A COMPOUND ON ANALYSIS GAVE THE FOOLOWING % COMPOSITION
Na=14.31% S= 9.97% H=6.22% O=69.50% calculate the molecular formula of the compound on the assumption that all the hydroden in the compound is present in composition with oxygen as water of crystrallization. molecular mass of compound is 322.
Asked by patidarshm | 11th Dec, 2015, 11:28: AM
Calculate empirical formula:
Element
Symbol
Percentage of element
At. Mass of element
Relative no. of atoms =
Percentage/At. Mass
Simplest atomic ratio
Simplest whole no. atomic ratio
Sodium
Na
14.31
23
14.31/23=0.62
0.62/0.31=2
2
Sulphur
S
9.97
32
0.31
1
1
Hydrogen
H
6.22
1
6.22
20
20
Oxygen
O
69.50
16
4.34
14
14
Thus, the empirical formula of the compound is = Na2SH20O14
The empirical formula mass of the compound = 2×23+1×32+20×1+14×16= 322
Now, calculate value of n = Molecular mass / Empirical mass = 322/322=1
Finally, calculate molecular formula of the compound = n × Empirical formula= 1 × Na2SH20O14
Hence, molecular formula of the compound = Na2SH20O14
Calculate empirical formula:
Element |
Symbol |
Percentage of element |
At. Mass of element |
Relative no. of atoms = Percentage/At. Mass |
Simplest atomic ratio |
Simplest whole no. atomic ratio |
Sodium |
Na |
14.31 |
23 |
14.31/23=0.62 |
0.62/0.31=2 |
2 |
Sulphur |
S |
9.97 |
32 |
0.31 |
1 |
1 |
Hydrogen |
H |
6.22 |
1 |
6.22 |
20 |
20 |
Oxygen |
O |
69.50 |
16 |
4.34 |
14 |
14 |
Thus, the empirical formula of the compound is = Na2SH20O14
The empirical formula mass of the compound = 2×23+1×32+20×1+14×16= 322
Now, calculate value of n = Molecular mass / Empirical mass = 322/322=1
Finally, calculate molecular formula of the compound = n × Empirical formula= 1 × Na2SH20O14
Hence, molecular formula of the compound = Na2SH20O14
Answered by Prachi Sawant | 13th Dec, 2015, 11:59: AM
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