How many formula units are contained in 1.34g CaO?

Mar 7, 2016

$\frac{1.34 \cdot g}{74.10 \cdot g \cdot m o {l}^{-} 1}$ $\times$ ${N}_{A}$, where ${N}_{A}$ $=$ $\text{Avogadro's number}$

Explanation:

There are $\frac{1.34 \cdot g}{74.10 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? " moles calcium hydride".

Now, it is a fact that $1$ $\text{mole}$ of stuff, there are $6.022 \times {10}^{23}$ individual items of that stuff.

You have $0.0181$ $m o l$. Therefore:

$0.0181$ $m o l$ $\times$ $6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ $=$ ??" formula units of calcium hydroxide".