# How to work out Empirical Formula

I have made some revision cards explaining how to work out empirical formula.
Hope they are useful!

Please comment and rate, thankyou :)

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## Relative Formula Mass

RFM stands for Relative Formula Mass

RAM (Relative Atomic Mass), this is the same number as the elements mass number. For example Carbon's RAM is 12 and Oxygen's is 16.

How to work RFM out;

What is the RFM of H2O?

So RAM's;    H=1   O=16

So there are two Hygrogens in the compound and one Oxygen atom

1+1+16=18

You use the RAM's of each of the atoms to work out the RFM.

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## Harder RFM example

What is the RFM of 2CaO?

RAM's;   Ca=40       O=16

What do we do?

There is only one calcium atom and one oxygen atom

so...      40+16=56

When doing the RFM we ignore the number used to balance the equation in this case it is 2.

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## Empirical Formula

Q) A painting was found from many many years ago, scientists decided to split the compounds up in the paint. They found small amounts of Calcium and Bromine. Calculate the empirical formula of the compiund composed of 20% Calcium and 80% Bromine.
RAM's; Ca=40      Br=80

A) 20  (The percentage amount of that element)
----  = 0.5      (Calcium)
40  (The RFM value of that same element)

80
--- =1 (Bromine)

80

Then you would divide each of the new values by the smallest value;

0.5/0.5= 1 and  1/0.5=2

So the empirical formula of this compund is CaBr2

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## Another worked example

Q) A compound was found in some ancient Grecian make-up it was composed of 75% carbon and 25% hydrogen. Calculate the empirical formula of this compound?
RAM's;   C=12   H=1

A)    75/12=6.25
25/1=25

6.25/6.25=1
25/6.25=4

The empirical formul of this compound is CH
We can also identify that this compound is infact Methane.

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## Another worked example

Q) Lead oxides have been used for hundreds of years as colours in paint. A smaple of red oxide used in paint was found to contain 6.21g of lead and 0.64g of oxygen. Calculate the empirical formula of this compound.
RAM's;    O=16    Pb=207

A)     6.21/207=0.03
0.64/16=0.04

0.03/0.03=1
0.04/0.03=1.3

Notice that the value I got for the oxide  is not a whole number, so you need to multiply both numbers by another number to get a whole number.

1 x 3 = 3
1.3 x 3 = 4  (the value for this number was 3.9 but we aren't going to get any closer to a whole number so we just round it )

So the empirical formula is Pb3O4

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## Explanation of the previous card

Notice that on the previous example the values for the amount of each element were in grams not %'s. It doesn't matter what measurement they are in always use the same method . A common mistake is that people try and convert the values back into %'s, you may not end up with the same results and it just takes longer doing this.

Also watch out for any values which are not whole numbers, always round to one decimal place or two at the most.

I hope my cards have helped!

:)

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Thanks helps a lot

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Thanks helps a lot

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I'm a bit confused by this example. Have the teaching methods changed since the past 5 years?

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