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This might be a basic question but I am a little confused.

I have some lecture notes that says calcium phosphate ($\ce{Ca3(PO4)2}$) reacts with calcium hydroxide ($\ce{Ca(OH)2}$) to form Hydroxyapatite ($\ce{Ca10(PO4)6(OH)2}$).

So, the chemical formula (even when unbalanced) should be:

$$\ce{Ca3(PO4)2 + Ca(OH)2 -> Ca10(PO4)6(OH)2}$$

I can't figure out why $\ce{Ca10(PO4)6(OH)2}$ is the product of $\ce{Ca3(PO4)2}$ and $\ce{Ca(OH)2}$ and I am having trouble balancing it (see below). On Sigma Aldrich's website, it says the linear Formula for hydroxyapatite is $\ce{3Ca3(PO4)2·Ca(OH)2}$. I copied and pasted this directly. I assume the dot between the calcium phosphate and calcium hydroxide means 'plus'.

On the Sigma Aldrich's website, the calcium phosphate is written/balanced as $\ce{3Ca3(PO4)2}$, whereas the lecture notes I have are just $\ce{Ca3(PO4)2}$. Even with the three in front as the coefficient, I get $\ce{9Ca + Ca -> Ca10}$, $\ce{(PO4)2}$ and $\ce{(OH)2}$ which is $\ce{Ca10(PO4)2(OH)2}$, instead of $\ce{Ca10(PO4)6(OH)2}$. If I use the formula in my lecture notes, I get $\ce{Ca3(PO4)2 + Ca(OH)2 -> Ca4(PO4)2(OH)2}$, which is even more off.

Should it be $\ce{3(Ca3(PO4)2)}$ instead of $\ce{3Ca3(PO4)2}$ so the coefficient 3 can be applied to the subscript 2 on the $\ce{PO4}$ to give $\ce{(PO4)6}$?

What am I missing?

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    $\begingroup$ Well, it seems that before studying chemistry you must know the notation which is passed around through a word of mouth. In 3Ca3(PO4)2, the "4" refers to O and nothing else, the "2" refers to PO4, the second "3" refers to Ca, but the first "3" refers to everything after it. Pretty obvious, huh? $\endgroup$ – Ivan Neretin Jul 18 '20 at 6:12
  • $\begingroup$ Stephanie's last question show that she does not master the basic convention for writing chemical formulas. I repeat her question here : "Should it be "3(Ca3(PO4)2)" instead of "3Ca3(PO4)2" ? " Well. Stephanie should absolutely be convinced that these two formulas are identical, before going ahead with apatite structures. $\endgroup$ – Maurice Jul 18 '20 at 9:17
  • $\begingroup$ @IvanNeretin I know this. But in my working out, I always (to make it clear to myself) always, always put everything in as many brackets as needed to separate the items (maths or otherwise) so I thought the 3 without the brackets was only for the Ca. Got it now. Thanks $\endgroup$ – Stephanie Chen-Xu Jul 19 '20 at 10:09
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First, let me state the standard formula of hydroxyapaptite:

The term "apatite" applies to a group of compounds with a general formula in the form $\ce{M10(XO4)6Z2}$, where $\ce{M^2+}$ is a metal and species $\ce{XO4^3-}$ and $\ce{Z-}$ are anions. The particular name of each apatite depends on the elements or radicals M, X and Z. In these terms, hydroxyapatite (HAp) has the molecular structure of apatite, where M is calcium ($\ce{Ca^2+}$), X is phosphorus ($\ce{P^5+}$) and Z is the hydroxyl radical ($\ce{OH-}$). This is known as stoichiometric hydroxyapatite and its atomic ratio Ca/P is ${1.67}$. Its chemical formula is $\ce{Ca10(PO4)6(OH)2}$, with $\ce{39 {%}}$ by weight of $\ce{Ca}$, $\ce{18.5 {%} P}$ and $\ce{3.38 {%}}$ of $\ce{OH}$.

The anion radical can be replaced with fluorine/chlorine to become fluoroapatite/chloroapatite. This formula can also be written as $\ce{3[Ca3(PO4)2]· Ca(OH)2}$ which indicates three moles of calcium phosphate and one mole of calcium hydroxide which is basically the same as the Sigma-Aldrich's formula. The confusion arises since they have omitted the brackets, you didn't considered calcium phosphate as single entity for which you had a wrong formula. Brackets are generally not given but you have consider them as a single entity or you can even use brackets conveniently for your purpose. But do note that this formula doesn't prove that it is a basic salt but is actually a product of polymerisation:

According to Bassett, the solid existing in stable equilibrium at $\pu{25 °C}$., with solutions of a range from faintly acid to nearly pure lime-water, is oxy- or hydroxyapatite, but, although its composition may be indicated by the formula $\ce{3Ca3(PO4)2.Ca(OH)2}$, it is not to be regarded as a basic salt, but rather as the salt of a polymeride of phosphoric acid, $\ce{H11P3O13}$ (or $\ce{3H3PO4.H2O}$), with one hydrogen atom un-neutralised (like orthophosphoric acid or pyrophosphoric acid). The slowness of formation of hydroxyapatite favours the view that it is produced by polymerisation.

Reference

  1. Hydroxyapatite-Based Materials: Synthesis and Characterization By Eric M. Rivera-Muñoz, 2011, DOI: 10.5772/19123
  2. http://calcium.atomistry.com/basic_calcium_phosphates.html
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  • $\begingroup$ Oh thank you, this is exactly what I needed. I always use (more) brackets (than typically needed) to make calculations clear for me, so the way it was written here confused me. Thank you so much. $\endgroup$ – Stephanie Chen-Xu Jul 19 '20 at 10:11

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