# What is the mass in 3.01 * 10^23 molecules of F_2?

Jan 16, 2016

The answer to this is $1.14 \cdot {10}^{25} g r a m s$

#### Explanation:

To convert mol ${F}_{2}$ to grams ${F}_{2}$, you simply have to multiply the given number of mol to the molar mass of the compound.

For this case:

$g {F}_{2} = \left(3.01 \cdot {10}^{23} m o l\right) \cdot \left(37.996 \frac{g}{m o l}\right)$

$g {F}_{2} = 1.1436796 \cdot {10}^{23}$

Using correct number of significant figures,
$g {F}_{2} = 1.14 \cdot {10}^{25} g r a m s$

note:
molar mass of ${F}_{2}$ means you have to multiply Fluorine's molar mass by 2.

molar mass of ${F}_{2} = 2 \cdot \left(18.992 \frac{g}{m o l}\right) = 37.996 \frac{g}{m o l}$