# Using the mole ratio to figure out the molecular formula [closed]

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 $$\ce{T}?$$

• Write reaction equations for various valences of M. Nov 6, 2021 at 7:19
• Metal T or metal M. Tomayto tomahto.
– Karsten
Nov 6, 2021 at 11:36
• @KarstenTheis Metal T obviously, I have forgotten the unusual designation of a metal. Nov 6, 2021 at 13:09
• @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). Nov 6, 2021 at 13:13
• @andselisk Such puzzles are often very tricky :-) Nov 6, 2021 at 13:40

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 $$\ce{T3(PO4)2}$$