Question

Determine the true formula of the compound whose molecular mass is $90.0$ grams and contains $40.0\%$ carbon, $6.67\%$ hydrogen, and $53.33\%$ oxygen.

• A. $C_2{H}_2{O}_4$
• B. $C{H}_2{O}_4$
• C. $C_3{H}_{6}O$
• D. $C_{3}H{O}_3$
• E. $C_3{H}_6{O}_3$

Answer 1

Answer: (E)

$C_3{H}_6{O}_3$

Mass of $C=\frac{40.0}{100}\times 90.0=36$ g
The atomic mass of C is 12 g/mol. The number of moles of C $=\frac{36}{12}=3$.
Mass of $H=\frac{6.67}{100}\times 90.0=6$ g
The atomic mass of H is 1 g/mol. The number of moles of H $=\frac{6}{1}=6$.
Mass of $O=\frac{53.33}{100}\times 90.0=48$ g
The atomic mass of O is 16 g/mol. The number of moles of O $=\frac{48}{16}=3$.
Hence, the true formula of the compound is $C_3{H}_6{O}_3$.