# How many total number of all the elements present in mole of (NH4)2 Cr2O7? Thanks a lot.

Nov 19, 2016

We use $\text{Avogadro's number, } {N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$, to quantify the numbers of individual atoms in a mole, i.e. a $\text{252.07 g}$ mass of ${\left(N {H}_{4}\right)}_{2} C {r}_{2} {O}_{7}$.......

#### Explanation:

In such a mass, there are $2 \times {N}_{A}$ $\text{nitrogen atoms}$, and

$8 \times {N}_{A}$ $\text{hydrogen atoms}$, and

$2 \times {N}_{A}$ $\text{chromium atoms}$, and

$7 \times {N}_{A}$ $\text{oxygen atoms}$.

Here we use ${N}_{A}$ as a counting number, i.e. as we would use the number $10$, or a $\text{million}$, or a $\text{dozen}$.