A metal, M, forms two fluorides containing 40.43% F and 50.44% F by mass. Identify the metal
You need to use the general concepts of the law of definite proportions and the law of multiple proportions to solve this:
Sample 1 is 59.57% $\ce{M}$ and 40.43% $\ce{F}$.
Sample 2 is 49.56% $\ce{M}$ and 50.44% $\ce{F}$.
Assume $\ce{M_{x}F_{y}}$ for sample 1 and $\ce{M_{a}F_{b}}$ for compound 2.
Let $m$ by the molar mass of $\ce{M}$. We have
$$\frac{19.00y}{mx + 19.00y}=\frac{40.43}{100}$$
and
$$\frac{19.00b}{ma + 19.00b}=\frac{50.44}{100}$$
We can rearrange the equations to:
$$m\frac{x}{y}=28.0$$
and
$$m\frac{a}{b}=18.7$$
Note that $$\frac{28.0}{18.7}\approx\frac{3}{2}$$
so we can start by trying:
$$m\frac{3}{1} = 28.0$$
and
$$m\frac{2}{1} = 18.7$$
This is weird because the compounds would be $\ce{M3F}$ and $\ce{M2F}$.
Fortunately, as long as the proportions work, it's fine, so divide both fractions by 6:
$$m\frac{1}{2} = 28.0$$
and
$$m\frac{1}{3} = 18.7$$
This gives $m=56.0$, fairly close to the molar mass of iron which is $55.9$.
Cadmium is a bad candidate, since that would be $\ce{CdF4}$ and $\ce{CdF6}$.