# Calculate the concentration of each of the following solutions in mass percent

Here's the problem:

Calculate the concentration of each of the following solutions in mass percent.

a). 13.5g C2H6O in 67.4g H2O

b). 104g KCl in 574g H2O

c) 38.8mg KNO3 in 2.58g H2O

now I now mass %= mass sloute/mass solution X 100%

so first I added both g of each substance...then divided the solute by the total gram-age then X100 but didn't get the right answers....

any ideas? thanks!!!

IAmLegendToo

Wed, 2008-11-05 14:23

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## Re: Another difficult problem....

Percent by Mass if I remember correctly is (m/m) or grams of solute/grams of solution.

But, I'm not quite sure how to approach this question.

My first instincts would be to believe that the first parts are the Solutes, and the H2Os are the Solvents. The addition of both I would think to be the Solution. So I would first try adding both together, which gives me tmy grams of solution. And then the grams of solute would be the grams of the other compound. Let's take for an example.

13.5g C2H6O + 67.4g H2O = 80.9g of Solution (C2H6O + H2O)

13.5g/80.9g = .16687. Convert to a percent by x 100.

I get 16.69% or simply 17%.

Assuming my theory is right, that would be the answer. We JUST started this in my class, so I don't have a great understanding of it, but using what I've learned so far, that's how I would FIRST ATTEMPT it. It at least gives me a slight idea as to what to do, or maybe not lol.

IAmLegendToo

Wed, 2008-11-05 14:25

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## Re: Another difficult problem....

LMAO, after re-reading your post, I understand that I just explained exactly what you first tried. So we are actually along the same lines here. I misread your post the first time lol. Oh well. At least now I've established we've encountered the same ideas.

jaymzjr1984

Thu, 2008-11-06 09:51

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## Re: Another difficult problem....

LOL no worries. I may have just had addition problems or entered it incorrectly in the computer.....atleast I now can be a little more confident I'm on the right track. THanks for your help again!!! :-)