# How many atoms are present in 48.60 g of Mg?

Jul 30, 2016

$1.204 \times {10}^{24} \text{atoms}$ of magnesium are present

#### Explanation:

In order to go from mass of magnesium to atoms of magnesium, we have to do two things:

1. Convert mass of Mg to moles of Mg using the molar mass of Mg as a conversion factor

2. Convert moles of Mg to atoms of Mg using Avogadro's number ($6.02 \times {10}^{23}$) as a conversion factor

color(indigo)("Step 1:"
Before we start, I should note that the molar mass of Mg is $24.31 \frac{g}{m o l}$. We can go from mass to moles using dimensional analysis. The key to dimensional analysis is understanding that the units that you don't need any more cancel out, leaving the units that are desired:

$48.60 \cancel{g} \times \frac{1 m o l}{24.31 \cancel{g}}$ $= 2.00 m o l$

color(gold)("Step 2:"
We'll use this relationship:

Using the moles of Mg that we just obtained, we can use Avogrado's number to perform dimensional analysis in order to cancel out units of $m o l$ to end up with atoms of Mg:

2.00cancel"mol"xx(6.02xx10^(23) "atoms")/(1cancel"mol") $= 1.204 \times {10}^{24} \text{atoms}$

color(brown)("Thus, 48.60g of Mg is equivalent to" $1.204 \times {10}^{24} \text{atoms}$