A compound contains 4.07% hydrogen,
24.27% carbon and 71.65% chlorine. Its
molar mass is 98.96 g. What are its
empirical and molecular formulas?
Answers
Answer:
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Explanation:
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EASY TO MAKE THIS ANIMATION
Answer:
The empirical formula is
CH
2
Cl
. The molecular formula is
C
2
H
4
Cl
2
.
Explanation:
Step 1. Calculate the empirical formula
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of
C
to
H
to
Cl
.
Your compound contains 24.27 %
C
, and 4.07 %
H
.
Assume that you have 100 g of sample.
Then it contains 24.27 g of
C
and 4.07 g of
H
.
Mass of O = (100 - 24.27 - 4.07) g = 71.66 g
Moles of C
=
24.27
g C
×
1 mol C
12.01
g C
=
2.021 mol C
Moles of H
=
4.07
g H
×
1 mol H
1.008
g H
=
4.038 mol H
Moles of Cl
=
71.66
g Cl
×
1 mol Cl
35.45
g Cl
=
2.021mol Cl
From this point on, I like to summarize the calculations in a table.
Element
m
Mass/g
X
l
Moles
X
l
l
Ratio
m
Integers
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
m
C
X
X
X
m
m
l
24.27
X
m
l
2.021
m
m
l
1
m
m
m
m
m
l
l
1
m
H
X
X
X
X
m
l
l
4.07
m
m
l
4.038
X
m
l
l
1.998
m
m
m
l
l
2
m
Cl
X
X
X
m
m
71.66
X
m
l
2.021
m
m
l
1.000
m
m
m
l
l
1
The empirical formula is
CH
2
Cl
.
Step 2. Calculate the molecular formula of the compound
The empirical formula mass of
CH
2
Cl
is 49.48 u.
The molecular mass is 98.96 u.
The molecular mass must be an integral multiple of the empirical formula mass.
MM
EFM
=
98.96
u
49.48
u
=
2.000
≈
2
The molecular formula must be twice the empirical formula.
MF
=
(
EF
)
2
=
(
CH
2
Cl
)
2
=
C
2
H
4
Cl
2
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