Verified Answer :
- For the first compound (40.0% C, 6.67% H, 53.8% O), calculations yield an empirical formula of \(CH_2O\).
- With a molecular mass of 180, the ratio \(n = 180 / 30.03 \approx 6\) leads to a molecular formula of \(C_6H_{12}O_6\).
- For the hydrocarbon (85.9% C, 14.1% H), calculations yield an empirical formula of \(CH_2\).
- Final answers: \(\boxed{CH_2O, C_6H_{12}O_6; CH_2}\).
Explanation
- Empirical Formula of the First Compound
To determine the empirical formula for the first compound, we assume a sample mass of \(100 \text{ g}\). Given the weight percentages \(C=40.0\%\), \(H=6.67\%\), and \(O=53.8\%\), we have masses of \(40.0 \text{ g}\) of carbon, \(6.67 \text{ g}\) of hydrogen, and \(53.8 \text{ g}\) of oxygen. Converting these to moles using atomic masses (\(C \approx 12.01 \text{ g/mol}\), \(H \approx 1.008 \text{ g/mol}\), \(O \approx 16.00 \text{ g/mol}\)):
- Moles of C: \(40.0 \text{ g} / 12.01 \text{ g/mol} \approx 3.331 \text{ mol}\)
- Moles of H: \(6.67 \text{ g} / 1.008 \text{ g/mol} \approx 6.617 \text{ mol}\)
- Moles of O: \(53.8 \text{ g} / 16.00 \text{ g/mol} \approx 3.363 \text{ mol}\)
Dividing each by the smallest value (\(\approx 3.331\)):
- C: \(3.331 / 3.331 \approx 1\)
- H: \(6.617 / 3.331 \approx 1.99 \approx 2\)
- O: \(3.363 / 3.331 \approx 1.01 \approx 1\)
The empirical formula is \(CH_2O\).
- Molecular Formula of the First Compound
The empirical formula mass of \(CH_2O\) is \((1 \times 12.01) + (2 \times 1.008) + (1 \times 16.00) \approx 30.03 \text{ g/mol}\). Given the molecular mass is \(180 \text{ g/mol}\), we find the integer ratio \(n\):
\(n = 180 / 30.03 \approx 6\).
Multiplying the subscripts of the empirical formula by \(n=6\), we get the molecular formula:
\(C_{(1 \times 6)}H_{(2 \times 6)}O_{(1 \times 6)} = C_6H_{12}O_6\).
- Empirical Formula of the Hydrocarbon
For the hydrocarbon, the percentage of hydrogen is \(100\% - 85.9\% = 14.1\%\). Assuming a \(100 \text{ g}\) sample:
- Moles of C: \(85.9 \text{ g} / 12.01 \text{ g/mol} \approx 7.152 \text{ mol}\)
- Moles of H: \(14.1 \text{ g} / 1.008 \text{ g/mol} \approx 13.988 \text{ mol}\)
Dividing by the smallest value (\(7.152\)):
- C: \(7.152 / 7.152 = 1\)
- H: \(13.988 / 7.152 \approx 1.956 \approx 2\)
The empirical formula is \(CH_2\).
Examples
Empirical and molecular formulas are fundamental in chemistry for identifying substances. For example, in the food industry, knowing the chemical formula of a sugar like glucose (\(C_6H_{12}O_6\)) allows scientists to understand its energy content and metabolic properties. Similarly, identifying hydrocarbons helps engineers understand the properties of fuels and plastics, ensuring they are designed and used safely and efficiently.