Answer:
\(CH_2O_3\)
Explanation:
Taking a calculation basis of 100 grams, you have:
\(m_{H}=3.25 g\)
\(m_{C}=19.36 g\)
\(m_{O}=77.39 g\)
Dividing by their atomic weights:
\(n_{H}=\frac{3.25 g}{1g}=3.25\)
\(n_{C}=\frac{19.36 g}{12g}=1.61\)
\(n_{O}=\fracc{77.39 g}{16g}=4.83\)
Dividing by the smallest:
\(n_{H}=\frac{3.25}{1.61}=2\)
\(n_{C}=\frac{1.61}{1.61}=1\)
\(n_{O}=\fracc{4.83}{1.61}=3\)
The empirical formula will be:
\(CH_2O_3\)
