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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].`
2 years ago

A M S ARUN KRISHNA
216 Points
```										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 Na2SH20O14Calculation of Molecular formulaEmpirical formula mass = (23 x 2) + 32 + (20 x 1) + (16 x 14) = 322Molecular mass                                322n = ----------------------------------------   = ------------  = 1            Empirical formula mass                   322 Hence molecular formula = Na2SH20O14Since 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 Na2SH20O14Calculation of Molecular formulaEmpirical formula mass = (23 x 2) + 32 + (20 x 1) + (16 x 14) = 322Molecular mass                                322n = ----------------------------------------   = ------------  = 1            Empirical formula mass                   322 Hence molecular formula = Na2SH20O14Since 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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2 years ago
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