C
H
%
82 17.8
/AR
6.83
17.8
/small
no.
1 2.5
Therefore empirical formula is C2H5.
To get the molecular formala
58 / ( 2X 12+ 5 X1 ) = 2
Therefore, molecular formula is 2 (C2H5) = C4H10
B is butane so A should be butene
Oh, so you have to multiply to get a whole number. HAHA. I
thought you have to round up or round down. Thanks. By the way,
have you had any idea what 3 other substances present in chimney
gas besides CO2, SO2 and oxides of nitrogen? I think CO is one of
them but what about other substances?
Thanks