Answer:
ok
Step-by-step explanation:
Element %composition Mol. mass No. of C atoms Simplest Rounding
ratio off
C 82.76 12 6.89 1× 2 2
H 17.24 1 17.24 2.5× 2 5
Number of carbon atoms present =
12
82.76
=6.89
Number of hydrogen atoms present =
1
17.24
=17.24
simplest ratio =
6.89
6.89
=1
simplest ratio =
6.89
17.24
=2.5
∴ Empirical fromula is C
2
H
5
Empirical formula mass= 2×12+5 = 29
∴ V.D. = 29
Mol. mass = 2 × V.D.
=58
n=
Empiricalformulamass
Molecularmass
=
29
58
=2
∴ Molecular formula = C
4
H
10
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