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

##### 1 Answer

#### Answer:

#### Explanation:

The problem wants you to go from *molecules* of iodine, *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

useAvogadro's numberto determine how manymolesof iodine you have in your sampleuse iodine'smolar massto determine how manygramswould 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

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 **every mole** of iodine has mass of

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)))|)#