Verified Answer :
- For Part 1, we determine the mole ratios of C, H, and O, leading to an empirical formula of \(\boxed{CH_2O}\).
- Given the molecular mass of 180, we multiply the empirical formula by the factor \(n=6\) to find the molecular formula \(\boxed{C_6H_{12}O_6}\).
- For Part 2, we calculate the moles of C and H in the hydrocarbon, yielding an empirical formula of \(\boxed{CH_2}\).
Explanation
- Analyzing and Solving Part 1: Empirical Formula
For Part (1), we have a compound with a composition by weight of \(C = 40.0 \%\), \(H = 6.67 \%\), and \(O = 53.8 \%\). To find the empirical formula, we assume a \(100 \text{ g}\) sample. This gives us \(40.0 \text{ g}\) of Carbon, \(6.67 \text{ g}\) of Hydrogen, and \(53.8 \text{ g}\) of Oxygen. Using the atomic weights \(C \approx 12.01 \text{ g/mol}\), \(H \approx 1.008 \text{ g/mol}\), and \(O \approx 16.00 \text{ g/mol}\), we convert these masses into moles:
- Moles of C: \(\frac{40.0 \text{ g}}{12.01 \text{ g/mol}} \approx 3.331 \text{ mol}\)
- Moles of H: \(\frac{6.67 \text{ g}}{1.008 \text{ g/mol}} \approx 6.617 \text{ mol}\)
- Moles of O: \(\frac{53.8 \text{ g}}{16.00 \text{ g/mol}} \approx 3.363 \text{ mol}\)
Dividing each of these by the smallest value (\(3.331\)), we get ratios of approximately \(1 : 1.987 : 1.010\), which round to the integer ratio \(1:2:1\). Thus, the empirical formula is \(CH_2O\).
- Analyzing and Solving Part 1: Molecular Formula
Now we calculate the molecular formula for Part (1). The empirical formula \(CH_2O\) has an empirical molar mass of \(12.01 + 2 \times 1.008 + 16.00 = 30.026 \text{ g/mol}\). Given the molecular mass is \(180 \text{ g/mol}\), we find the multiplier \(n\) by dividing the molecular mass by the empirical mass:
\(n = \frac{180 \text{ g/mol}}{30.026 \text{ g/mol}} \approx 5.995\).
Rounding \(n\) to the nearest integer, we get \(n = 6\). Therefore, the molecular formula is \(6 \times (CH_2O) = C_6H_{12}O_6\).
- Analyzing and Solving Part 2: Hydrocarbon Empirical Formula
For Part (2), a hydrocarbon contains \(85.9 \%\) Carbon by weight. Since a hydrocarbon contains only Carbon and Hydrogen, the percentage of Hydrogen is \(100\% - 85.9\% = 14.1\%\). Assuming a \(100 \text{ g}\) sample, we have \(85.9 \text{ g}\) of Carbon and \(14.1 \text{ g}\) of Hydrogen. Converting to moles:
- Moles of C: \(\frac{85.9 \text{ g}}{12.01 \text{ g/mol}} \approx 7.152 \text{ mol}\)
- Moles of H: \(\frac{14.1 \text{ g}}{1.008 \text{ g/mol}} \approx 13.988 \text{ mol}\)
Dividing by the smallest molar value (\(7.152\)):
- C: \(\frac{7.152}{7.152} = 1\)
- H: \(\frac{13.988}{7.152} \approx 1.956\)
Rounding to the nearest whole numbers, we get the ratio \(1:2\). Thus, the empirical formula is \(CH_2\).
Examples
Understanding empirical and molecular formulas is crucial in chemistry for identifying unknown substances. For example, in a lab setting, when a researcher isolates a new organic compound, they first perform elemental analysis to determine the percentage by mass of each element. This data allows them to deduce the compound's structure, which is essential for synthesizing drugs or creating new materials. It serves as the foundation for identifying molecular components in everything from pharmaceuticals to household products.