Answer:
Empirical Formular:
When given the percentage by mass of the constituents in a compound, we can figure out the ratios of the atoms to each other and the ratio is a definite whole number. By comparing the molar mass of the empirical formula and the molar mass of the molecular formula, we determine the actual formular of the compound or molecule.
Answer and Explanation:
Using the percentages for 100g of the molecule, we have:
Element
C
H
N
Mass
39.97
g
13.41
g
46.62
g
Mole Calculation
=
m
A
r
39.97
g
12.01
g
13.41
g
1.01
g
46.62
g
14.00
g
Mole Ratio
3.33
13.3
3.33
Dividing by 3.33
1
4
1
The empirical formular is
C
H
4
N
and the molecular formular is
C
x
H
4
x
N
x
.
The molecular mass M if the sum of the constituent atoms, i.e.
M
=
x
A
r
(
C
)
+
4
x
A
r
(
H
)
+
x
A
r
(
N
)
=
x
(
12.0
g
/
m
o
l
)
+
4
x
(
1.0
g
/
m
o
l
)
+
x
(
14.0
g
/
m
o
l
)
=
30
x
g
/
m
o
l
.
We are given
M
=
60
g
/
m
o
l
. Equantity, we get:
30
x
g
/
m
o
l
e
=
60.06
g
/
m
o
l
x
≈
2
The molecular formular is thus for x =2, i.e.
C
2
H
8
O
2
.
