Calculation of water of crystallization

[lolbits]

Homework Statement

We did an experiment at school. We heated $$Na_2SO_4*10H_2O$$ so that the water evaportated. We started with 2 gram of the substance, and ended up with 1.8 grams. Unfortunately, not all the water did evaporate because of the equipment we used at school. So the question is, a) how many moles of water evaporated and b) what would the new formula be?

Homework Equations

mol = m(g) / M_m

The Attempt at a Solution

a) Moles of water evaporated:
Know that 0.2 grams of water evaporated.
0.2/18 = 0.011 mol H2O evaporated.
That would be about 2 moles for each mol Na2SO4.

b) 2grams $$Na_2SO_4*10H_2O$$ $$\rightarrow$$ $$Na_2SO_4*xH_2O$$

$$M_m$$ of $$Na_2SO_4*10H_2O$$ is $$322g/mol$$

$$\frac{2g}{322g/mol}$$ gives $$0.0062 mol$$

$$M_m$$ of $$Na_2SO_4$$ = $$142 g/mol$$

$$\frac{1.8g}{0.0062 mol}=290.32 g/mol$$

$$290.32g/mol - 142g/mol = 148g/mol$$

$$M_m$$ of $$H_2O=18$$

$$\frac{142 g/mol}{18g/mol} = 8$$

So the new formula would be:

$$Na_2SO_4*8H_2O$$

Is this right?

 

[lolbits]

No one? This should not be hard..

 

[Borek]

Calculations are almost OK, there is one weak point - you can't be sure you started with decahydrate, as amount of water may depend on many factors (including sample history - was it kept in open/closed bottle, what was humidity at the time and so on). But you can't check it, so your assumption is not bad.

The only thing I don't like is that you ignore accuracy - that is, some numbers are presented with guard digits, while other are presented already rounded down in a rather unpredictable way.

--

 

[lolbits]

Calculations are almost OK, there is one weak point - you can't be sure you started with decahydrate, as amount of water may depend on many factors (including sample history - was it kept in open/closed bottle, what was humidity at the time and so on). But you can't check it, so your assumption is not bad.

The only thing I don't like is that you ignore accuracy - that is, some numbers are presented with guard digits, while other are presented already rounded down in a rather unpredictable way.

--
www.titrations.info, www.chemistry-quizzes.info, www.ph-meter.info

Thank you for replying!

What I wrote was just quick. I will be more accurate next time. But again, thanks!

 

[DeeNos]

Yes, your solution is correct. It is important to note that the new formula for the hydrated compound would be Na2SO4*8H2O, as the original compound had 10 water molecules attached to it. This means that 2 water molecules evaporated during the experiment, and the remaining 8 water molecules are still attached to the compound. This new formula reflects the loss of 2 moles of water in the experiment. Great job!