# 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.

• $\ce{Na}$'s molar mass is $23$ g/mol not $11$ g/mol Mar 20, 2016 at 12:24
• Oh yeah totally forgot about that! Thanks for noticing my silly mistake. I'll change it. Mar 20, 2016 at 12:28
• Also, why do you have $n(\ce{H_2O})=1$ ? It should be the same as all the others, $n=\frac mM=\frac{55.8}{2+16}=3.1$. That gives you $3.1/.3\approx 10$ molecules of $\ce{H_2O}$ Mar 20, 2016 at 12:29
• I was thinking that it was just one mole. Mar 20, 2016 at 12:30
• Your percentages add up to 91. You are missing something. Mar 20, 2016 at 12:44

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 $\ce{Na_2SO_4.10H_2O}$

• Thanks for the assist! My teacher must have made a mistake when she was coming up with the percentages. I'll tell her about it when I get to school tomorrow. I will also correct the molar mass for sodium. Thanks again! :) Mar 20, 2016 at 13:02