Question

What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass?

Answer 1

CH2O3

C=19.36\12    =  1.61
H=3.25\1.008 =3.22
O=77.39\16=4.83

then dived the moles with the smallest mole 

C=1.61\1.61=1
H=3.22\1.61=2
O=4.83\1.61=3

E.F
CH2O3

Answer 2

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:

$CH_2O_3$