[fcb]

If epsom salt, MgSO₄._xH₂O is heated to 250°C all the water of hydration is lost. After heating a 2.024g sample of the hydrate,0.989g of anhydrous MgSO₄ remains

Calculate how many molecules of water there are per formula unit of MgSO₄

Would you minus the 0.989-2.024 and then use Avogadro's number to find out the molecules?? IInfact i think its wrong by a mile, I am reading ahead in class therefore i am going to cover this topic in the next week or so. So i thought i may have a go at this question.

[AWK]

use molar scale - molar mas of hydrate (120+18x) and molar mass of anhydrous salt (120)

[fcb]

Now I am just confused. Molar mass scale?

[fcb]

I foundthe percentage of water and this is my process. Can someone tell emwhether this is right or not

water = 51.13%
anhydrous = 48.89%

48.87/100=2.024/_xg
_x=4.141

4.141-2.024=2.117

2.117/18.02=0.11748 moles of water

convert to molecules

[AWK]

I foundthe percentage of water and this is my process. Can someone tell emwhether this is right or not

water = 51.13%
anhydrous = 48.89%

48.87/100=2.024/_xg
_x=4.141

4.141-2.024=2.117

2.117/18.02=0.11748 moles of water

convert to molecules


Convert percentage to moles
water = 51.13%/18
anhydrous = 48.89%/120

[fcb]

I foundthe percentage of water and this is my process. Can someone tell emwhether this is right or not

water = 51.13%
anhydrous = 48.89%

48.87/100=2.024/_xg
_x=4.141

4.141-2.024=2.117

2.117/18.02=0.11748 moles of water

convert to molecules


Convert percentage to moles
water = 51.13%/18
anhydrous = 48.89%/120

so is water 2.83moles? People at school were attempting it and I heard a few answers. There was one that came around a bit and that was approx. 7 or so molecules.

[Borek]

Would you minus the 0.989-2.024 and then use Avogadro's number to find out the molecules??

Not bad (although looks like you got it reversed and you will have negative mass of water), just do it separately for MgSO₄ that was left after heating and for water that left the compound - that will tell what was their ratio in the original compound.

Or - instead of going through molecules - go through moles. Don't calculate percentages, just go for mass of anhydrous sulfate and mass of water removed, convert them to moles and calculate ratio. Same final result.

And yes, 7 is a correct one.

[fcb]

Would you minus the 0.989-2.024 and then use Avogadro's number to find out the molecules??

Not bad (although looks like you got it reversed and you will have negative mass of water), just do it separately for MgSO₄ that was left after heating and for water that left the compound - that will tell what was their ratio in the original compound.

Or - instead of going through molecules - go through moles. Don't calculate percentages, just go for mass of anhydrous sulfate and mass of water removed, convert them to moles and calculate ratio. Same final result.

And yes, 7 is a correct one.

Thanks for confirming 7 is the correct answer.

Can you please show me your working and i promise not to make stupid threads like this again. I dont know how to get the answer, So come test time, I will fail

[AWK]

http://chem.lapeer.org/Chem1Docs/HydrateLab2.html
http://chem.lapeer.org/Chem1Docs/HydrateLab1.html

[fcb]

i think i have this down and dusted

0.989/120.4 = 8.2464x10^-3
8.2464x10^-3 _x 18.02 (H₂O) = 0.1484
1.035/0.1484 = 6.974

Thanks so much to everyone on the board that helped me, And to those who clicked on this thread.
Regards,

[Borek]

i think i have this down and dusted

I am not sure. You got the correct answer, but you have found it in so convoluted way I am not sure you got it right, and not by accident.

0.989/120.4 = 8.2464x10^-3

Correct - number of moles of sulfate.

8.2464x10^-3 _x 18.02 (H₂O) = 0.1484

Hmm. This is mass of water. Can you elaborate what water it is - how it should be understood?

[fcb]

i think i have this down and dusted

I am not sure. You got the correct answer, but you have found it in so convoluted way I am not sure you got it right, and not by accident.

0.989/120.4 = 8.2464x10^-3

Correct - number of moles of sulfate.

8.2464x10^-3 _x 18.02 (H₂O) = 0.1484

Hmm. This is mass of water. Can you elaborate what water it is - how it should be understood?

for the last question, isn't itthe amount of water used up during this process

[Borek]

No, but you can easily calculate mass of the water that was liberated, and number of moles of water that was liberated.

I have a feeling I am repeating myself.