Empirical Formule?

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Can anyone explain empirical formule? 
I missed the lesson when my teacher taught us this

Posted: 11-05-11 09:35 by Emily Sarah

An Empirical Formula is a chemical formula showing the ratio of elements in a compound rather than the total number of atoms. For instance, lets do an example:

Suppose 3.2g of sulfur reacts with oxygen to produce 6.4g of sulfur oxide. What is the formula of the oxide? Use the fact that the Ar of sulfur is 32 and the Ar of oxygen is 16.

Find the mass of each element. Conservation of mass tells us that the mass of oxygen = the mass of sulfur oxide - the mass of sulfur.

The mass of oxygen reacted = 6.4 - 3.2 = 3.2g

So we have 3.2g of sulfur and 3.2g of oxygen.

Now divide the mass of each element by its Ar value.

sulfur: 3.2 ÷ 32 = 0.1

oxygen: 3.2 ÷ 16 = 0.2

Finally, find the ratio of the elements.

You can do this by dividing the results by the smallest of the numbers to give you the number of atoms of each element in the compound.

In this case the smallest value is 0.1, so divide both results by that.

S = 0.1 ÷ 0.1 = 1

O = 0.2 ÷ 0.1 = 2

(If one of the numbers ends in 0.5, multiply all the numbers by 2 - this is because you cannot have half-atoms in a compound.)

So the ratio of sulfur to oxygen is 1:2

The number of atoms tells you that the formula for sulfur oxide is SO2

stepactionSO 1 find masses 3.2 3.2 2 look up given Ar values 32 16 3 divide masses by Ar 0.1 0.2 4 find the ratio 1 2

Posted: 13-05-11 16:45 by jess

This is a really easy topic:
its five easy steps:

1) List all the element in the compoud (usually only two or three)

2) Underneath them, write their experimental massess or percentages

3) Divide each mass or percentage by the Relative Atomic Mass Number for that particular element

4) Turn the numbers into nice simple ratios by mulitplying or dividing them by well chosen numbers

5) get the ratio in simplest form, and that tells you the emperical formula of the compound

E.G find the E.F of the iron oxide produced when 44.8g of iron react with 19.3g of oxygen ( Relative Atomic Mass Number for Oxygen=16 and Iron-56)

1)  Fe O

2) 44.8 19.2

3) 44.8/56 19.2/16

0.8 1.2

4) Multiply by 10...then divide by 2

8 12 (x10)

4 6 (/2)

2 3 (/2)

5) so the simplist formula is 2 atoms of Fe to 3 atoms of O i.e. Fe2 O3

and thats it. :)

Posted: 14-05-11 14:43 by Shahera