Question

An organic compound contains 10.4 percent of carbon and 0.84 percent of hydrogen and 89.1 2% of chlorine derive the molecular formula and empirical formula molecular mass of the compound is 119 amu

Answer 1

To make the calculations easier, you can pick a

100.0-g

sample of chloroform and use its percent concentration by mass to find that it contains

89.1 g

→

chlorine

0.84 g

→

hydrogen

10.06 g

→

carbon

Next, use the molar masses of the three elements to find how many moles of each you have in this sample

For Cl:

89.1

g

⋅

1 mole Cl

35.453

g

=

2.513 moles C

For H:

0.84

g

⋅

1 mole H

1.00794

g

=

0.8334 moles H

For C:

10.06

g

⋅

1 mole C

12.011

g

=

0.8376 moles C

To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one

For Cl:

2.513

moles

0.8334

moles

=

3.0154

≈

3

For H:

0.8334

moles

0.8334

moles

=

1

For C:

0.8376

moles

0.8334

moles

=

1.005

≈

1

The compound's empirical formula, which tells you what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be

C

1

H

1

Cl

3

⇒

CHCl

3

Answer 2

Answer:

Empirical formula = CHCl3

Molecular formula = CHCl3

Explanation:

Given:

1. Carbon-10%
2. Hydrogen-0.84%
3. Chlorine-89.12%
4. Volume-122ml
5. Weight of the compound-0.6 g

Solution:

Element. %. Atomic mass. Mole ratio. Simple ratio.

C. 10. 12. 10/12=0.83. 1

H. 0.84. 1. 0.84/1=0.84. 1

Cl. 89.12. 35.5. 89.12/35.5=3.02. 3

So Emprical formula = CHCl3

Emprical weight = 12+1+(35.5*3)=119.5

Molecular weight = w*22400

-------------

v

= 0.6*22400

------------------

122

=120

n=Molecular mass

--------------------

Emprical mass

=120

---------

119.5

=1 approx.

Molecular formula = n*(Emprical formula)

=1*CHCl3

Answer=CHCl3

The volume in the question given may be wrong. Answer is possible only when volume is 122 ml.

Thanks for checking out this answer.

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