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Posted (edited)

Question: A sample of a compound of N and O reacts with excess H2 (g) to give 0.123g of NH3 and 0.325g of H2O. Determine the empirical formula of this compound.



Been trying to solve it for a few days. My chem teacher wants the class (Grade 11 chem) to solve the questions on our own for prep for the unit test but no one else knows how to do this. I tried finding % composition to get to Empirical Formula but I got nowhere with it. I also tried doing mole ratios but I font have molecular mass of the unknown compound to find the mass of it. I think I need to get the balanced chemical equation but I don't know how to.



Can anyone help me? Any hints? or where to start.


Edited by amoong
Posted

If you divide 0.123 g by mass of Ammoniac,

and divide 0.325 g by mass of Water,

you have moles.

Divide one by another,

and you see there is ratio 1:2.5, or 2:5 (we can't have fractions).

Draw 2 molecules of Ammonic and 5 molecules of Water,

discard Hydrogen,

and start joining remaining atoms,

remembering rules how they join with other atoms (f.e. oxidation state they could exist).

What do you get?

 

There is one compound which matches it perfectly.

Actually you cannot join it with water, because it will create Nitric Acid (Anhydride of Nitric Acid).

 

Posted

Sensei, you have a knack for making these things seem confusing.

 

OP: consider the unbalanced reaction first and foremost.

 

NxOy + H2 --> H2O + NH3 (x and y are the numbers of each atom respectively)

 

Based on your compound, you should be able to see that all of the nitrogen and all of the oxygen in the product came from the compound. What you need to work out is the molar ratio of nitrogen to oxygen. For instance, if you discover you have 2 mole of nitrogen and 2 of oxygen, the ratio of nitrogen to oxygen would be 1:1 and the empirical formula is NO.

 

As Sensei mentioned, your first task is to work out the moles of water and of ammonia. You have the masses of each, so this is simply a matter of using an equation you should be familiar with, n = m/MW (n is number of moles, m is mass in g and MW is molar mass in g / mol).

 

This then allows you to work out how many moles of nitrogen and oxygen you have, since for every mole of NH3 and H2O, you have 1 mole of N and O, respectively.

 

It is then just a matter of working out the ratio into whole numbers, which you can do by dividing your smallest number by itself and the other by the same number. For example, if you work out you have 0.0334 moles of N and 0.1002 moles of oxygen, you would divide both by the smallest number (0.0334), giving you 1 for nitrogen and 3 for oxygen. This makes the ratio of N:O 1:3 and the formula NO3.

 

Hope that helps.

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