An unknown liquid is composed of 5.75% H, 28.01% Cl, and 66.42%C. The molecular mass found by mass spectrometry is 126.58 amu. What is the molecular formula of the compound?

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Assuming these are mass percentages.

Mass ratio of unknown compound -

H:Cl:C = 5.75:28.01:66.42

To calculate the mole ratio we can divide each by respective molar weights.

Molar weight of H - 1

Molar weight of Cl - 35.5

Molar weight of C - 12

Therefore molar ratios are,

H:Cl:C...

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Assuming these are mass percentages.

Mass ratio of unknown compound -

H:Cl:C = 5.75:28.01:66.42

To calculate the mole ratio we can divide each by respective molar weights.

Molar weight of H - 1

Molar weight of Cl - 35.5

Molar weight of C - 12

Therefore molar ratios are,

H:Cl:C = (5.75/1):(28.01/35.5):(66.42/12)

H:Cl:C = 5.75:0.789:5.535

Dividing by 0.789

H:Cl:C = 7.283: 1: 7

 

Therefore for approximate intgers, molar ratio of H:Cl:C is

H:Cl:C = 7:1:7

Therefore the empirical formula of the molecule is (C7H7Cl)n

Now we have to calculate n. This can be done by calculating molecular mass.

Molecular mass of unknown liquid fron spectrometry is 126.58

According to formula,

n x(7x12+7x1+1x5.5) =126.58

n x 126.5 = 126.58

n = 1 approximately.

Thefore the molecular formula of this unknown liquid is C7H7Cl

 

So it can be an aromatic compound, (C6H5)CH2Cl (With a benzene ring). It is actualy Benzyl chloride which is liquid at room temperature (with a boiling temperature of 173 C).

 

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