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Combustion analysis of an unknown compound containing only carbon and hydrogen produced 4.554 g of CO₈ and 2.322 g of H₈O. What is the empirical formula of the compound?

-CH₈
-C₄H₁₀
-C₅H₈
-C₈H₅

User HeyZiko
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Final answer:

To find the empirical formula of the compound from the combustion analysis, we calculate the moles of carbon and hydrogen from the produced CO₂ and H₂O and then determine the simplest whole number ratio of elements. The result shows a CH₃ empirical formula.

Step-by-step explanation:

To determine the empirical formula of a compound from combustion analysis, first, we calculate the molar amounts of carbon and hydrogen using the provided masses of carbon dioxide (CO₂) and water (H₂O). Each mole of CO₂ contains one mole of carbon, and each mole of H₂O contains two moles of hydrogen.

Given that the sample produced 4.554 g of CO₂ and 2.322 g of H₂O:

  • Moles of carbon = Mass of CO₂ / Molar mass of CO₂ = 4.554 g / 44.01 g/mol ≈ 0.1035 mol
  • Moles of hydrogen = (Mass of H₂O x 2) / Molar mass of H₂O = (2.322 g x 2) / 18.02 g/mol ≈ 0.2577 mol

To find the simplest whole number ratio, divide the moles of each element by the smallest number of moles calculated:

  • Carbon to hydrogen ratio = 0.1035 mol C / 0.1035 mol : 0.2577 mol H / 0.1035 mol = 1 : 2.49 ≈ 1 : 2.5

Since we are looking for the simplest whole number ratio, we will round 2.5 to the nearest whole number, which is 3. This gives us a carbon to hydrogen ratio of 1:3, meaning for every one carbon atom, there are three hydrogen atoms. Therefore, the empirical formula is CH₃.

User Jatawn
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