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Question

I am sure there are some out there that will find this easy, i have spent all summer on this work and this is 1 of 4 I just can not get, and I have a test coming up!

A sample of a mixture containing only sodium chloride and potassium chloride has a mass of 4.000 g. When this sample is dissolved in water and excess silver nitrate is added, a white solid (silver chloride) forms. After filtration and drying, the solid silver chloride has the mass 8.5904 g. Calculate the mass percent of each mixture complaint.

Answer 1

I'm not sure what a mixture complaint is, but I think you want to know how much NaCl and KCl are in the 4.000 g sample. If not, never mind.

Collect the molar masses to use later:

AgCl = 143.321 g/mol

NaCl = 58.443 g/mol

KCl = 74.551 g/mol

Find moles of AgCl formed, since addition of AgNO3 to the original sample will pretty much completely precipitate out all the chloride:

8.5904/143.321 = 0.059938 mol AgCl, = mol Cl in original sample

Since NaCl and KCl both have 1 mole of Cl per mole of compound, we know that 4.000 g of the original sample has 0.059938 mol, which consists of a mixture of NaCl and KCl. So we can let x = moles of NaCl, and 0.059938-x = moles of KCl. Then,

4.000 g = 58.443 g/mol (x) + 74.551 g/mol (0.059938-x)

4.000 = -16.108 x + 4.4684

16.108 x = 0.4684

x = 0.02908 mol NaCl --> (0.02908 mol)(58.443 g/mol) = 1.700 g NaCl

This leaves the remainder of the sample, 2.300 g, as KCl

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Added later, after your added info:

Yes, you got it.  You're very welcome.  Good luck on your test, or whatever is coming up with this stuff.