# What is the mass of 3.7 * 10^20 molecules of Rubidium Phosphate? (Rb_3PO_4)

May 28, 2018

Approximately $2.16 \cdot {10}^{-} 1 \setminus \text{g"=0.216 \ "g}$

#### Explanation:

We first convert to moles of $R {b}_{3} P {O}_{4}$. In a mole of a substance, there'll be $6.02 \cdot {10}^{23}$ units of that substance.

So, here we have:

(3.7*10^20color(red)cancelcolor(black)(Rb_3PO_4 \ "molecules"))/(6.02*10^23color(red)cancelcolor(black)(Rb_3PO_4 \ "molecules")"/mol")~~6.15*10^-4 \ "mol"

Now to find the mass, we multiply by its molar mass, which is $351.3748 \setminus \text{g/mol}$.

$6.15 \cdot {10}^{-} 4 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol"*(351.3748 \ "g")/(color(red)cancelcolor(black)"mol")~~2.16*10^-1 \ "g}}}}$