How many grams is #3.30 * 10^25# molecules of #I_2#?

1 Answer
Stefan V.
Mar 6, 2016

#"13,900 g"#

Explanation:

The problem wants you to go from molecules of iodine, #"I"_2#, to grams of iodine, which means that you're going to have to go through moles of iodine first.

More specifically, your strategy here will be to

  • use Avogadro's number to determine how many moles of iodine you have in your sample
  • use iodine's molar mass to determine how many grams would contain that many moles

So, Avogadro's number tells you how many molecules you get in one mole of a given substance. In order to have one mole of iodine, you need to have #6.022 * 10^(23)# molecules of iodine.

This means that your sample will be equivalent to

#3.30 * 10^(25)color(red)(cancel(color(black)("molecules I"_2))) * overbrace("1 mole I"_2/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules I"_2)))))^(color(brown)("Avogadro's number")) = "54.8 moles I"_2#

Now, molecular iodine has a molar mass of #"253.81 g mol"^(-1)#, which means that every mole of iodine has mass of #"253.81 g"#.

In your case, the sample you're working with will have a mass of

#54.8 color(red)(cancel(color(black)("moles I"_2))) * overbrace("253.81 g"/(1color(red)(cancel(color(black)("mole I"_2)))))^(color(purple)("molar mass of I"_2)) = "13909 g"#

Rounded to three sig figs, the number of sig figs you have for the number of molecules of iodine, the answer will be

#m_(I_2) = color(green)(|bar(ul(color(white)(a/a)"13,900 g I"_2color(white)(a/a)))|)#

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