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Empirical Formula

A compound contains 36.5% of sodium and 25.5% of sulphur by mass, the rest being oxygen.

Use this information to show that the empirical formula of the compound is Na2SO3.


Can someone explain this to me please step by step? It'll be much appreciated :smile:
Reply 1
Original post by RubyTay
A compound contains 36.5% of sodium and 25.5% of sulphur by mass, the rest being oxygen.

Use this information to show that the empirical formula of the compound is Na2SO3.


Can someone explain this to me please step by step? It'll be much appreciated :smile:


Hi there, I will try my best to explain it :smile:

The first step is to find the percentage of oxygen in the compound, so you have it for reference:
100% - 36.5% - 25.5% = 38%
So there is 38% oxygen in the compound.

Now we divide each percentage of each compound by its atomic mass so we can work out the ratio of each compound.
Sodium: 36.5 / 23 = 1.586956522
Sulphur: 25.5 / 32.1 = 0.794392523
Oxygen: 38 / 16 = 2.375

As you can see, the number for sulphur is the smallest (0.79...). Now to work out the ratio of each compound you divide each number shown above by the smallest number out of the three (in this case, sulphur's). I hope that makes sense!

Sodium: 1.586... / 0.794... = 1.99... -> rounds up to 2
Sulphur: 0.794... / 0.794... = 1
Oxygen: 2.375 / 0.794... = 2.98... -> rounds up to 3

So the overall ratio is Na2SO3. Empirical formula is the simplest whole number ratio of a compound, and the ratio shown cannot be simplified anymore than it already is.

I hope that is helpful, if there is something I haven't made clear feel free to ask :smile:
Reply 2
Original post by zef1995
Hi there, I will try my best to explain it :smile:

The first step is to find the percentage of oxygen in the compound, so you have it for reference:
100% - 36.5% - 25.5% = 38%
So there is 38% oxygen in the compound.

Now we divide each percentage of each compound by its atomic mass so we can work out the ratio of each compound.
Sodium: 36.5 / 23 = 1.586956522
Sulphur: 25.5 / 32.1 = 0.794392523
Oxygen: 38 / 16 = 2.375

As you can see, the number for sulphur is the smallest (0.79...). Now to work out the ratio of each compound you divide each number shown above by the smallest number out of the three (in this case, sulphur's). I hope that makes sense!

Sodium: 1.586... / 0.794... = 1.99... -> rounds up to 2
Sulphur: 0.794... / 0.794... = 1
Oxygen: 2.375 / 0.794... = 2.98... -> rounds up to 3

So the overall ratio is Na2SO3. Empirical formula is the simplest whole number ratio of a compound, and the ratio shown cannot be simplified anymore than it already is.

I hope that is helpful, if there is something I haven't made clear feel free to ask :smile:


That is great thank you so much, what about writing balanced equations for these four:

1. complete combustion of ethane
2. reaction of magnesium with nitric aid
3. reaction between potassium carbonate and sulfuric acids
4. the thermal decomposition of copperiicarbonate

that'd be a fantastic help :biggrin:
Reply 3
Original post by RubyTay
That is great thank you so much, what about writing balanced equations for these four:

1. complete combustion of ethane
2. reaction of magnesium with nitric aid
3. reaction between potassium carbonate and sulfuric acids
4. the thermal decomposition of copperiicarbonate

that'd be a fantastic help :biggrin:


1. Balance this for the first one.

Unparseable latex formula:

\mbox{C}_{2}H_{6}+xO_{2} \rightarrow y\mbox{C}O_{2}+zH_{2}O



2. Remember the formulae for Mg ions and Nitrate ions are Mg2+Mg^{2+} and NO3NO_{3^{}}^{-} and that the general solution is Acid + Metal \rightarrow Salt + Hydrogen

3. Finish this expression
Unparseable latex formula:

K_{2}\mbox{C}O_{3}+H_{2}\mbox{S}O_{4} \rightarrow


Remembering the general solution for an acid + carbonate reaction is Acid + Base \rightarrow Salt + Water + Carbon Dioxide

4. The formula for Copper(II) Carbonate is
Unparseable latex formula:

\mbox{C}u\mbox{C}O_{3}


Recall that Carbon Dioxide is liberated during this decompostion, you should be able to work it out.
(edited 10 years ago)
Reply 4
Original post by RubyTay
That is great thank you so much, what about writing balanced equations for these four:

1. complete combustion of ethane
2. reaction of magnesium with nitric aid
3. reaction between potassium carbonate and sulfuric acids
4. the thermal decomposition of copperiicarbonate

that'd be a fantastic help :biggrin:


You're welcome!
Looks like alow has covered that for you :biggrin:

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