# A sample of chlorine gas has a mass of 39.2 grams. How many Cl_2 molecules are there in the sample?

May 16, 2017

See below.

#### Explanation:

This problem is a classic stoichiometry problem. To solve this problem, you want to have your periodic table (to find the molar mass (g/mol) of Cl2), calculator, and Avogadro's number handy. The approach is outlined below:

$39.2 \text{ g Cl2" ((1 " mole of Cl2")/(70.9 " g Cl2"))((6.022 * 10^23 "molecules Cl2")/(1 " mole Cl2")) = 3.33 * 10^23 " molecules Cl2}$.

Hence, the answer is $3.33 \cdot {10}^{23}$ molecules of Cl2. Note that $6.022 \cdot {10}^{23}$ is also known as Avogadro's number, and it can be referred to as the number of molecules in one mole of that substance. Also, I calculated $70.9$ because $35.45 \text{ g/mol}$ (the mass of one Chlorine atom) times $2$ is $70.9 \text{ g/mol}$ (Cl2 has two Chlorine atoms).

I hope that helps!

May 16, 2017

There are $3.33 \times {10}^{23}$ molecules of $\text{Cl"_2}$ in the sample.

#### Explanation:

There are $6.022 \times {10}^{23}$ items in one mole of anything, including molecules. You have to convert the mass of $\text{Cl"_2}$ into moles, then convert the moles into molecules. You do that by multiplying the given mass by the inverse of its molar mass, then multiply the moles by $6.022 \times {10}^{23} \textcolor{w h i t e}{.} \text{molecules}$/mol. The molar mass is the subscript in the formula of $\text{Cl"_2}$ multiplied by the atomic weight of chlorine on the periodic table in grams/mole, or g/mol.

Given mass: $\text{39.2 g}$
Molar mass: (2xx35.45"g/mol")="70.9 g/mol"

Determine the moles $\text{Cl"_2}$.

39.2color(red)cancel(color(black)("g Cl"_2))xx(1"mol Cl"_2)/(70.9color(red)cancel(color(black)("g Cl"_2)))+"0.553 mol Cl"_2"

Determine molecules of $\text{Cl"_2}$.

$0.553 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol Cl"_2))xx(6.022xx10^23"molecules")/(1color(red)cancel(color(black)"mol Cl"_2}}}}$$=$$3.33 \times {10}^{23} \textcolor{w h i t e}{.} \text{molecules Cl"_2}$