Question #2612f

1 Answer
anor277
Jul 31, 2017

I think you have #CH_2#.....

Explanation:

#2.65*g# of a hydrocarbon, #C_xH_y# was burnt in air to give #8.33*g# #CO_2# and #3.41*g# water.........

#"Moles of carbon dioxide"=(8.33*g)/(44.01*g*mol^-1)=0.189*mol#

#"Moles of water"=(3.41*g)/(18.01*g*mol^-1)=0.189*mol#, i.e. #0.189*molxx2H#

The empirical formula is thus #CH_2#......

We need a molecular weight determination in order to assess the molecular formula.

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