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

Mar 6, 2016

$\text{13,900 g}$

#### Explanation:

The problem wants you to go from molecules of iodine, ${\text{I}}_{2}$, 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 $6.022 \cdot {10}^{23}$ 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 ${\text{253.81 g mol}}^{- 1}$, which means that every mole of iodine has mass of $\text{253.81 g}$.

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

${m}_{{I}_{2}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{13,900 g I}}_{2} \textcolor{w h i t e}{\frac{a}{a}}}} |}$