{g} - (1.903 , \text{g} + 0.255 , \text{g}) \approx 4.083 - 2.158 = 1.925 , \text{g} \]

5. Convert the mass of lead to moles:

   • Molar mass of Pb = 207.2 g/mol \[ \text{Moles of Pb} = \frac{1.925 , \text{g}}{207.2 , \text{g/mol}} \approx 0.0093 , \text{mol} \]

6. Calculate the mole ratio:

   • Moles of C = 0.1585
   • Moles of H = 0.2532
   • Moles of Pb = 0.0093

7. Divide by the smallest number of moles:

   • Smallest number of moles is approximately \(0.0093\): \[ \text{Ratio of C} = \frac{0.1585}{0.0093} \approx 17.01 \approx 17 \] \[ \text{Ratio of H} = \frac{0.2532}{0.0093} \approx 27.22 \approx 27 \] \[ \text{Ratio of Pb} = \frac{0.0093}{0.0093} = 1 \]

8. Empirical formula: The empirical formula can thus be written as: \[ \text{C}{17}\text{H}{27}\text{Pb} \]

Hence, the empirical formula of the gasoline additive is C17H27Pb.