What is the empirical formula of a compound containing 40.00% carbon, 6.67% hydrogen, and 53.33%
What is the empirical formula of a compound containing 40.00% carbon, 6.67% hydrogen, and 53.33% oxygen by mass?
- A. C\(_4\)H\(_8\)O\(_4\)
- B. CH\(_2\)O
- C. C\(_2\)H\(_4\)O\(_2\)
- D. C\(_3\)H\(_6\)O\(_3\)
Correct Answer: B. CH\(_2\)O
Explanation
To determine the empirical formula, we first need to convert the percentages to moles by assuming a convenient mass for the sample. Let's assume we have 100 grams of the compound.
Mass of carbon = 40.00 grams
Mass of hydrogen = 6.67 grams
Mass of oxygen = 53.33 grams
Now, we calculate the moles of each element:
Moles of carbon = 40.00 g / molar mass of carbon = 40.00 g / 12.01 g/mol ≈ 3.33 mol
Moles of hydrogen = 6.67 g / molar mass of hydrogen = 6.67 g / 1.01 g/mol ≈ 6.60 mol
Moles of oxygen = 53.33 g / molar mass of oxygen = 53.33 g / 16.00 g/mol ≈ 3.33 mol
Now, we find the smallest whole-number ratio of the moles:
C ≈ 3.33 / 3.33 ≈ 1
H ≈ 6.60 / 3.33 ≈ 2
O ≈ 3.33 / 3.33 ≈ 1
Thus, the empirical formula is CH2O.
To determine the empirical formula, we first need to convert the percentages to moles by assuming a convenient mass for the sample. Let's assume we have 100 grams of the compound.
Mass of carbon = 40.00 grams
Mass of hydrogen = 6.67 grams
Mass of oxygen = 53.33 grams
Now, we calculate the moles of each element:
Moles of carbon = 40.00 g / molar mass of carbon = 40.00 g / 12.01 g/mol ≈ 3.33 mol
Moles of hydrogen = 6.67 g / molar mass of hydrogen = 6.67 g / 1.01 g/mol ≈ 6.60 mol
Moles of oxygen = 53.33 g / molar mass of oxygen = 53.33 g / 16.00 g/mol ≈ 3.33 mol
Now, we find the smallest whole-number ratio of the moles:
C ≈ 3.33 / 3.33 ≈ 1
H ≈ 6.60 / 3.33 ≈ 2
O ≈ 3.33 / 3.33 ≈ 1
Thus, the empirical formula is CH2O.