[windom626]

Hi!

Im having trouble with understanding one answer from the back of my chemistry textbook.

So. You Have 5.71g of mixutre containing NaNO3 and Na2SO4. The total mass of sodium is 1.61g.
The question is - what is the mass percetage of NaNO3 in this mixture.

1. What I did:

If we have 1.61g of total sodium this means we have 0.07 mol of sodium in this mixture (n=m/M; 1.61/23).

The ratio of Sodium from diffrent compounds is 1:2 (NaNO3 and Na2SO4)- this gives me one equation:
x+2x =0.07 (mol) where X is the amount of mol of the specific sodium.
x=0.023 (mol).

Now I calculate if we have 0.023 mol of sodium in NaNO3 - and we asume that each of this 0.023 is in compound - this gives us a total weight of NaNO3 - 0.023mol x 85g/mol = 1.955g

1.955g/5.37g *100% = 36.04%
So my answer would be that 36.04% of the mixture weight comes from NaNO3.

BUT...
In the textbook there is 45.05% and I have no idea what is wrong. I've been trying to do this for 1h, as my late hour chemistry practice and I got so frustrated.

Please help me. Thanks!

[Hunter2]

Wrong is you assume that the ratio is 1:2. You don't know how much NaNO3 and Na2SO4 was mixed.  So the ratio of sodium is valid for the molecules, but not for the mixture.

[windom626]

Ok but then give me a small hint

[Hunter2]

You have to develop two equations.
One is m(NaNO3) + m Na2SO4) = 5 81 g
Express this in m = n*M
The Difference of the mass of the mixture with the mass of sodium gives the mass of the Anions.
From this you can develop the second equation.
Express this also with m =n*M

[Hunter2]

5 81 g is misstyping  instead it has to be 5.71 g

[windom626]

So:

5.37-1.61 = 3.76g (weight of NO3- + SO4(2-))

85x+142y=5.317   (molar mass (NaNO3) * number of moles (NaNO3) + molar mass (Na2SO4)* number of moles Na2SO4)

62x+96y=3.76 (molar mass (NO3-)*number of moles (NaNO3) + molar mass (SO4(2-))*number of moles (Na2SO4)

Am I right?

[windom626]

Still does not add up...what the heck!

[Borek]

TBH, not sure what you are doing.

Many ways to skin that cat, but no need to go for anions, it is actually kinda like grabbing your left ear with your right hand.

Assume there are x g of NaNO₃ and y g of Na₂SO₄.

What is the total mass of the sample, x+y?

Now, from molar masses - what fraction of the mass x is Na? What fraction of the mass y? And what do these sum up to?

You have two unknowns x and y, you have two simple equations. Just solve.

[Borek]

OK, you were trying to express mass in terms of number of moles. Should work.

Book answer doesn't look correct OK to me (that's not what I got).

[windom626]

Well I triedy your version - correct me if im wrong.

X = mass of NaNO3 in grams
Y = mass of Na2NO3 in grams

x+y = 5.317g

Mole mass of NaNO3 = 85g/mole  -> 27% of mass comes from sodium (0.27)
Mole mass of Na2SO4= 142g/mole -> 32% of mass comes from sodium (0.32)

0.27x+0.32y=1.61

Which gives us X=0.5715 and y= 4.745 as mass of each compund in this mixture...still it is not 45.05%. Jesus...

 

[Borek]

Assuming 45% of NaNO₃ there should be 1.71 g of Na in the sample, looks like a typo in the problem.

[Borek]

x+y = 5.317g

Probably typo, but idea looks OK.

0.27x+0.32y=1.61

Yep.

[windom626]

Straight from text book.

"Mixture of NaNO3 and Na2SO4 weights 5.37g and contains 1.61g of sodium. What is the mass percentage of NaNO3 in this mixture?"

[windom626]

There was typo true, when i wrote it in the begining, however it still does not solve the problem.

[Hunter2]

Why not follow what I wrote.

You have m1 + m2 = 5.37 g

Then you convert

n1* M1 + n2* M2 = 5,37 g

The same you do with the anions.

Because n(NaNO3) = n(NO3-) and n(Na2SO4) = n(SO4 2-)

You get two equation with n1 and n2

Can you solve now.