Experimentally is was determined that compound contains 39.72% , 1.67% and 58.61% of .
Find the empirical formula of compound.
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Answers
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
Answer:
Example
An unknown compound contains
85.63 % C
and
14.37 % H
. Its experimental molar mass is
56 g/mol
. What is its molecular formula?
Solution
Assume we have
100 g
of the compound. Then we have
85.63 g of C
and
14.37 g of H
.
Moles of C
=
85.63
g C
×
1 mol C
12.01
g C
=
7.130 mol C
Moles of H
=
14.37
g H
×
1 mol H
1.008
g H
=
14.26 mol H
Moles of C
Moles of H
=
7.130
mol
14.26
mol
=
1
2.000
≈
1
2
The empirical formula is
CH
2
.
The empirical formula is the simplest formula of a compound.
The actual formula is an integral multiple of the empirical formula.
If the empirical formula is
CH
2
, the actual formula is
(
CH
2
)
n
or
C
n
H
2
n
, where
n
=
1
,
2
,
3
,
…
.
Our job is to determine the value of
n
.
The empirical formula mass of
CH
2
is
14.03 u
. The molecular mass of
56 u
must be some multiple of this number.
n
=
56
u
14.03
u
=
4.0
≈
4
∴ The molecular formula is
C
n
H
2
n
=
C
4
H
8
.
Explanation:
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