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BITSAT 2019 Slot 2 Solved Paper
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© examsnet.com
Question : 78
Total: 150
In a compound
C
,
H
and
N
are present in
9
:
1
:
3.5
by weight. If molecular weight of compound is 108 , the molecular formula of compound is
C
2
H
6
N
2
C
3
H
4
N
C
6
H
8
N
2
C
9
H
12
N
3
Validate
Solution:
Element
Ratio (by weight)
Atomic weight
Mol Ratio
Simplest Mol Ratio
C
9
12
9
12
=
0.75
0.75
0.25
=
3
H
1
1
1
1
=
1.00
1.00
0.25
=
4
N
3.5
14
3.5
14
=
0.25
0.25
0.25
=
1
∴
Empirical formula
=
C
3
H
4
N
The empirical formula mass
=
3
×
12
+
4
×
1
+
14
=
54
g
Given, molecular mass
=
108
g
Thus,
n
=
108
54
=
2
Hence, molecular formula
=
(
Empirical formula
)
×
2
(
C
3
H
4
N
)
×
2
=
C
6
H
8
N
2
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