To determine the empirical formula of the gasoline additive containing lead (Pb), carbon (C), and hydrogen (H), we need to perform the following steps:
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Calculate the moles of CO2 and H2O produced:
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For CO2: \[ \text{Molar mass of CO2} = 12.01 , (\text{C}) + 2 \times 16.00 , (\text{O}) = 44.01 , \text{g/mol} \] \[ \text{Moles of CO2} = \frac{4.444 , \text{g}}{44.01 , \text{g/mol}} \approx 0.1010 , \text{mol} \]
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For H2O: \[ \text{Molar mass of H2O} = 2 \times 1.008 , (\text{H}) + 16.00 , (\text{O}) = 18.016 , \text{g/mol} \] \[ \text{Moles of H2O} = \frac{2.274 , \text{g}}{18.016 , \text{g/mol}} \approx 0.1265 , \text{mol} \]
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Calculate the moles of carbon and hydrogen in the products:
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For carbon from CO2: Since each mole of CO2 contains 1 mole of C: \[ \text{Moles of C} = \text{Moles of CO2} = 0.1010 , \text{mol} \]
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For hydrogen from H2O: Since each mole of H2O contains 2 moles of H: \[ \text{Moles of H} = 2 \times \text{Moles of H2O} = 2 \times 0.1265 , \text{mol} = 0.2530 , \text{mol} \]
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Determine the total mass of carbon and hydrogen:
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Mass of carbon: \[ \text{Mass of C} = \text{Moles of C} \times \text{Molar mass of C} = 0.1010 , \text{mol} \times 12.01 , \text{g/mol} \approx 1.213 , \text{g} \]
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Mass of hydrogen: \[ \text{Mass of H} = \text{Moles of H} \times \text{Molar mass of H} = 0.2530 , \text{mol} \times 1.008 , \text{g/mol} \approx 0.255 , \text{g} \]
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Calculate the mass of lead (Pb):
The total mass of the sample is 4.083 g. Therefore, the mass of lead can be determined by subtracting the masses of carbon and hydrogen from the total mass:
\[ \text{Mass of Pb} = 4.083 , \text{g} - (1.213 , \text{g} + 0.255 , \text{g}) \approx 4.083 , \text{g} - 1.468 , \text{g} \approx 2.615 , \text{g} \]
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Calculate the moles of lead:
The molar mass of lead is approximately 207.2 g/mol. \[ \text{Moles of Pb} = \frac{2.615 , \text{g}}{207.2 , \text{g/mol}} \approx 0.0126 , \text{mol} \]
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Determine mole ratios:
Now we have the following moles:
- C: 0.1010 mol
- H: 0.2530 mol
- Pb: 0.0126 mol
To find the simplest whole number ratio, we divide each by the smallest number of moles (0.0126 mol):
- For C: \[ \frac{0.1010}{0.0126} \approx 8.02 \approx 8 \]
- For H: \[ \frac{0.2530}{0.0126} \approx 20.08 \approx 20 \]
- For Pb: \[ \frac{0.0126}{0.0126} = 1 \]
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Empirical formula:
Using the ratios calculated, we get the empirical formula: \[ \text{Empirical formula} = \text{C}8\text{H}{20}\text{Pb} \]
Thus, the empirical formula of the gasoline additive is \( \text{C}8\text{H}{20}\text{Pb} \).