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It takes $\pu{8.0 cm^3}$ of aqueous sodium phosphate to fully react with $\pu{12.0 cm^3}$ of an aqueous solution of the nitrate of metal $\ce{T}$. If both solutions have a concentration of $\pu{1 mol dm^-3},$ what's the formula of the phosphate of metal $\ce{T}?$

The answer: As both solutions have the same concentration, the mole ratio is the same as the volume ratio which is $12:8$ or $3:2.$ Thus, the formula is $\ce{T3(PO4)2}.$

I get how the mole ratio is $3:2,$ but how do they deduce the formula as being $\ce{T3(PO4)2}$ from it? Isn't the mole ratio just a way of quantifying the coefficients of the chemical equation? How do they, from the mole ratio, figure out the molecular formula of the phosphate of the mystery metal $\ce{T}?$

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    $\begingroup$ Write reaction equations for various valences of M. $\endgroup$
    – Poutnik
    Nov 6, 2021 at 7:19
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    $\begingroup$ Metal T or metal M. Tomayto tomahto. $\endgroup$
    – Karsten ♦
    Nov 6, 2021 at 11:36
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    $\begingroup$ @KarstenTheis Metal T obviously, I have forgotten the unusual designation of a metal. $\endgroup$
    – Poutnik
    Nov 6, 2021 at 13:09
  • $\begingroup$ @Poutnik Speaking of unusual notations, keep in mind there are emojis used to teach math: YouTube — Emoji Maths Puzzle. I actually find it pretty effective for teaching (and horrible for IRL or field applications). $\endgroup$
    – andselisk ♦
    Nov 6, 2021 at 13:13
  • $\begingroup$ @andselisk Such puzzles are often very tricky :-) $\endgroup$
    – Poutnik
    Nov 6, 2021 at 13:40

1 Answer 1

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There is $0.008$ moles of $\ce{PO4^{3-}}$ ions. So the total number of negative charges is $3·0.008 = 0.024$ moles. This number must be equal to the number of positive charges. Let's check it.

As there is also $0.012 $ mol $\ce{T^{x+}}$, the total number of positive charges is : $\pu{x · 0.012 = 0.024}$ moles. As a consequence x = $2$, and the unknown ion $\ce{T^{x+}}$ is doubly charged, like $\ce{Ca^{2+}}$ or $\ce{Zn^{2+}}$ . And now we know the formula of the ions forming the precipitate. They are : $\ce{T^{2+}}$ and $\ce{PO4^{3-}}$.

In order to obtain a neutral precipitate (total charges = zero), it is necessary to combine at least $\ce{2 PO4^{3-}}$ and $\ce{3 T^{2+}}$. This would make $2·3 = 6$ positive charges neutralized by $3·2 = 6$ negative charges. The precicitate must have the formula $\ce{T3(PO4)2}$

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