Question #87de9

1 Answer
P dilip_k
Oct 1, 2016

#color(red)(C_2H_4O)#

Explanation:

Atomic mass of #H=1" g/mol"#
Atomic mass of #C=12" g/mol"#
Atomic mass of #O=16" g/mol"#
Molar mass of #CO_2=44" g/mol"#
Molar mass of #H_2O=18" g/mol"#

Mass of C in 7.59g #CO_2=(7.59g)/(44"g/mol")xx12"g/mol"=2.07g#

Mass of H in 3.11g #H_2O=(3.11g)/(18"g/mol")xx2"g/mol"=0.345g#

These two amounts of C and H have been obtained from 3.8g compound containing C ,H and O.

So the remaining (3.80-2.07-0.35)g=1.38g is the mass of Oxygen present in 3.8 g compound.

Hence dividing these masses of elements with respective atomic masses of elements and taking their ratio we get the ratio of number of atoms of C, H and O in the molecule of the compound as below

#C:H:O=2.07/12:0.345/1:1.38/16#
#=0.1725:0.345:0.08625#

#=(0.1725/0.08625):(0.345/0.08625):(0.08625/0.08625)#
#=2:4:1#

Hence empirical formula of the compound is #color(red)(C_2H_4O)#

Related questions
See all questions in Empirical and Molecular Formulas
Impact of this question
14335 views around the world
You can reuse this answer
Creative Commons License