Question

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

Answer 1

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 " 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 * 10^23` molecules of Cl2. Note that `6.022 * 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 " g/mol"` (the mass of one Chlorine atom) times `2` is `70.9 " g/mol"` (Cl2 has two Chlorine atoms).

I hope that helps!

Answer 2

There are `3.33xx10^23` molecules of `"Cl"_2"` in the sample.

Explanation:

There are `6.022xx10^23` items in one mole of anything, including molecules. You have to convert the mass of `"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 `6.022xx10^23color(white)(.)"molecules"`/mol. The molar mass is the subscript in the formula of `"Cl"_2"` multiplied by the atomic weight of chlorine on the periodic table in grams/mole, or g/mol.

Given mass: `"39.2 g"`
Molar mass: `(2xx35.45"g/mol")="70.9 g/mol"`

Determine the moles `"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 `"Cl"_2"`.

`0.553color(red)cancel(color(black)("mol Cl"_2))xx(6.022xx10^23"molecules")/(1color(red)cancel(color(black)"mol Cl"_2"))` `=` `3.33xx10^23color(white)(.)"molecules Cl"_2"`