Question

A compound on analysis gave the percentage composition Na = 14.31 %, H = 6.22 %, S=9.97% & O= 69.5%

Calculate the molecular formula of the compound on the assumption that all hydrogen atoms are present in combination with oxygen as water of crystallisation. The molecular mass of compound is 322 (At-mass : Na=23,5=32,H = 11 & 0=16)

No irrelevant answers please

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Answer 1

Answer:

Na2SO4.10H2O (Sodium Sulfate Decahydrate)

Explanation:

percentage composition is given, we know that,

(molecular mass of atoms/total molecular weight)*100 is percentage composition, so,

let's find the x value of each atom,

for Sodium atom,

23/x = 14.31%

23/x = 14.31/100

23/x = 0.1431

x = 160.727

similarly for Hydrogen,

1/x = 6.22%

x = 1/0.0622

x = 16.077

for Sulphur,

x = 32/0.0997

x = 320.963

for Oxygen,

x = 16/0.695

x = 23.021

now,

which x has highest number?

for Sulphur right?, so sulphur is one here,

now, divide each x value with highest value

for Hydrogen,

320.963/16.077

= 20

for Sodium,

320.963/160.727

= 2

for Oxygen,

320.963/23.021

= 14

so, there are following number of atoms,

Na - 2

H -20

O - 14

S - 1

and the compound is,

Na2SO4.10H2O (Sodium Sulfate Decahydrate) whose molecular weight is 322g/mol.