Answered

  1. A compound has the following composition by weight: $C=40.0\%$, $H=6.67\%$, $O=53.3\%$. Calculate the empirical formula of the compound. If its molecular mass is 180, find its molecular formula.


  1. Calculate the empirical formula of the hydrocarbon containing $85.9\%$ carbon.

Verified Answer :

  • For compound (1), the molar ratio of C, H, and O is approximately 1:2:1, resulting in an empirical formula of \(CH_2O\).
  • The molecular mass of 180 divided by the empirical formula mass of 30 gives a multiplier of 6, leading to a molecular formula of \(\boxed{C_6H_{12}O_6}\).
  • For compound (2), the hydrocarbon has a molar ratio of C to H of approximately 1:2, leading to an empirical formula of \(\boxed{CH_2}\).

Explanation

  1. Compound (1) Empirical and Molecular Formulas

For part (1), we have a compound with 40.0% C, 6.67% H, and 53.3% O by weight. To find the empirical formula, we assume a 100 g sample:

  • Mass of C = 40.0 g, Mass of H = 6.67 g, Mass of O = 53.3 g.
  • Moles of C = \(40.0 / 12.01 \approx 3.33\) mol.
  • Moles of H = \(6.67 / 1.008 \approx 6.62\) mol.
  • Moles of O = \(53.3 / 16.00 \approx 3.33\) mol.

Dividing these mole values by the smallest value (3.33), we get the molar ratio:

  • C : H : O = \(1 : 1.99 : 1 \approx 1 : 2 : 1\).

Thus, the empirical formula is \(CH_2O\). The empirical formula mass is \(12.01 + 2(1.008) + 16.00 = 30.03\) g/mol. With a molecular mass of 180, the ratio is \(180 / 30.03 \approx 6\). Therefore, the molecular formula is \(6 \times CH_2O = C_6H_{12}O_6\).

  1. Compound (2) Empirical Formula

For part (2), we have a hydrocarbon with 85.9% C. Since it is a hydrocarbon, the remainder is H:

  • Percentage of H = \(100\% - 85.9\% = 14.1\%\).

Assuming a 100 g sample:

  • Mass of C = 85.9 g, Mass of H = 14.1 g.
  • Moles of C = \(85.9 / 12.01 \approx 7.15\) mol.
  • Moles of H = \(14.1 / 1.008 \approx 13.99\) mol.

Dividing by the smallest value (7.15):

  • Ratio of C = \(7.15 / 7.15 = 1\).
  • Ratio of H = \(13.99 / 7.15 \approx 1.96 \approx 2\).

Thus, the empirical formula is \(CH_2\).

Examples

Empirical formulas are crucial in chemical analysis for identifying unknown substances. For example, in environmental chemistry, scientists analyze pollutants collected from air filters. By determining the percentage composition of elements through mass spectrometry, they can calculate the empirical formula to identify the pollutant and assess its potential environmental impact, helping regulatory agencies establish safety guidelines. Understanding this concept is fundamental for quality control in pharmaceuticals and food safety.

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