1. Advanced Higher Unit 3 Elemental microanalysis

2. Elemental Microanalysis Elemental microanalysis (or combustion analysis) is used to determine the masses of the elements in a sample of an organic compound in order to calculate its empirical formula. The carbon, hydrogen and nitrogen contents are measured using an instrument known as a CHN analyser.

3. The Process Step 1 -An accurate mass of sample (~2 mg) is weighed out. Step 2 -Sample is then placed in a sealed oxygen atmosphere and combusted at ~1000 °C. Step 3 -CO 2, H 2 O and N x O y are produced and passed through a CHN analyzer (a different but similar analyzer is used for S) Step 4 -Other elements, except oxygen, are determined by alternative methods. Step 5 -Samples are analysed in dup;icates to give more reliable results.Results are calculated and expressed as a percentage, by mass. Now watch the RSC video on ‘Elemental Microanalysis’

4. Empirical Formula Calculations C H O 72 12 16 12 1 16 6 12 1 6  1 =6 12  1 =12 1  1 =1 6 12 1 ElementMass Present RAMMoles present  by smallest Whole number ratio Calculate the empirical mass of a compound that contains 72 % carbon, 12 % hydrogen, and 16 % oxygen Empirical Formula is C 6 H 12 O

5. Calculate the empirical formula of the following two examples a) C = 73.17 % H = 7.32 % O = 19.51 % b)C = 69.77 % H = 11.63 % O = 18.60 % C5H6OC5H6O C 5 H 10 O

6. A sample of an organic compound with a mass of 1.224g was completely burned in oxygen and found to produce 2.340g of CO 2 and 1.433g of H 2 O. Calculate the empirical formula of the organic compound. Mass of carbon = 12.0/44.0 x 2.340 = 0.638g Mass of hydrogen = 2.0/18.0 x 1.433 = 0.159g Mass of oxygen = 1.224 – (0.638 + 0.159) =0.427g

7. Number of moles Mole ratio C0.638/12 = 0.053 0.053/0.0267 = 2 H0.159/1 = 0.159 0.159/0.0267 = 6 O0.427/16=0.02670.0267/0.0267 = 1 Empirical formula = C 2 H 6 O

8. Exercise Now try the exercise on page 3 of your Unit 3(d) notes.