Empirical formula of Glauber's salt

I'm not too sure on how to get the empirical formula for Glauber's salt because water ($\ce{H2O}$) is included.


The question is:



> 
> Glauber's salt contains $14.3\%$ sodium, $10\%$ sulphur, $10.9\%$ oxygen (not in water) and $55.8\%$ water. Find this substance's empirical formula.
> 
> 
> 


This is what I did. I got $\ce{Na4SO.H2O}$ as my answer.


[![My working out for the question](https://i.stack.imgur.com/VJapR.jpg)](https://i.stack.imgur.com/VJapR.jpg) 




Your approach seems correct, however you seemed to have made a few computation errors. First of Sodium's molar mass is $ 23 \ce{g/mol}$ and not 11. 


So assuming a total mass of 100g, we can re-interpret the given percentages as follows: 


mass of sodium = 14.3 g


mass of sulphur = 10 g


mass of oxygen = 19.9 g 


(*note: the percentage you provided seems incorrect, the percentages add up to 91, not a 100. I've changed this data point to get the correct result.*)


mass of water = 55.8 g


Dividing these by their respective molar masses to get no. of moles:


no. of moles of sodium = 0.62


no. of moles of sulphur = 0.31


no. of moles of oxygen = 1.2 


no. of moles of water = 3.1


The ratio we get is 2:1:4:10. Thus, the empirical formula is $\ce{Na\_2SO\_4.10H\_2O}$


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