Question

Considering 100% pure, which of the following has a maximum percentage of Pb?
(A) $PbS$
(B) $PbS{O_4}$
(C) $PbC{O_3}$
(D) $PbO$

Answer 1

As we know that substances are 100 % pure. So we can calculate Molecular mass of each component. After we get molecular mass, then to find the percentage of Pb, we can divide the Mass of Pb (Lead) to the total molecular mass.

Complete Step by step answer: It is given in question that the substances are 100% pure, so we can calculate the percentage of Pb in each molecule.
The percentage of Pb, can be calculated in the following manner.
${\text{Percentage of Pb = }}\dfrac{{{\text{Atomic mass of Pb}}}}{{{\text{Molecular mass of Molecule}}}}$ --- equation 1
Now, let us assume mass of Pb to be ‘x’
(A) PbS
Molecular mass of PbS = Atomic mass of Pb + Atomic mass of S
$\therefore {\text{Molecular mass of PbS = x + 32}}$
Substitute the values in equation 1
${\text{Percentage of Pb}} = \dfrac{x}{{x + 32}}$
(B) $PbS{O_4}$
Molecular mass of $PbS{O_4}$ = Atomic mass of Pb + Atomic mass of S +$4 \times {\text{Atomic mass of oxygen}}$

$${\text{Molecular mass of PbS}}{{\text{O}}_{\text{4}}}{\text{ = x + 32 + 4}} \times {\text{16}} \\\ {\text{Molecular mass of PbS}}{{\text{O}}_{\text{4}}}{\text{ = x + 32 + 64}} \\\ \therefore {\text{Molecular mass of PbS}}{{\text{O}}_{\text{4}}}{\text{ = x + 96}} \\\$$

Substitute the values in equation 1
${\text{Percentage of Pb}} = \dfrac{x}{{x + 96}}$
(C) $PbC{O_3}$
Molecular mass of $PbC{O_3}$ = Atomic mass of Pb + Atomic mass of C + $4 \times {\text{Atomic mass of oxygen}}$

$${\text{Molecular mass of PbC}}{{\text{O}}_3}{\text{ = x + 12 + 3}} \times {\text{16}} \\\ {\text{Molecular mass of PbC}}{{\text{O}}_3}{\text{ = x + 12 + 48}} \\\ \therefore {\text{Molecular mass of PbC}}{{\text{O}}_3}{\text{ = x + 60}} \\\$$

Substitute the values in equation 1
${\text{Percentage of Pb}} = \dfrac{x}{{x + 60}}$

(D) $PbO$
Molecular mass of $PbO$ = Atomic mass of Pb + Atomic mass of Oxygen
$\therefore {\text{Molecular mass of PbO = x + 16}}$
Substitute the values in equation 1
${\text{Percentage of Pb}} = \dfrac{x}{{x + 16}}$
Now, we can observe that the numerator is the same in all 4 options, but the denominator will decide the percentage.
Higher the denominator, lower will be the value of percentage of Pb. Now we can see that in case (B), the denominator is highest.

Thus, the correct option is (B) $PbS{O_4}$.

Note: If purity is varying in all cases, then that should be taken into consideration, but here all substances are 100 % pure. To calculate molecular mass, we should know the atomic mass of individual common elements, so that we can find molecular mass. If one can remember the mass of Pb (Lead) then that can be substituted in case of ‘x’, but even without substitution of atomic mass of Pb, we can calculate the percentage, as the question says to compare all 4 salts given.