October 09, 2024, 09:17:46 AM
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Topic: Help with chemistry and empirical formulas  (Read 777 times)

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Offline Katerinne

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Help with chemistry and empirical formulas
« on: September 17, 2024, 03:37:07 AM »
Hello just landed here and need help please. I need to find the ratio as in how many moles of water is associated with single atom of anhydrate by dividing moles of water by moles of anhydrate. So the moles of water I got was 0.00699 bc I divided grams of water, 0.126g,over molar mass of water, 18.02g/mol. and the moles of anhydrate was 0.01108 bc I divided mass of anhydrate, 1.509g, over molar mass of calcium sulfate CaSO4(salt I'm doing)which is 136.14. And when dividing 0.01108/0.00699 I get 1.58... So idk if I can round to make it 2 because the empirical formula should be CaSO4*2H2O or if I'm just wrong in everything. I left it at 1.58 because of significant figures, not sure if they count here and all my answers were unrounded so far. If anyone could help that would be nice ty.

Offline Borek

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Re: Help with chemistry and empirical formulas
« Reply #1 on: September 17, 2024, 05:45:18 AM »
Your post is a bit chaotic, it is not clear what you did and where you got these numbers from, and that leaves a lot to guessing (never helps).

However: 1.58 instead of 2 in the context of determining composition of hydrates doesn't look very bad.

First of all, hydrates rarely have exact composition, many of them contain well defined number of water molecules only in a narrow range of temperature and humidity, so not getting a good integer value is nothing unusual. Some compounds create series of hydrates, stable in different conditions, samples of unknown history often contain a mixture of these, so the hydration number determined experimentally can be a bit random. That being said, quick googling for thermogravimetric curve of calcium sulfate dihydrate seems to suggest its composition should be reasonably stable near the STP.

Second - what is the accuracy of the numbers you are using? Without knowing what is the experimental error it is hard to say what 1.58 really means, can be 2 is in the ±error range.
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