Solution Found!
Answer: Leaded gasoline contains an additive to prevent
Chapter 3, Problem 148P(choose chapter or problem)
Leaded gasoline contains an additive to prevent engine "knocking." On analysis, the additive compound is found to contain carbon, hydrogen, and lead \((\mathrm{Pb})\) (hence, "leaded gasoline"). When \(51.36 \mathrm{~g}\) of this compound are burned in an apparatus such as that shown in Figure \(3.6,55.90 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(28.61 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. Determine the empirical formula of the gasoline additive.
Questions & Answers
QUESTION:
Leaded gasoline contains an additive to prevent engine "knocking." On analysis, the additive compound is found to contain carbon, hydrogen, and lead \((\mathrm{Pb})\) (hence, "leaded gasoline"). When \(51.36 \mathrm{~g}\) of this compound are burned in an apparatus such as that shown in Figure \(3.6,55.90 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(28.61 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. Determine the empirical formula of the gasoline additive.
ANSWER:Step 1 of 2
The goal of the problem is to find the empirical formula of the gasoline additive.
Given:
Mass of the gasoline \(=51.36 \mathrm{~g}\)
Mass of \(\mathrm{CO}_{2}=55.90 \mathrm{~g}\) of \(\mathrm{CO}_{2}\)
Mass of \(\mathrm{H}_{2} \mathrm{O}=28.61 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\)
To determine the empirical formula, we should know the mass or number of moles in each element. So, let's Calculate the mass of \(\mathrm{C}\) in \(\mathrm{CO}_{2}\), and the mass of \(\mathrm{H}\) in \(\mathrm{H}_{2} \mathrm{O}\).
Mass of \(\mathrm{C}\) in \(\mathrm{CO}_{2} \mathrm{~F}\)
Molar mass of \(\mathrm{CO}_{2}=44.01 \mathrm{~g}\)
Molar mass of \(\mathrm{C}=12.01 \mathrm{~g}\)
\(=55.90 \mathrm{~g} \mathrm{CO}_{2} \times \frac{1 \mathrm{molCO}_{2}}{44.01 \mathrm{gCO}_{2}} \times \frac{1 \mathrm{molC}}{1 \mathrm{molCO}_{2}} \times \frac{12.015 \mathrm{gC}}{1 \mathrm{molC}}\)
Mass of \(\mathrm{C}\) in \(\mathrm{CO}_{2}:\)
Molar mass of \(\mathrm{H}_{2} \mathrm{O}=18.015 \mathrm{~g}\)
Molar mass of \(\mathrm{H}=1.0079 \mathrm{~g}\)
\(\begin{aligned}=28.61 \mathrm{~g} \mathrm{CO}_{2} \times \frac{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}{18.015 \mathrm{gH}_{2} \mathrm{O}} \times \frac{2 \mathrm{~mol} \mathrm{H}}{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}} \times \frac{1.0079 \mathrm{gH}}{1 \mathrm{~mol} \mathrm{H}} \\=3.201 \mathrm{~g} \mathbf{H}\end{aligned}\)
Given that the compound contains \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{Pb}\). We know the mass of \(\mathrm{C}\) and \(\mathrm{H}\). The mass of \(\mathrm{Pb}\) is calculated by taking the difference.
\(\begin{aligned}\Rightarrow \text { mass } \mathrm{C}+\text { mass } \mathrm{H}+\text { mass of } \mathrm{Pb}=51.36 \mathrm{~g} \\\Rightarrow 15.254 \mathrm{~g}+3.201 \mathrm{~g}+\text { mass of } \mathrm{Pb}=51.36 \mathrm{~g} \\ \Rightarrow \text { mass of } \mathrm{Pb}=51.36 \mathrm{~g}-18.455 \mathrm{~g}\end{aligned}\)
\(\Rightarrow \text { mass of } \mathrm{Pb}=32.905 \mathrm{~g} \mathrm{~Pb}\)