Question

An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be

• A. $CH_3 O$
• B. $CH_2 O$
• C. CHO
• D. $CH_4 O$

Answer 1

The Correct Option is A

To determine the empirical formula of the organic compound, we need to use the given percentages of carbon and hydrogen, and deduce the percentage of oxygen. The steps are as follows: 

1. Calculate the percentage of oxygen in the compound:
   • The given percentages are: Carbon (C) = 38.71%, Hydrogen (H) = 9.67%.
   • Since the compound contains only carbon, hydrogen, and oxygen, the percentage of oxygen (O) is obtained by subtracting the sum of carbon and hydrogen percentages from 100%:
2. Convert these percentages to grams (assuming 100 g of the compound):
   • Carbon: 38.71 g
   • Hydrogen: 9.67 g
   • Oxygen: 51.62 g
3. Determine the number of moles of each element:
   • Moles of Carbon: \(\frac{38.71}{12.01} \approx 3.22 \text{ moles}\) (using the atomic mass of carbon: 12.01 g/mol)
   • Moles of Hydrogen: \(\frac{9.67}{1.008} \approx 9.59 \text{ moles}\) (using the atomic mass of hydrogen: 1.008 g/mol)
   • Moles of Oxygen: \(\frac{51.62}{16.00} \approx 3.23 \text{ moles}\) (using the atomic mass of oxygen: 16.00 g/mol)
4. Determine the simplest whole-number ratio of the moles of each element:
   • The ratio of moles of C : H : O is approximately 3.22 : 9.59 : 3.23.
   • Divide each by the smallest number of moles (which is approximately 3.22):
     • Carbon: \(\frac{3.22}{3.22} = 1\)
     • Hydrogen: \(\frac{9.59}{3.22} \approx 3\)
     • Oxygen: \(\frac{3.23}{3.22} \approx 1\)
5. The empirical formula based on these whole-number ratios is \(CH_3O\).

Hence, the empirical formula of the compound is \(CH_3 O\).