A M S ARUN KRISHNA
Last Activity: 9 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 = Na
2SO
4.10H
2O.
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.
