A compound contains 12.76% carbon,21.3% of hydrogen and 85.11% bromine. what is molecular formula ?
Answers
Explanation:
As always with these problems, it is useful to assume
100
⋅
g
of unknown compound....and thus find an empirical formula...
Moles of carbon
=
12.8
⋅
g
12.011
⋅
g
⋅
m
o
l
−
1
=
1.066
⋅
m
o
l
.
Moles of hydrogen
=
2.1
⋅
g
1.00794
⋅
g
⋅
m
o
l
−
1
=
2.084
⋅
m
o
l
.
Moles of bromine
=
85.1
⋅
g
79.90
⋅
g
⋅
m
o
l
−
1
=
1.066
⋅
m
o
l
.
And we divide thru by the LOWEST molar quantity to give an empirical formula of.....
C
H
2
B
r
But we gots a molecular mass, and we know that the molecular formula is a whole number multiple of the empirical formula.
And so
187.9
⋅
g
⋅
m
o
l
−
1
=
n
×
(
12.011
+
2
×
1.00794
+
79.9
)
⋅
g
⋅
m
o
l
−
1
And thus
n
=
2
, and the molecular formula is
C
2
H
4
B
r
2
...
I prefer questions that quote actual microanalytical data.
C
2
H
4
B
r
2
is a liquid, and you would rarely be able to get combustion data on a liquid (the analyst would probably laugh at you!).