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Sagot :
The empirical formula is KCO2
Explanation:
As with all these problems, we assume a
100
⋅
g
mass of unknown compound, and then we work out the molar quantity:
Moles of potassium
=
47.0
⋅
g
39.10
⋅
g
⋅
m
o
l
−
1
=
1.20
⋅
m
o
l
Moles of carbon
=
14.5
⋅
g
12.011
⋅
g
⋅
m
o
l
−
1
=
1.21
⋅
m
o
l
Moles of oxygen
=
38.5
⋅
g
16.0
⋅
g
⋅
m
o
l
−
1
=
2.41
⋅
m
o
l
We divide thru by the smallest molar quantity to give the empirical formula:
K
C
O
2
.
Now the molecular formula is always a whole number of the empirical formula:
i.e.
molecular formula
=
n
×
empirical formula
And thus with the molecular mass, we can solve for
n
.
166.2
⋅
g
⋅
m
o
l
−
1
=
n
×
(
39.1
+
12.011
+
2
×
16.00
)
⋅
g
⋅
m
o
l
−
1
166.2
⋅
g
⋅
m
o
l
−
1
=
n
×
(
83.1
)
⋅
g
⋅
m
o
l
−
1
Clearly,
n
=
2
, and the
molecular formula
=
K
2
C
2
O
4
The compound is LIKELY the potassium salt of oxalic acid,
K
+
−
O
(
O
=
)
C
−
C
(
=
O
)
O
−
K
+
, i.e.
potassium oxalate.
Answer linkExplanation:
As with all these problems, we assume a
100
⋅
g
mass of unknown compound, and then we work out the molar quantity:
Moles of potassium
=
47.0
⋅
g
39.10
⋅
g
⋅
m
o
l
−
1
=
1.20
⋅
m
o
l
Moles of carbon
=
14.5
⋅
g
12.011
⋅
g
⋅
m
o
l
−
1
=
1.21
⋅
m
o
l
Moles of oxygen
=
38.5
⋅
g
16.0
⋅
g
⋅
m
o
l
−
1
=
2.41
⋅
m
o
l
We divide thru by the smallest molar quantity to give the empirical formula:
K
C
O
2
.
Now the molecular formula is always a whole number of the empirical formula:
i.e.
molecular formula
=
n
×
empirical formula
And thus with the molecular mass, we can solve for
n
.
166.2
⋅
g
⋅
m
o
l
−
1
=
n
×
(
39.1
+
12.011
+
2
×
16.00
)
⋅
g
⋅
m
o
l
−
1
166.2
⋅
g
⋅
m
o
l
−
1
=
n
×
(
83.1
)
⋅
g
⋅
m
o
l
−
1
Clearly,
n
=
2
, and the
molecular formula
=
K
2
C
2
O
4
The compound is LIKELY the potassium salt of oxalic acid,
K
+
−
O
(
O
=
)
C
−
C
(
=
O
)
O
−
K
+
, i.e.
potassium oxalate.
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