Question

Question #cb68a

Answer 1

Butane. `C_4H_10`

Explanation:

We use the reaction equation to find the moles of each substance from the given masses.
`C_xH_n + O_2 → xCO_2 + (n/2)*H_2O`

`CO_2 = 9.6g`; `H_2O = 4.9g` ; `(9.6/44) = 0.22 mol CO_2`; `(4.9/18) = 0.27 mol H_2O`
So, the molar ratios of x and n are 0.22 and 0.54 respectively, for relative molar ratios of 0.29 and 0.71. That ration means that `n = (0.71/0.29)*x ; n = 2.45*x`
The hydrocarbon formula is thus `(12*x) + (1*n) = 58`; `(12*x) + (1*(2.45*x)) = 58`
`(14.45*x) = 58`; `x = 4`; `n = 10`
The formula is thus `C_4H_10`.
This is the common form of an alkane `C_nH_(2n+2)`, so it is butane.