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1) A compound on analysis gave the following percentage composition: Na=14.31%, S = 9.97%, H = 6.22%, O = 69.5%, calculate the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32, H = 1, 0 = 16].

1)A compound on analysis gave the following percentage composition: Na=14.31%, S = 9.97%,              H = 6.22%, O = 69.5%, calculate the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32, H = 1, 0 = 16].

Grade:9

1 Answers

A M S ARUN KRISHNA
216 Points
8 years ago
Solution :- Calculation of empirical formula
 
Element
Percentage
Relative number of moles
Simple ratio moles
Simplest whole number ratio
Na
14.31
14.31
------- = 0.62
  23
0.62
------  = 2
0.31
2
S
9.97                      
19.97
-------- = 0.31
  32
0.31
------  = 1
0.31
1
H
6.22
6.22
------ = 6.22
  1
6.22
-----  = 20
0.31
20
O
69.5
  69.5
-------- = 4.34
   16
4.34
------ = 14
0.31
14
 
The empirical formula is Na2SH20O14
Calculation of Molecular formula
Empirical formula mass = (23 x 2) + 32 + (20 x 1) + (16 x 14) = 322
Molecular mass                                322
n = ----------------------------------------   = ------------  = 1
            Empirical formula mass                   322
 
Hence molecular formula = Na2SH20O14
Since all hydrogens are present as H2O in the compound, it means 20 hydrogen atoms must have combined. It means 20 hydrogen atoms must have combined with 10 atoms of oxygen to form 10 molecules of water of crystallisation. The remaining (14 - 10 = 4) atoms of oxygen should be present with the rest of the compound.
Hence, molecular formula = Na2SO4.10H2O.
Solution :- Calculation of empirical formula
 
Element
Percentage
Relative number of moles
Simple ratio moles
Simplest whole number ratio
Na
14.31
14.31
------- = 0.62
  23
0.62
------  = 2
0.31
2
S
9.97                      
19.97
-------- = 0.31
  32
0.31
------  = 1
0.31
1
H
6.22
6.22
------ = 6.22
  1
6.22
-----  = 20
0.31
20
O
69.5
  69.5
-------- = 4.34
   16
4.34
------ = 14
0.31
14
 
The empirical formula is Na2SH20O14
Calculation of Molecular formula
Empirical formula mass = (23 x 2) + 32 + (20 x 1) + (16 x 14) = 322
Molecular mass                                322
n = ----------------------------------------   = ------------  = 1
            Empirical formula mass                   322
 
Hence molecular formula = Na2SH20O14
Since all hydrogens are present as H2O in the compound, it means 20 hydrogen atoms must have combined. It means 20 hydrogen atoms must have combined with 10 atoms of oxygen to form 10 molecules of water of crystallisation. The remaining (14 - 10 = 4) atoms of oxygen should be present with the rest of the compound.
Hence, molecular formula = Na2SO4.10H2O.

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