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What is the empirical formula of a compound containing 40% carbon, 6.7% hydrogen

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Question: What is the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen?

Options:

  1. CH2O
  2. C2H4O2
  3. C3H6O3
  4. C4H8O4

Correct Answer: CH2O

Solution: 

Converting percentages to moles gives C: 40/12 = 3.33, H: 6.7/1 = 6.7, O: 53.3/16 = 3.33. The simplest ratio is 1:2:1, giving the empirical formula CH2O.

What is the empirical formula of a compound containing 40% carbon, 6.7% hydrogen

Practice Questions

Q1
What is the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen?
  1. CH2O
  2. C2H4O2
  3. C3H6O3
  4. C4H8O4

Questions & Step-by-Step Solutions

What is the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen?
  • Step 1: Start with the percentages of each element in the compound: 40% Carbon (C), 6.7% Hydrogen (H), and 53.3% Oxygen (O).
  • Step 2: Convert the percentages to grams. Assume you have 100 grams of the compound. This means you have 40 grams of C, 6.7 grams of H, and 53.3 grams of O.
  • Step 3: Convert grams to moles for each element using their atomic masses: Carbon (C) has an atomic mass of about 12 g/mol, Hydrogen (H) is about 1 g/mol, and Oxygen (O) is about 16 g/mol.
  • Step 4: Calculate the moles of each element: For Carbon, 40 grams / 12 g/mol = 3.33 moles; for Hydrogen, 6.7 grams / 1 g/mol = 6.7 moles; for Oxygen, 53.3 grams / 16 g/mol = 3.33 moles.
  • Step 5: Find the simplest ratio of the moles by dividing each mole value by the smallest number of moles calculated. The smallest number is 3.33.
  • Step 6: Divide each mole value by 3.33: C: 3.33/3.33 = 1, H: 6.7/3.33 ≈ 2, O: 3.33/3.33 = 1.
  • Step 7: Write the empirical formula using the simplest ratio: C1H2O1, which is commonly written as CH2O.
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