The atomic weight of phosphorus is 30.974 u. What is the mass of a phosphorus sample which contains 0.585 moles of phosphorus atoms?

Nov 1, 2015

$\text{18.1 g}$

Explanation:

You know that the atomic weight of phosphorus is equal to $\text{30.794 u}$, where $u$ represent the unified atomic mass unit.

The unified atomic mass unit is equivalent to $\text{1 g/mol}$, but let's take the long road and prove that identity.

Now, the unified atomic mass unit is defined as $\frac{1}{12} \text{th}$ of the mass of a single unbound carbon-12 atom in its ground state and is equivalent to

$\text{1 u" = 1.660539 * 10^(-24)"g}$

This means that the mass of one phosphorus atom will be

30.974color(red)(cancel(color(black)("u"))) * (1.660539 * 10^(-24)"g")/(1color(red)(cancel(color(black)("u")))) = 5.14335 * 10^(-23)"g"

You know that one mole of any element contains exactly $6.022 \cdot {10}^{23}$ atoms of that element - this is known as Avogadro's number.

Well, if you know the mass of one phosphorus atom, you can use Avogadro's nubmer to determine what the mass of one mole of phosphorus atoms

$5.14335 \cdot {10}^{- 23} \text{g"/color(red)(cancel(color(black)("atom"))) * (6.022 * 10^(23)color(red)(cancel(color(black)("atoms"))))/"1 mole" = "30.974 g/mol}$

Finally, if one mole of phosphorus atoms has a mass of $\text{30.974 g}$, then $0.585$ moles will have a mass of

0.585color(red)(cancel(color(black)("moles"))) * "30.974 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("18.1 g")