Question

1.5 gm of $CdC{{l}_{2}}$ was found to contain 0.9 gm of Cd. Calculate the atomic weight of Cd.

Answer 1

Hint: The required amount of an atom in a molecule is calculated with respect to the given amount of the combining atom in the same. Both are responsible to form a molecule in a specific ratio.

Complete step by step solution:
The given problem can be solved by simple knowledge of atoms, molecules, atomic weights, molecular weights, stoichiometry, etc. Let us study them before solving the illustration;
-Atom is an element in the modern periodic table having a specific atomic mass and a symbol to represent itself.
-The molecule is formed by the combination of numerous atoms, where their oxidation states and valencies are satisfied while forming the same. It is represented by a specific molecular formula.
-Atomic mass is the mass of an atom with the reference to the mass of Carbon-12.
-Molecule has a molecular weight calculated by considering the atomic weights of atoms (who are combined while forming the molecule) and their ratios too.
-Stoichiometry is the concept where we can say that we balance both the sides of the reaction taking place, to satisfy the respective equation. We balance the equation by adding stoichiometric coefficients and not changing the subscripts of a molecule.
Now, let us move forward with the basic ideas to solve the given problem,
Illustration-
Given that,
Weight of $CdC{{l}_{2}}$ = 1.5 gm
Weight of Cd = 0.9 gm
Thus, we can say that,
1.5 gm of $CdC{{l}_{2}}$ contain = (1.5 – 0.9) = 0.6 gm of chlorine.
By stoichiometry,
0.6 gm of chlorine combines with 0.9 gm of Cd.
Thus, 71 gm of chlorine will combine with = $\dfrac{0.9}{0.6}\times 71gm=106.5gm$ Cd.

Therefore, the atomic weight of Cd is 106.5 gm.

Note: We can calculate the same by considering the aggregate of 100% and then calculating the percentages of the atoms in the molecule (ratio calculations). But go with the steps which would be easy and can avoid confusion.