Question #cb68a

1 Answer
Apr 22, 2017

Answer:

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.