Molar mass of Al = 27.0 g/mol Molar mass of S = 32.1 g/mol Molar mass of Aluminium sulfide = 27.0 x 2 + 32.1 x 3 = 150.3 g/mol % of Al (there are 2 Al atoms) in Aluminium sulfide = [tex] \frac{54.0}{150.3} * 100 = 35.9%[/tex] % of S (there are 3 S atoms) in Aluminium sulfide = [tex] \frac{96.3}{150.3} * 100 = 64.1%[/tex] Assume the percentages to be masses, so: In 100.0 g of Aluminium sulfide, n (Al) = [tex] \frac{35.9}{27.0} = 1.33 mol[/tex] In 100.0 g of Aluminium sulfide, n (S) = [tex] \frac{64.1}{32.1} = 2.00 mol[/tex] Convert values to a molar ratio, 1.33 : 2.00 Divide by smallest ratio value, 1 : 1.50 Multiply each ratio value by a factor of 2 to eliminate the decimal, 2 : 3
Hence why the empirical formula of Aluminium sulfide is [tex] Al_{2} S_{3} [/tex]
