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

`CH_2O_3`

Step-by-step 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)=(3.25 g)/(1g)=3.25`

`n_(C)=(19.36 g)/(12g)=1.61`

`n_(O)=\fracc{77.39 g}{16g}=4.83`

Dividing by the smallest:

`n_(H)=(3.25)/(1.61)=2`

`n_(C)=(1.61)/(1.61)=1`

`n_(O)=\fracc{4.83}{1.61}=3`

The empirical formula will be:

`CH_2O_3`