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

2 Answers
May 16, 2017

Answer:

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!

May 16, 2017

Answer:

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"#