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12 Chemistry 2.3 gravimetric analysis CR 07 Empirical formulae The Empirical Formula (EF) is the ratio of all elements in a Compound It is the smallest/simplest.

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Presentation on theme: "12 Chemistry 2.3 gravimetric analysis CR 07 Empirical formulae The Empirical Formula (EF) is the ratio of all elements in a Compound It is the smallest/simplest."— Presentation transcript:

1 12 Chemistry 2.3 gravimetric analysis CR 07 Empirical formulae The Empirical Formula (EF) is the ratio of all elements in a Compound It is the smallest/simplest ratio of elements Eg. C 6 H 12 O 6 EF is CH 2 O

2 12 Chemistry 2.3 gravimetric analysis CR 07 You can work out what an unknown substance is by Calculating %age composition of each element and then working out the Empirical Formula

3 12 Chemistry 2.3 gravimetric analysis CR 07 %age comp  EF 1.Assume 100g of sample so %age of each element is number of g of each element 2.Calculate number of moles for each element 3.Simplify the ratio of moles of each element (divide by smallest number)

4 12 Chemistry 2.3 gravimetric analysis CR 07 4.Make sure that the ratios are whole numbers 5.Write the empirical formula from the ratios

5 12 Chemistry 2.3 gravimetric analysis CR 07 Worked example Analysis of a clear liquid shows 5.9 % hydrogen and 94.1 % oxygen 1.100g so 5.9g H and 94.1g O 2.n(H)=5.9/1=5.9 n(O)=94.1/16= /5.88=1.003, 5.9/5.9=1 4.H:O = 1:1 5.EF is HO (ie 1 H for every O atom)

6 12 Chemistry 2.3 gravimetric analysis CR 07 EF is HO Possible molecular formula (actual ratio of atoms) are: HO, H 2 O 2, H 3 O 3… Molar mass17, 34, 51

7 12 Chemistry 2.3 gravimetric analysis CR 07 EF  MF MF =  x EF -where  = molar mass/ mass of EF

8 12 Chemistry 2.3 gravimetric analysis CR 07 Unknown formula % composition Empirical Formula Molecular mass Molecular Formula


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