Correct Answer - A
`underset(0.858g)([C,H])+O_(2)tounderset(2.63)(CO_(2))+underset(1.28)(H_(2)O)`
%age C`= (12)/(44)xx(2.63)/(0.853)xx100=83.59`
`%age H = (2)/(18)xx(2.63)/(0.858)xx100=16.57`
`:. E.F. = C_(3)H_(7)`
`:.` E.F. mass (lowest mol mass) `= 12xx3+7xx1 = 43`