Question

Experimentally it was determined that compound contains 39.72% C, 1.67% H

and 58.61% of Cl. Find the empirical formula of compound.​

Answer 1

Answer:

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

Explanation:

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

.