Question #343ff

1 Answer
Sep 27, 2016

Let the molecular formula of HC be #C_xH_y#

And the balanced equation of the combustion reaction of the HC in oxygen is

#C_xH_y+(x+y/4)O_2(g)->xCO_2(g)+y/2H_2O(l)#

So by this equation the stochiometric ratio of #CO_2# and #H_2O# produced on combustion is
#=(xmol)/(y/2 mol)=(44xg)/(18/2yg)=(44x)/(9y)#
#"Where " 44g/"mol"" is the molar mass of "CO_2 and 18g/"mol" " is the molar mass of "H_2O#

But by the given data this ratio is

#(321mg)/(64mg)=321/64~~5#

So equating these two we get

#(44x)/(9y)=5#

#=>x/y=45/44~~1/1#

Since the ratio of number of atoms C and H in the HC molecule is #1:1# then we can easily say the Empirical formula of HC is #color(red)(CH)#