Answer
A sample with 110.5g of this ore contains 9.57g of lead.
Work Step by Step
- Purity of the ore: 100% ore has 10.0% $PbS$ by mass.
- Molar mass for lead sulfide: $ 207.2* 1 + 32.07* 1 = 239.3g/mol (PbS)$
- Molar mass for lead: $207.2 g / mol (Pb)$
- Number of lead atoms in $PbS$: Each lead sulfide has 1 $Pb$ atom. The ratio is 1 mol(Pb) to 1 mol(PbS).
Use these informations as conversion factors:
$110.5 g (ore) \times \frac{10.0\%(PbS)}{100\%(ore)} \times \frac{1mol(PbS)}{239.3g (PbS)} \times \frac{1mol(Pb)}{1mol(PbS)} \times \frac{207.2g(Pb)}{1mol(Pb)} = 9.57g(Pb)$
