Question #68923

a pure gass which contains 14.3% hydrogen and 85.7% carbon of total mass.its density is 2.5 gram per leter at 0 degree centigrate and 1 atm.what is it molecular formula?

Expert's answer

From the equation of state for perfect gas

PV=m/M RT

We obtain the expression for the density of gas

ρ=m/V=PM/RT.

Therefore the molar mass of gas is

M=ρRT/P=(2.5×8.3×273)/(1.01×〖10〗^5 )=56 g/mol.

Because

M(H)=1 g/mol,M(C)=12 g/mol.

We obtain

14.3/85.7×12=2.

So the number of hydrogen atoms is twice of carbon.

Because the molar mass of gas is 56, we obtain equation for the number of carbon atoms x

2×1×x+12×x=56 ⇒ x=4.

So the molecular formula of the gas is

C_4 H_8

Answers: C_4 H_8

PV=m/M RT

We obtain the expression for the density of gas

ρ=m/V=PM/RT.

Therefore the molar mass of gas is

M=ρRT/P=(2.5×8.3×273)/(1.01×〖10〗^5 )=56 g/mol.

Because

M(H)=1 g/mol,M(C)=12 g/mol.

We obtain

14.3/85.7×12=2.

So the number of hydrogen atoms is twice of carbon.

Because the molar mass of gas is 56, we obtain equation for the number of carbon atoms x

2×1×x+12×x=56 ⇒ x=4.

So the molecular formula of the gas is

C_4 H_8

Answers: C_4 H_8

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