An organic compound contains 20.2% carbon, 6.50% hydrogen, 46.60% nitrogen and 26.709 the empirical formula of the
Answers
Answer:
With all these empirical formula problems, we assume that there are
100
⋅
g
of unknown compound, and work out the empirical formula appropriately:
Moles of carbon:
=
20.0
⋅
g
12.011
⋅
g
⋅
m
o
l
−
1
=
1.67
⋅
m
o
l
.
Moles of hydrogen:
=
6.66
⋅
g
1.00794
⋅
g
⋅
m
o
l
−
1
=
6.61
⋅
m
o
l
.
Moles of nitrogen:
=
47.33
⋅
g
14.01
⋅
g
⋅
m
o
l
−
1
=
3.38
⋅
m
o
l
.
And the balance of mass was oxygen:
Moles of oxygen:
=
(
100
−
20.0
−
6.66
−
47.33
)
⋅
g
15.999
⋅
g
⋅
m
o
l
−
1
=
1.63
⋅
m
o
l
.
And now we divide thru by the SMALLEST molar quantity, that of oxygen, to get the empirical formula:
C
:
1.67
⋅
m
o
l
1.63
⋅
m
o
l
≅
1
H
:
6.61
⋅
m
o
l
1.63
⋅
m
o
l
≅
4
N
:
3.38
⋅
m
o
l
1.63
⋅
m
o
l
≅
2
O
:
1.63
⋅
m
o
l
1.63
⋅
m
o
l
=
1
We thus get an empirical formula of
C
H
4
N
2
O
(and possibly, this is
urea
,
O
=
C
(
N
H
2
)
2
).
(Note that I approve of this problem (for what that's worth!), because an analyst WOULD never quote the
%
O
content. He/she would give you
%
C
,
%
H
,
%
N
,
and sometimes
%
X
, but oxygen analysis is not routinely performed.)
Answer link
Explanation:
CH2N is the empirical formula