March 28, 2024, 04:35:55 PM
Forum Rules: Read This Before Posting


Topic: Empirical formula  (Read 5764 times)

0 Members and 1 Guest are viewing this topic.

Offline chilli

  • Regular Member
  • ***
  • Posts: 36
  • Mole Snacks: +0/-1
Empirical formula
« 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?
« Last Edit: March 14, 2017, 09:36:28 PM by chilli »

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3471
  • Mole Snacks: +526/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Empirical formula
« Reply #1 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.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline chilli

  • Regular Member
  • ***
  • Posts: 36
  • Mole Snacks: +0/-1
Re: Empirical formula
« Reply #2 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

Offline sjb

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 3652
  • Mole Snacks: +222/-42
  • Gender: Male
Re: Empirical formula
« Reply #3 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?

Offline chilli

  • Regular Member
  • ***
  • Posts: 36
  • Mole Snacks: +0/-1
Re: Empirical formula
« Reply #4 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!
« Last Edit: March 15, 2017, 04:03:30 AM by chilli »

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27633
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Empirical formula
« Reply #5 on: March 15, 2017, 04:20:18 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?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline chilli

  • Regular Member
  • ***
  • Posts: 36
  • Mole Snacks: +0/-1
Re: Empirical formula
« Reply #6 on: March 15, 2017, 04:26:26 AM »
1?
I mean the ratio is 1:1:3
« Last Edit: March 15, 2017, 04:46:18 AM by chilli »

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27633
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Empirical formula
« Reply #7 on: March 15, 2017, 07:34:53 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?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline chilli

  • Regular Member
  • ***
  • Posts: 36
  • Mole Snacks: +0/-1
Re: Empirical formula
« Reply #8 on: March 15, 2017, 08:44:00 AM »
One

Offline sjb

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 3652
  • Mole Snacks: +222/-42
  • Gender: Male
Re: Empirical formula
« Reply #9 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.

Offline chilli

  • Regular Member
  • ***
  • Posts: 36
  • Mole Snacks: +0/-1
Re: Empirical formula
« Reply #10 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)

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2048
  • Mole Snacks: +296/-12
Re: Empirical formula
« Reply #11 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.

Offline chilli

  • Regular Member
  • ***
  • Posts: 36
  • Mole Snacks: +0/-1
Re: Empirical formula
« Reply #12 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?

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2048
  • Mole Snacks: +296/-12
Re: Empirical formula
« Reply #13 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.

Offline chilli

  • Regular Member
  • ***
  • Posts: 36
  • Mole Snacks: +0/-1
Re: Empirical formula
« Reply #14 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

Sponsored Links