Question

How many atoms are in 3456 grams of (NH4)3PO3?

Answer 1

Firstly we will convert this mass to moles then form moles to atoms:

$\begin{gathered} moles=\frac{mass}{molar\text{ }mass} \\ _n(NH_4)_3PO_3=\frac{3456\text{ }g}{149.09\text{ }gmol^{-1}} \\ _n(NH_4)_3PO_3=23.18mol \end{gathered}$

We will now convert moles to atoms:

$\begin{gathered} Avogadro^{\prime}s\text{ }Number:6.02\times10^{23}atoms \\ 1mol(NH_4)_3PO_3=6.02\times10^{23}atoms \\ 23.18\text{ }mol(NH_4)_3PO_3=x\text{ }atoms(NH_4)_3PO_3 \\ 23.18\text{ }mol(NH_4)_3PO_3\times\frac{6.02\times10^{23}atoms}{1mol(NH_4)_3PO_3}=1.40\times10^{25}atoms \end{gathered}$

Answer: 1.40x10^25 atoms (NH4)3PO3