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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?

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  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. $\endgroup$ – Loong Dec 26 '16 at 21:27
  • $\begingroup$ You need to use the law of definite proportions and the law of multiple proportions. You're looking for a simple ratio between the moles of F to moles of the metal. $\endgroup$ – Zhe Dec 26 '16 at 21:32
  • $\begingroup$ If you have two accounts that you would like to join together, please sign into either account, visit the contact form and select ‘I need to merge user profiles’. $\endgroup$ – Loong Dec 27 '16 at 12:39
  • $\begingroup$ Assuming that your calculation is correct, which oxidation states sound better? Fe-(II)/(III) or Cd-(IV)/(VI)? You'll agree that's OK for iron, what about cadmium? Have one look into the periodic table, and you know the answer. $\endgroup$ – Karl Dec 27 '16 at 19:18
  • $\begingroup$ I'd like to thank you all for your help. I understand that Cd(VI) or even Cd(IV) is quite unlikely, making Fe the better answer. $\endgroup$ – LCHL Dec 27 '16 at 20:00
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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 multiple 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 $\ce{CdF6}$.

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  • $\begingroup$ Why do you say that the compounds would be $\ce{M3F}$ and $\ce{M2F}$? I see that those are the proportions by mass, but not by mole. Is that what you meant and if so, why is it weird? Also I don't see where you would have lost a significant figure anywhere. Your choice of rounding for F warrants rounding to 4 in any final answer. Aside from those questions I think it is a + answer. $\endgroup$ – Joseph Hirsch Dec 27 '16 at 16:51
  • $\begingroup$ So I don't see where you would have lost a sig fig, but I do agree with Li Zhi that if you were limited to 3, the rounding should not have happened at the step that yielded 27.99 and 18.67 $\endgroup$ – Joseph Hirsch Dec 27 '16 at 16:59
  • $\begingroup$ I see now that you should have lost one in the summation (subtration) that yielded 18.67. $\endgroup$ – Joseph Hirsch Dec 27 '16 at 17:42
  • $\begingroup$ Also @ Zhe you divided both fractions by 6. 18.67 not 16.67. $\endgroup$ – Joseph Hirsch Dec 27 '16 at 18:32
  • $\begingroup$ Thanks @JosephHirsch. Fixed that and the rounding issues while I was at it. $\endgroup$ – Zhe Dec 27 '16 at 18:38
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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.

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  • $\begingroup$ So, you're proposing $\ce{CdF6}$? $\ce{CdF4}$ instead of $\ce{FeF2}$ and $\ce{FeF3}$? Also, I mentioned the issue with rounding. I was just too lazy to fix it after the fact... $\endgroup$ – Zhe Dec 27 '16 at 16:12
  • $\begingroup$ Where did Zhe claim he should have rounded "as an intermediate step"? He claimed that his final. Actually, in manipulating the formula, he has an intermediate step where he takes 1900. - 768.17. Even though the 1900 has 4 sig figs, when subtracting or adding in an intermediate step, you have to round immediately (in this case to the 1's place). You can't carry 1131.83 to the end because it is a summation so it should have been rounded to 1132, but it only changes the final answer to 28.00 $\endgroup$ – Joseph Hirsch Dec 27 '16 at 17:20
  • $\begingroup$ In the other manipulation we have 1900. - (50.44)19.00. Once again this summation requires immediate rounding to the 1's place in this case giving 942 and 18.7 Again, you have to round to place immediately after summation. In this case you lose 1 sig fig. This gives an m=56.9 using his final step which is pretty far off. (or 56.1 and 56.0 if you use 19.00 to get the the denominators). $\endgroup$ – Joseph Hirsch Dec 27 '16 at 17:37
  • $\begingroup$ That should be 56.1 and 56.00 above^^ $\endgroup$ – Joseph Hirsch Dec 27 '16 at 17:45
  • $\begingroup$ You know, if you had registered you would probably have had enough reputation to comment by now … $\endgroup$ – Jan Dec 27 '16 at 21:08

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