Answer:
The empirical formulae is C6H12S02
Explanation:
1. First we need to obtain the mass of each element in the sample and compound formed
Carbon = (62.637 mg * 12.011 g/mol / 44.009 g/mol) = 17.094 mg of Carbon
Hydrogen = ( 25.641 mg * (2 *1..008 g/mol) / 18.015 g/mol) = 2.869 mg of Hydrogen
Sulphur = (13.54 mg * 32.066 g/mol / 64.066 g/mol) = 6.777 mg of Sulphur
2. Next is to determine the percentage composition. Here we divide the respective mass by the mass of the sample
Carbon = 17.094 / 35.161 * 100 = 48.62 %
Hydrogen = 2.869/ 35.161 *100 = 8.16 %
Sulphur = 6.777/ 31.321 *100 = 21.64 %
Oxygen = (100 - (48.62 + 8.16 + 21.64)) = 21.58 %
3. Next is to divide the mass assuming there are 100 mg by the respective atomic masses to obtain the number of moles
Carbon = 48.62 / 12.011 = 4.048 mol
Hydrogen = 8.16 / 1.008 = 8.095 mol
Sulphur = 21.64 / 32.066 = 0.675 mol
Oxygen = 21.58 / 16.000 = 1.348 mol
Next is to divide by the smallest value
Carbon = 4.048/ 0.675 =5.997 = 6
Hydrogen = 8.095 / 0.675 =11.993 =12
Sulphur = 0.675/ 0.675 = 1
Oxygen = 1.348 / 0.675 = 1.997 = 2
So therefore the empirical formulae of the sample is C6H12SO2
