How many tin (II) ions, Sn^+2, are present in 3.5 grams of Sn3(PO4)2?

Question

Help!!! Please!!! And thank you!!!!!!!!!!!!!!!!!!

Answer 1

Hello, Sophia,

The place to start is to determine the molar mass of Sn₃(PO₄)₂. I'll assume you can calculate this based on adding the atomic masses of the individual elements to generate the total molecular weight of the molecule, which is also it's molar mass, g/mole.

I find 546.07 g/mole.

Convert 3.5 grams into moles by dividing it by the molar mass:

3.5g/546.1g/mole =0.006409 mole

One mole = 6.023x10²³ molecules, in this cse. (A mole can be anything, atoms, molecules, bananas, pencils, ants, etc. - it is a term used as a unit of measure, just as in "a dozen" of anything).

So we have 0.00641 moles*(6.023x10²³ molecules/mole) or 3.87x10²⁴ molecules of Sn₃(PO₄)₂

But note that each molecule has 3 Sn (tin) atoms. That means we need to multiply by 3, which brings us to 1.16x10²⁵ tin(II) ions.

I hope this helps,

Bob