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I have recently stumbled upon a question I can't seem to solve :

Chlorine can also be used to synthesise a bromine-containing oxoanion that is mainly used in continuous or batch dyeing processes. When chlorine is passed through an aqueous solution containing potassium hydroxide and potassium bromide, the bromide ions are oxidised to the bromine-containing oxoanion. Careful addition of aqueous lead nitrate to the resulting solution precipitates 4.17g of $PbCl_2$. When this precipitate is filtered off and the resulting filtrate is evaporated, 0.835g of a white crystalline solid is obtained. The white solid has the following composition by mass : $K$, 23.4%; $Br$, 47.8%; $O$, 28.8%. $\mathbb \ Determine\ the\ formula\ of\ the\ white\ crystalline\ solid.$

I attempted by first finding the empirical formula of the solid using by dividing the percentage compositions by their respective elements' mass number, and ended up getting $KBrO_3$ as the empirical formula, which would make my second step to find $n$, where the molecular formula of the white crystalline solid is $K_nBr_nO_{3n}$. However, how can I go about doing this without being given how many moles of the white crystalline solid was precipitated? I can't seem to construct the equations that could potentially lead me to the mole ratio between $PbCl_2$ and $K_nBr_nO_{3n}$. If possible, can anyone tell me how to get a better grip on forming equations in scenarios such as this? I have been struggling in this topic for a while and this is likely to be one of the key reasons why. Sorry for the lengthy question!

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  • $\begingroup$ The short answer is that you can't. Not with the information given, at least. $\endgroup$ – orthocresol Jan 2 at 13:56
  • $\begingroup$ How many moles is 4.17g of PbCl2? $\endgroup$ – Waylander Jan 2 at 13:58
  • $\begingroup$ @Waylander 0.014989 moles, but how do I get the relationship between PbCl2 and the white crystalline? I am really weak in this :( $\endgroup$ – noobchemudjcidixjxjistry Jan 2 at 14:02
  • $\begingroup$ @orthocresol I feel like there might be a chance? This is all the information given in the question. Assuming (i wonder how this is justifiable) all the KCl (when chlorine is oxidised) turns into PbCl2? I'm really not sure but is it really not possible? $\endgroup$ – noobchemudjcidixjxjistry Jan 2 at 14:04
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    $\begingroup$ An alternative (maybe easier) way of looking at it is that the equation is $$ \ce{KBr + 6 KOH + 3 Cl2 -> $\frac{1}{n}$ K_nBr_nO_{3n} + 6 KCl + 3 H2O},$$ so no matter what the value of $n$ is, the same amount of $\ce{KCl}$ will be produced, which leads to the same amount of $\ce{PbCl2}$ being precipitated. Those two equations I've written are the same, just multiplied/divided through by $n$. Also, when you react $\ce{Pb(NO3)2}$ with $\ce{KCl}$, what happens to the $\ce{KNO3}$? Has it just vanished? This problem is problematic... $\endgroup$ – orthocresol Jan 2 at 14:11
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Ionic solids like NaCl, KOH, KBr, PbCl2 and KBrO3 have ratios of individual elements, but the "molecule", if you could use that term to describe it, would be the whole crystal, because all the ions are attracted to nearby ions, ad infinitum.

The white solid could be written as "K23.4% Br47.8% O28.8%", and every time this experiment is run, the same product with the same composition will be obtained. And it will behave exactly the same as the commercial product called potassium bromate with a label saying KBrO3. We prefer the simpler formula.

The numbers given for the weight of PbCl2 and the product KBrO3 do not define a specific reaction, because other reaction products and excess reactants are not specified. And to be realistic, it seems that the filtrate was not evaporated to dryness, because all these other components are not obtained. It seems that this question is posed not just to get you to divide 3 percentages by 3 atomic weights, but to decide what numbers are important to answering the question. The mole ratio between PbCl2 and KBrO3 is unimportant (it's 3) - you got the answer at KBrO3 and should stop there.

Remember this riddle: What has 4 legs and a tail, barks like a dog and concrete? Answer: A dog. (I just threw in the concrete to make it hard.)

A common test technique of teachers is to throw in extra information to see if you know how to select the correct information to use.

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    $\begingroup$ We can't always simplify formulae of ionic solids; as a counterexample, consider $\ce{K2S2O8}$ which we don't write as $\ce{KSO4}$, or $\ce{Na2O2}$, or $\ce{Hg2Cl2}$. Obviously, with chemical intuition, we know that the bromate ion is indeed $\ce{BrO3-}$ and not $\ce{Br2O6^2-}$, $\ce{Br3O9^3-}$, or any other possibility, and so we write $\ce{KBrO3}$. But no information in the question allows us to deduce this. $\endgroup$ – orthocresol Jan 2 at 15:38
  • $\begingroup$ @ orthocresol: The question was: π·π‘’π‘‘π‘’π‘Ÿπ‘šπ‘–π‘›π‘’ π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘€β„Žπ‘–π‘‘π‘’ π‘π‘Ÿπ‘¦π‘ π‘‘π‘Žπ‘™π‘™π‘–π‘›π‘’ π‘ π‘œπ‘™π‘–π‘‘. (not the ions). There are 3 common definitions for (chemical) formula: the empirical formula, the molecular formula and structural formula. (Wikipedia) Structural formula (crystal structure) and molecular formula are not what was meant (since the compound does not consist of molecules). Wikipedia: In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound. The only correct answer must be KBrO3. $\endgroup$ – James Gaidis Feb 2 at 21:21

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