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 3.5.9/5.88=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