Question

Rubies are essentially alumina Al2O3 with a contaminant of chromium in the form of Cr2O3. A lab made ruby weighing 1.994 g is ground, digested in hot nitric acid, and the chromium precipitated as Cr(OH)3 If the mass of the Cr(OH)3 is .0135 g, compute the percentage of chromium in the ruby. Also compute the percentage of Cr2O3 in the ruby.

Answer 1

Mass of Cr(OH)₃ = 0.0135 g

moles Cr(OH)₃ = (mass Cr(OH)₃) / (molar mass Cr(OH)₃)

moles Cr(OH)₃ = (0.0135 g) / (103.02 g/mol)

moles Cr(OH)₃ = 1.31 x 10^-4 mol

moles Cr = moles Cr(OH)₃

moles Cr = 1.31 x 10^-4 mol

mass Cr = (moles Cr) * (molar mass Cr)

mass Cr = (1.31 x 10^-4 mol) * (51.996 g/mol)

mass Cr = 0.00681 g

percentage of chromium in ruby = (mass Cr / mass of sample) * 100

percentage of chromium in ruby = (0.00681 g / 1.994 g) * 100

percentage of chromium in ruby = 0.3417 %

moles Cr₂O₃ = (moles Cr) / 2

moles Cr₂O₃ = (1.31 x 10^-4 mol) / 2

moles Cr₂O₃ = 6.55 x 10^-5 mol

mass Cr₂O₃ = (moles Cr₂O₃) / (molar mass Cr₂O₃)

mass Cr₂O₃ = (6.55 x 10^-5 mol) * (152 g/mol)

mass Cr₂O₃ = 0.009956 g

percentage of Cr₂O₃ in ruby = (mass Cr₂O₃ / mass of sample) * 100

percentage of Cr₂O₃ in ruby = (0.009956 g / 1.994 g) * 100

percentage of Cr₂O₃ in ruby = 0.4993 %