Question

Calculate the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

• A. \( \mathrm{CH_2O} \)
• B. \( \mathrm{C_2H_4O} \)
• C. \( \mathrm{C_3H_6O_3} \)
• D. \( \mathrm{CH_4O} \)

Hint:

To find empirical formula: 1. Convert % to grams. 2. Divide each by atomic mass. 3. Divide by smallest value to get ratio. 4. Use ratio to get subscripts.

Answer 1

The Correct Option is A

The empirical formula of a compound is determined by converting the percentage composition of each element to moles, then simplifying the ratio to the smallest whole numbers. Here are the steps to calculate the empirical formula for a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen. 

1. Convert percentages to grams: Assume 100 g of compound. Thus, you have 40 g of carbon, 6.7 g of hydrogen, and 53.3 g of oxygen.
2. Convert grams to moles:
   • Carbon: \( \frac{40}{12.01} = 3.33 \, \text{moles} \)
   • Hydrogen: \( \frac{6.7}{1.01} = 6.63 \, \text{moles} \)
   • Oxygen: \( \frac{53.3}{16.00} = 3.33 \, \text{moles} \)
3. Determine the simplest mole ratio: Divide the number of moles of each element by the smallest number of moles calculated:
   • Carbon: \( \frac{3.33}{3.33} = 1 \)
   • Hydrogen: \( \frac{6.63}{3.33} \approx 2 \)
   • Oxygen: \( \frac{3.33}{3.33} = 1 \)
4. Write the empirical formula: Using the ratios obtained, the empirical formula is \( \mathrm{CH_2O} \).

Answer 2

Step 1: Assume 100 g of the compound. Then, the masses of elements are: Carbon = 40 g
Hydrogen = 6.7 g
Oxygen = 53.3 g Step 2: Convert mass to moles: \[ \text{C: } \frac{40}{12} = 3.33 \quad \text{mol} \] \[ \text{H: } \frac{6.7}{1} = 6.7 \quad \text{mol} \] \[ \text{O: } \frac{53.3}{16} = 3.33 \quad \text{mol} \] Step 3: Divide by smallest mole value (3.33): \[ \text{C: } \frac{3.33}{3.33} = 1 \] \[ \text{H: } \frac{6.7}{3.33} \approx 2 \] \[ \text{O: } \frac{3.33}{3.33} = 1 \] Empirical formula: \( \mathrm{CH_2O} \)