A metal, M, forms two fluorides containing 40.43% F and 50.44% F by mass. Identify the metal

Question

A metal, $\ce{M}$, forms two fluorides containing 40.43 % $\ce{F}$ and 50.44 % $\ce{F}$ by mass. Identify the metal.

I get that you can find out how much of the metal reacts with 1 mole of fluorine atoms in the 40.43 % one. This can be calculated to be 27.99 g.

Then from that you can work out that 28.49 g of fluorine reacts with the metal in the 50.44 % one and that that is about 1.5 moles of $\ce{F}$.

I'm a bit lost after that. How might you continue from here?

So the law of multiple proportions conveys that the fluorides contain fluorine in a 1.5:1 or 3:2 ratio. This means the possibilities are

$\ce{MF2}$ and $\ce{MF3}$, or
$\ce{MF4}$ and $\ce{MF6}$

Since there is twice as much fluorine, the mass of $\ce{M}$ in $\ce{MF2}$ must be 56 g. The next one is 112 g. If there are no errors in what was said, the metal must be iron or cadmium. What makes one a better candidate over the other?

Answer 1

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 $\ce{CdF6}$.

Answer 2

Another approach would be to assume that one of the compounds is of the type $\ce{MF_n}$ with $n$ a small integer. Using a spreadsheet (or just building a table) you use the small integer multiples of $18.9984$ (say across the top 6 of the columns $n=1,2,3,4,5,6$ — hexavalent compounds are fairly common, higher valences are possible but much less likely. For the rows' first column you choose the values of the % of F. This allows you to calculate the mass (which we can (but don't have to) assume is the atomic weight (atomic mass)) of the metal). That calculation is that the mass of the F atoms is the respective % of that row, so the total mass is the Fs mass ÷ %F. Subtract the mass of Fs from this and you get the mass of the M. (Yes, you could have done the same thing using the $100-\%\ \ce{F}$ values, maybe even simpler, IDK).

Then, you look at the entries, and see two (and only two) pairs that are "interesting" - meaning they're close to one another. Under 2F ($37.9968$) is $55.984$ and under 3F is $56.001$ and under 4F is $111.970$ and under 6F is $112.002$. These suggest that Fe at $55.845$ or Cd at $112.414$ are good matches. In fact if you compute the difference between their theoretical %s and the measured values, the sum of the square of the differences are $\ce{Fe} = 0.41$ and $\ce{Cd} = 0.00$, which indicates that statistically Cd is the "better" answer. You then should compare the values and determine if the difference between them is meaningful given the measured results. Since the %s were given to 4 places, I think that Cd is the "more correct" answer, the error for Fe is simply too large, IMHO.

This approach has another advantage: you can inspect the table to see if there are any values which are small multiples of an actual atomic weight of a metal (atomic mass). This would allow you to evaluate the assumption that the formula is $\ce{MF_n}$ and not $\ce{M2F_n}$ or $\ce{M3F_n}$, depending on how good your mental arithmetic is.