Question

A 2.70-g geological sample, containing the mineral Boulangerite (Pb5Sb4S11) as the only lead-bearing compound, is subjected to chemical treatment in which the lead is quantitatively recovered as solid PbCl2. The mass of PbCl2 obtained is 1.64 g. What was the percentage (by mass) of Pb5Sb4S11 in the sample?

Answer 1

Molar mass of PbCl2,

MM = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

mass(PbCl2)= 1.64 g

number of mol of PbCl2,

n = mass of PbCl2/molar mass of PbCl2

=(1.64 g)/(278.1 g/mol)

= 5.897*10^-3 mol

moles of Pb = 5.897*10^-3 mol

One mol of Pb5Sb4S11 has 5 moles of Pb

so, moles of Pb5Sb4S11 = 5.897*10^-3 / 5 = 1.1794*10^-3 mol

Molar mass of Pb5Sb4S11,

MM = 5*MM(Pb) + 4*MM(Sb) + 11*MM(S)

= 5*207.2 + 4*121.8 + 11*32.07

= 1875.97 g/mol

mass of Pb5Sb4S11,

m = number of mol * molar mass

= 1.179*10^-3 mol * 1875.97 g/mol

= 2.213 g

% = mass of Pb5Sb4S11*100/mass of sample

= 2.213*100/2.70

= 82.0 %

Answer: 82.0 %