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Empirical Formulas (GCSE level) I need help!!

I'm doing GCSE OCR Chem tomorrow and I'm revising calculations. I'm stuck on the empirical formulas (since we didn't actually cover them in school...I'm just trying to understand this CGP revision guide).

Does anyone here have an easy way of calculating empirical formulas? The method I'm using here goes like this -
1) Divide each elements mass or percentage given in the question by the relative atomic mass of each element.
2) Turn the numbers into 'nice simple' ratios by x or / by well-chosen numbers.
3) Get the ratios in it's simplest form.

I can do step 1 easily but am having problems getting it into a simple ratio. For example,

Q2: Find the empirical formula of a compound of 82% nitrogen and 18% hydrogen.

Nitrogen (RAM 14):
82%/14 = 5.86 (3sf)

Hydrogen (RAM 1):
18/1 = 18

so the ratio is 5.86:18 - but obviously they've both got to be whole numbers.

How do I change it into a whole number ratio? Can I enter 5.86/18 into my calculator and turn the answer into a fraction (using the fraction button)??? This would make 41/126 which would mean the empirical formula was N41H126 . Is that correct? 41 and 126 seem a bit high...

Would anyone that is really good at chemistry help me...possibly work out the empirical formula in the question and see if it does come to N41H126?

I know this is a really long post so thanks for reading it!! I appreciate it and i hope someone can help me!

MissSurfer
Umm have you tried using the Mass = Moles X RAM formula I know on the AQA syllabus you can use that to work out ratios on there, not too sure with OCR though....
MissSurfer
I'm doing GCSE OCR Chem tomorrow and I'm revising calculations. I'm stuck on the empirical formulas (since we didn't actually cover them in school...I'm just trying to understand this CGP revision guide).

Does anyone here have an easy way of calculating empirical formulas? The method I'm using here goes like this -
1) Divide each elements mass or percentage given in the question by the relative atomic mass of each element.
2) Turn the numbers into 'nice simple' ratios by x or / by well-chosen numbers.
3) Get the ratios in it's simplest form.

I can do step 1 easily but am having problems getting it into a simple ratio. For example,

Q2: Find the empirical formula of a compound of 82% nitrogen and 18% hydrogen.

Nitrogen (RAM 14):
82%/14 = 5.86 (3sf)

Hydrogen (RAM 1):
18/1 = 18

so the ratio is 5.86:18 - but obviously they've both got to be whole numbers.

How do I change it into a whole number ratio? Can I enter 5.86/18 into my calculator and turn the answer into a fraction (using the fraction button)??? This would make 41/126 which would mean the empirical formula was N41H126 . Is that correct? 41 and 126 seem a bit high...

Would anyone that is really good at chemistry help me...possibly work out the empirical formula in the question and see if it does come to N41H126?

I know this is a really long post so thanks for reading it!! I appreciate it and i hope someone can help me!

MissSurfer



to work out the simple ratio 1st u have 2 divide both numbers by the smallest value-
in ur case u divide 5.68/5.68 =1
and 18/5.68 = 3 (round it up)
so ur formula is NO3(small 3!)
Reply 3
You will be told how much of each element you have. Check out this example:

E.g if they said: Find the formual of Iron Oxide when 44.8g or Iron react with 19.2g of Oxygen.

1. (get the moles) Moles = mass/RAM

So : Fe ------- O
Mass : 44.8 ------- 19.2
RAM : 56 ------- 16

Moles : 0.8 ------- 1.2


2. So youve got the moles. Next bit is to divide both numbers by the smallest number. E.g divide 0.8 and 1.2 by the smallest number (which is 0.8)

If you do that you get

Fe O
1 : 1.5

x2

2 : 3

(now you cant have 1.5, so double both numbers and you get the answer)

Fe2O3 ( NB- The O is oxygen, not a zero)