Question

An organic compound contains 37.5% C, 12.5% H and the rest oxygen. What is the empirical formula of the compound?

• A. \( {CH}_4{O} \)
• B. \( {C}_2{H}_3{O} \)
• C. \( {CH}_3{O}_2 \)
• D. \( {C}_2{H}_4{O} \)
• E. \( {CH}_3{O} \)

Hint:

When determining the empirical formula, first convert the percentage composition to moles, then divide each by the smallest number of moles to get the simplest whole number ratio.

Answer 1

The Correct Option is A

The molecular composition of the compound is given as percentages of C, H, and oxygen. Let's first assume that we have 100 g of the compound.
Therefore:
- The mass of C = 37.5 g
- The mass of H = 12.5 g
- The mass of oxygen will be the remaining percentage, which is \( 100 - (37.5 + 12.5) = 50 \, {g} \)
Next, we calculate the moles of each element: \[ {Moles of C} = \frac{37.5}{12} = 3.125 \, {mol} \] \[ {Moles of H} = \frac{12.5}{1} = 12.5 \, {mol} \] \[ {Moles of O} = \frac{50}{16} = 3.125 \, {mol} \] Now, we divide each by the smallest number of moles (3.125): \[ {C:} \frac{3.125}{3.125} = 1 \] \[ {H:} \frac{12.5}{3.125} = 4 \] \[ {O:} \frac{3.125}{3.125} = 1 \] 
Thus, the empirical formula of the compound is \( {CH}_4{O} \).