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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?

Respuesta :

alreem4
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
Giangiglia

Answer:

[tex]CH_2O_3[/tex]

Explanation:

Taking a calculation basis of 100 grams, you have:

[tex]m_{H}=3.25 g[/tex]

[tex]m_{C}=19.36 g[/tex]

[tex]m_{O}=77.39 g[/tex]

Dividing by their atomic weights:

[tex]n_{H}=\frac{3.25 g}{1g}=3.25[/tex]

[tex]n_{C}=\frac{19.36 g}{12g}=1.61[/tex]

[tex]n_{O}=\fracc{77.39 g}{16g}=4.83[/tex]

Dividing by the smallest:

[tex]n_{H}=\frac{3.25}{1.61}=2[/tex]

[tex]n_{C}=\frac{1.61}{1.61}=1[/tex]

[tex]n_{O}=\fracc{4.83}{1.61}=3[/tex]

The empirical formula will be:

[tex]CH_2O_3[/tex]

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