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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: chilli on March 14, 2017, 08:53:18 PM

Title: Empirical formula
Post by: chilli on March 14, 2017, 08:53:18 PM
Hi,
I have been on this problem for a while now... I know N=0.115 mol, but still cant understand how to find NaNO3 total number of moles
Would appreciate any help,
Thx!

A mixture of NaNO3 and Na2SO4 with a mass of 5.37g contains 1.61 Nitrogen.
What is the percentage mass of NaNO3 in the mixture?
Title: Re: Empirical formula
Post by: Corribus on March 14, 2017, 09:45:25 PM
It is a forum policy to show work before you can receive assistance. So, please show what you have done/tried so far, and take care to include units.
Title: Re: Empirical formula
Post by: chilli on March 14, 2017, 09:57:43 PM
Hmm, I cant show much work because I'm stuck on how to find NaNO3 total number of moles and its pretty crucial to the rest of the question
I don't need an answer, just an explanation on how to find the total moles of a compound given only one substance mass or moles
Title: Re: Empirical formula
Post by: sjb on March 15, 2017, 03:25:09 AM
If there are 1.61 (what, moles, atoms, grams, pounds?) of nitrogen in the mixture, how many moles is that?

If there are x moles of nitrate, what is the mass of that?
If there are y moles of sulfate, what is the mass of that?

Can you relate x and y to the number you first calculated?
Title: Re: Empirical formula
Post by: chilli on March 15, 2017, 03:48:01 AM
Hi,
Sorry I forgot to put grams there, 1.61g
so 1.61g/14g/mol= 0.115mol of N
and then Im just stuck, cant seems to find a way to figure out how to use this information in order to find NaNO3 number of moles

And yes if I had number of NaNO3 (x) moles then:
x = m/Mm or m = xMm
If I had its mass I could find its percentage
Same for sulfate

Thx!
Title: Re: Empirical formula
Post by: Borek on March 15, 2017, 04:20:18 AM
Quote from: chilli on March 15, 2017, 03:48:01 AM
so 1.61g/14g/mol= 0.115mol of N
and then Im just stuck, cant seems to find a way to figure out how to use this information in order to find NaNO3 number of moles

Look at the formula. How many moles of N per 1 mole of NaNO3?
Title: Re: Empirical formula
Post by: chilli on March 15, 2017, 04:26:26 AM
1?
I mean the ratio is 1:1:3
Title: Re: Empirical formula
Post by: Borek on March 15, 2017, 07:34:53 AM
Quote from: chilli on March 15, 2017, 04:26:26 AM
1?

If you are guessing seems like you don't know why.

Quote
I mean the ratio is 1:1:3

Yes, that the correct molar ratio between elements in the molecules, but that's not what I am asking about.

Imagine having 1 mole of N atoms. How many moles of NaNO3 can you produce, assuming every atom of N was used?
Title: Re: Empirical formula
Post by: chilli on March 15, 2017, 08:44:00 AM
One
Title: Re: Empirical formula
Post by: sjb on March 15, 2017, 10:52:23 AM
So if you have 1.61g/14g/mol= 0.115mol of N in your mixture, how many moles of nitrate are there?

Apologies, I misread the initial text and thought you had a mass of sodium (even though I typed nitrogen) - things are a bit more involved in that case.
Title: Re: Empirical formula
Post by: chilli on March 15, 2017, 11:15:52 AM
I want to say 0.115 mol NaNO3
But if I calculate its mass it doesn't make sense:
m (NaNO3) = 0.115mol x 85g/mol = 9.775g
Which is way more than the overall weight of the mixture (5.37g)
Title: Re: Empirical formula
Post by: mjc123 on March 15, 2017, 11:50:09 AM
You are correct. If there is 1.61 g nitrogen, there must be 9.76 g NaNO3. Have you read the question correctly? If not, there is a mistake in the question.
Title: Re: Empirical formula
Post by: chilli on March 15, 2017, 12:08:27 PM
Yes, this is the question, I have read it over and over and double checked everything
The final answer in the book is that NaNO3% in the mixture is 45.1%
Oh wow, I spent so much time banging my head against the wall, Is there really a mistake there?
Title: Re: Empirical formula
Post by: mjc123 on March 15, 2017, 01:27:02 PM
The answer looks like what you would get if there were 1.61g sodium. Are you sure it said N, not Na? It might be a typo.
Title: Re: Empirical formula
Post by: chilli on March 15, 2017, 01:38:45 PM
Omg yes, it is Na, so sorry!
Thx everyone for your guidance and patience, I'll go try and solve it now