Question

Calculate the experimental value for percent sulfate in alum.

Calculations for Part 2 1. Experimental value for percent sulfate in alum

Answer 1

a) Percent of water of hydration in the alum :

Experimental mass percent of water = mass of alum before - mass of alum after/mass of alum before) x 100 = (1.121 g - 0.5098 g)/1.121 g ) x 100 = 0.6112/1.121) x 100 = 54.5 %

Experiment mass percent of sulphate :

Mass of barium sulphate = 0.1703 g

Molar mass of sulphate ion (SO₄^-2) = 96.07 g/mol

Molar mass of barium sulphate (BaSO₄) = 233.4 g/mol

233.4 ---> 96.07 ----> 100 %

0.1703 ---> ?? ---> x %

Experimental percent mass of sulphate = (mass of barium sulphate x molar mass of sulphate/molar mass of barium sulphate) x 100 = (0.1703 g x 96.07 g/mol/233.4 g/mol) x 100 = 16.360721 x 100/233.4 = 1636.0721/233.4 = 7.01 %

Theoretical percent of sulphate : (molar mass of sulphate/molar mass of barium sulphate) x 100 = (96.07/233.4) x 100 = 41.16 %

percent error = ([exp percent - theoretical percent])/theoretical percent) x 100 = ([7.01-41.16])/41.16) x 100 = 34.15/41.16) x 100 = 82.97 %