# Question 6a19b

Jul 18, 2017

$3.0 \times {10}^{23}$ $\text{atoms}$

#### Explanation:

We're asked to find the number of atoms in $1.6$ ${\text{g CH}}_{4}$.

To do this, we can first find the number of moles of ${\text{CH}}_{4}$ present, using the molar mass of methane ($16.04$ $\text{g/mol}$):

1.6cancel("g")((1color(white)(l)"mol CH"_4)/(16.04cancel("g CH"_4))) = 0.0997 ${\text{mol CH}}_{4}$

Now, we can use Avogadro's number to find the number of molecules of ${\text{CH}}_{4}$ present:

0.0997cancel("mol CH"_4)((6.022xx10^23color(white)(l)"molecules CH"_4)/(1cancel("mol CH"_4)))

$= 6.01 \times {10}^{22}$ ${\text{molecules CH}}_{4}$

Now, we ask ourselves, "How many atoms are in one molecule of methane?"

Well, there is $1$ $\text{C}$ and $4$ $\text{H}$ , which totals to color(red)(5. We can use this relationship to find the number of atoms:

6.01xx10^22cancel("molecules CH"_4)((5color(white)(l)"atoms")/(1cancel("molecule"))) = color(blue)(3.0xx10^23 color(blue)("atoms"#

rounded to $2$ significant figures.