Answer
$144.3\text{ g/mol}$.
Work Step by Step
Mass of $O_{2}$ reacted = 1.683g - 1.443g = 0.240g
Moles of $O_{2}$ reacted = 0.240g/32$gmol^{-1}$ = $ 7.5 \times 10^{-3}mol$
The reaction between M and oxygen can be taken as,
$2M + 3O_{2} → M_{2}O_{3}$
We determine the number of moles of $O$:
$\frac{0.240}{15.999}\approx 0.015\text{ mols}$
In $M_2O_3$ there are $3$ moles of $O$ per $2$ moles of M, so the number of moles of M is:
$\frac{2}{3}\times 0.015=0.01\text{ mol}$
Now we calculate the atomic mass of M:
$\frac{1.443}{0.01}=144.3\text{ g/mol}$
The metal is Neodymium (Nd).
