# finding the % water of crystallisation in hydrated iron(II) sulfate

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#1
im doing AS chemistry and we did this experiment:

Method:
1. Using a balance that weighs to two decimal places, weigh a crucible. Add between 1.30 – 1.50 g of hydrated iron II sulphate crystals. Record the masses.
2. Place the crucible containing the hydrated iron(II) sulphate crystals on the pipe-clay triangle and gently heat for about two minutes. The heating should be carried out in either a fume cupboard or a well-ventilated room.
3. Allow to cool and weigh the crucible and the iron(II) sulphate.
4. Repeat steps 2 and 3.
5. Record all of the masses.

basically the aim was to find out what x was for this equation:

FeSO4.xH2O

these were my results from the experiment:
crucible weight + iron sulfate before = 25.77g (i used exactly 1.5g iron sulfate)
after= 25.11g
which led me to work out ther is 44% water
moles=mass/Mr was 0.037

how do i work out what the value of x is from this???
0
10 months ago
#2
Hi! Did you ever figure this out?
0
10 months ago
#3
(Original post by Ally1249562)
Hi! Did you ever figure this out?
Eight years late, I doubt you'll get a reply from number13zd.

Where you trying to help them? or do you need help yourself?
0
10 months ago
#4
Ohh I didn’t even see that🤣🤣 nah I’m struggling myself
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10 months ago
#5
(Original post by Ally1249562)
Ohh I didn’t even see that🤣🤣 nah I’m struggling myself
The OP's working out is not helpful.
From their data you should be able to work out the mass of water that was lost during heating. With a bit of work, you can work out the mass of the iron(II) sulfate after heating (by subtracting off the mass of the crucible).
You can convert both of these masses to amounts (in mol).
You can then simplify the ratio of these two amounts, which will give you the simplest whole number ratio of iron(II) sulfate:water, i.e. the # of WoC.
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