What Is the Weight Percentage of KCr(SO4)2.12H2O in the Original Mixture?

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SUMMARY

The weight percentage of KCr(SO4)2.12H2O in the original mixture can be calculated using the provided masses and molar masses. The original mass of the mixture is 1.6336 g, and after heating, the mass is reduced to 1.4659 g, indicating that 0.1677 g of water was lost. The moles of KCr(SO4)2.12H2O can be determined as 0.0007755, leading to a mass of 0.38728 g. The weight percentage of KCr(SO4)2.12H2O is then calculated as (0.38728 g / 1.6336 g) * 100, resulting in approximately 23.7%.

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lorka150
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I did this a bunch of times and did not understand what to do. Hopefully someone can shed some light. Our prof gave us hints, but they aren't really helping.

A mixture of KCr(SO3)2 and its hydrate KCr(SO4)2.12H2O has a mass of 1.6336 g. After heating (driving off water), the mass is 1.4659 g. What is the weight % of KCr(SO4)2.12H2O in the original mixture?

Molar mass of KCr(SO4)2 is 283.22 g/mol
Molar mass of KCr(SO4)2.12H2O is 499.40 g/mol.


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This is what I tried: I figured the mass of water was 0.1677g, and therefore the moles were 0.0009306 mol.

Then, the mols of KCr(SO4)2.12H2O is 0.0007755. The mass was that x 499.40, which gave me 0.38728 g.

And then I was lost. I don't know how to utlize the molar mass of KCr(SO4)2 in this question, either.

Thanks!
 
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You need to utilize the information for the water in this problem, by finding the moles of water, you can find calculate the moles of the hydrate compound, and then the grams.
 

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