What is the percentage by mass of lead sulphate?

Hint: We very well know that the percentage composition of a molecule or a compound or a complex is the amount of each element or atom divided by the total amount of individual elements or atoms in that particular molecule or compound and this is then multiplied with 100.
Formula used:
We will use the following formula:-
Percentage composition of each atom = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$

Complete answer:
Let us begin with discussion of percentage composition as follows:-
Percentage composition: It refers to the ratio of an amount of each element to the total amount of individual elements in a particular molecule, compound or complex which is further multiplied with 100. Mathematically it is represented as follows:-
Percentage composition of each atom = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
-The mass of 1 mole of lead sulphate ($PbS{{O}_{4}}$):-
Molar mass of Pb = 207 g/mol
Molar mass of S = 32 g/mol
Molar mass of O = 16g/mol
So total molar mass of$PbS{{O}_{4}}$= [(207) + (32) + 4(16)]g/mol = 303 g/mol
-Percentage composition of lead (Pb) in lead sulphate ($PbS{{O}_{4}}$):-
Total mass of lead present in the molecule of $PbS{{O}_{4}}$= 202 g/mol
So percent composition of Pb = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
$\begin{align}
  & \Rightarrow \dfrac{207g/mol}{303g/mol}\times 100\% \\
 & \Rightarrow 68.31\% \\
\end{align}$
-Percentage composition of sulphur (S) in lead sulphate ($PbS{{O}_{4}}$):-
Total mass of sulphur present in the molecule of $PbS{{O}_{4}}$= 32 g/mol
So percent composition of S = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
$\begin{align}
  & \Rightarrow \dfrac{32g/mol}{303g/mol}\times 100\% \\
 & \Rightarrow 10.56\% \\
\end{align}$
-Percentage composition of oxygen (O) in lead sulphate ($PbS{{O}_{4}}$):-
Total mass of oxygen present in the molecule of $PbS{{O}_{4}}$= 3(16) g/mol = 48 g/mol
So percent composition of O = $\dfrac{\text{Molar mass of the atom present}}{\text{Molar mass of the molecule}}$
$\begin{align}
  & \Rightarrow \dfrac{64g/mol}{303g/mol}\times 100\% \\
 & \Rightarrow 21.12\% \\
\end{align}$

Note:
-Remember that the sum of mass percentage of all the elements present in a molecule or compound is always equal to a hundred.
-Also try to remember and learn atomic numbers and molar masses of important elements from the periodic table for time saving.