20) Anorganic compound is found to contain c= 52.14% 24=13.18%
& rest being 02. Its molecular man is 46 glma. find its empirical
& molecular formulas.
Answers
Explanation:
Explanation:
AS with all these problems we assume a starting mass of
100
⋅
g
with respect to the unknown.
We then calculate the molar composition with respect to its constituent elements:
Moles of carbon
=
52.14
⋅
g
12.011
⋅
g
⋅
m
o
l
−
1
=
4.34
⋅
m
o
l
Moles of hydrogen
=
13.15
⋅
g
1.00794
⋅
g
⋅
m
o
l
−
1
=
13.04
⋅
m
o
l
Moles of oxygen
=
34.73
⋅
g
16.00
⋅
g
⋅
m
o
l
−
1
=
2.17
⋅
m
o
l
Note that NORMALLY an analyst would NEVER give you percentage oxygen, as its analysis is non-routine. Normally you would be expected to work it out by difference. An analyst would provide
%
C
;
%
H
;
%
N
, whose analysis is routine, and the missing percentage would be assumed to be due to oxygen.
And now we normalize these molar quantities by DIVIDING THRU by the smallest molar quantities, i.e. dividing thru by
moles of oxygen
....
And so we gets the
empirical formula
, the simplest ratio defining constituent atoms in a species as...
C
4.34
⋅
m
o
l
2.17
⋅
m
o
l
H
13.04
⋅
m
o
l
2.17
⋅
m
o
l
O
2.17
⋅
m
o
l
2.17
⋅
m
o
l
=
C
2
H
6
O
.
Now it is a fact that the
molecular formula
is always a whole number multiple of
empirical formula
.
And thus
molecular formula
≡
{
empirical formula
}
×
n
.
And so we use the quoted data and the atomic masses of each constituent element....and solve for
n
.
46.06
⋅
g
⋅
m
o
l
−
1
=
n
×
{
2
×
12.011
+
6
×
1.00794
+
16.00
}
⋅
g
⋅
m
o
l
−
1
Clearly
n
=
1
, and the
molecular formula
is the same as the
empirical formula
, i.e.
C
2
H
6
O
...
NOTE:-
Note that normally, if you presented a liquid to an analyst (and this LOW molecular mass species would definitely be a liquid; dimethyl ether, or ethanol), he would tell you to take a running jump, and might even use ruder terms. The question is not chemically reasonable.