Weight percentage / stoichiometry

[lorka150]

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!

 

[GCT]

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.

 

[Blueshift5]

First, it's important to understand that weight percentage and stoichiometry are related concepts. Weight percentage is a measure of the amount of a particular substance in a mixture, while stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction.

In this problem, we are given a mixture of KCr(SO3)2 and its hydrate KCr(SO4)2.12H2O, and we are asked to find the weight percentage of the hydrate in the original mixture.

To do this, we need to use stoichiometry to determine the amount of KCr(SO4)2.12H2O in the original mixture. We can do this by considering the mass of the mixture before and after heating, as well as the molar masses of both compounds.

First, we can calculate the mass of KCr(SO3)2 in the original mixture by subtracting the mass of KCr(SO4)2.12H2O (1.4659 g) from the total mass of the mixture (1.6336 g). This gives us a mass of 0.1677 g for KCr(SO3)2.

Next, we can use the ratio of molar masses to determine the moles of KCr(SO3)2 and KCr(SO4)2.12H2O in the mixture. Since the molar mass of KCr(SO4)2 is 283.22 g/mol and the molar mass of KCr(SO3)2 is 270.19 g/mol, we can set up the following equation:

0.1677 g KCr(SO3)2 / 270.19 g/mol = x g KCr(SO4)2 / 283.22 g/mol

Solving for x, we get a mass of 0.1606 g for KCr(SO4)2 in the original mixture.

To find the weight percentage of KCr(SO4)2.12H2O in the original mixture, we can divide the mass of KCr(SO4)2.12H2O (0.1606 g) by the total mass of the mixture (1.6336 g) and multiply by 100%. This gives us a weight percentage of approximately 9.8%.

In summary, by using stoichiometry and the molar masses of the compounds involved,