Question

Formula of phosphoric acid is ${H_3}P{O_4}$. A metal M forms chloride of the formula $MC{l_2}$. What is the formula of its phosphate?

Answer 1

Hint: The chemical formula of a compound can be written by using the symbols of the constituent elements or ions and the valency of the element or charge on the ion.
The formula of phosphoric acid: ${H_3}P{O_4}$.
The formula of metal chloride: $MC{l_2}$.

Complete step by step answer:
We know that the compounds are formed by elements when they combine in definite proportions and this can be represented by a chemical formula. For example, one molecule of water which has a formula of ${H_2}O$ is formed by 2 atoms of hydrogen and 1 atom of oxygen whereas in case 2 atoms of hydrogen combine with 2 atoms of oxygen, we get ${H_2}{O_2}$ which is a different compound.
So, let’s have a look at how we can write these useful chemical formulae. While writing a chemical formula, we use the symbol of the constituent elements and their valency. There are some rules that we have to keep in mind while writing a chemical formula and these can be listed as follows:
The charge neutrality principle must be followed.
In case metal and non-metal, both are constituents then the symbol/name of the metallic element is written first followed by that of the non-metal.
In case constituent ions are polyatomic, they are enclosed in parenthesis and their number is written outside of that.
Now, let’s have a look at the given metal chloride with formula $MC{l_2}$. So, we can infer from here that the symbol used for metal (written first) is $M$ and that of chlorine is $Cl$. Let’s assume that the valency of the metal is x and we know that valency of $Cl$ is $-1$. So, we can write as per charge neutrality principle:
$\left( {x \times 1} \right) + \left\{ {\left( { - 1} \right) \times 2} \right\} = 0$
$x - 2 = 0$
$x = + 2$
So, now we know that the valency of the metal is $+ 2$.
Let’s have a look at phosphoric acid now which has formula ${H_3}P{O_4}$. The phosphate ion originating from it can be written as:
${H_3}P{O_4} \to 3{H^ + } + PO_4^{3 - }$
So, we can see that phosphate ion is a polyatomic one with a charge of $- 3$. Now we can use the symbols of the metal $\left( M \right)$ along with its valency $\left( { + 2} \right)$ and that of phosphate ion $\left( {PO_4^{3 - }} \right)$ and its charge $\left( { - 3} \right)$ to write the formula by criss-crossing them in accordance with charge neutrality principle to give \[{M_3}{\left( {P{O_4}} \right)_2}\].
Hence, the formula of the metal phosphate is determined to be \[{M_3}{\left( {P{O_4}} \right)_2}\].

Note: We know that phosphoric acid contains three ${H^ + }$ and removal of these one by one can give different anions but here we have to use the phosphate ion.