What is the mass in $3.01 \cdot 10^{23}$ $3.01 * 10^23$ molecules of $F_{2}$ $F_2$ ?

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Answer 1 · 2016-01-16T15:21:34

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

To convert mol $F_{2}$ $F_2$ to grams $F_{2}$ $F_2$ , you simply have to multiply the given number of mol to the molar mass of the compound.

For this case:

$g F_{2} = (3.01 \cdot 10^{23} m o l) \cdot (37.996 \frac{g}{m o l})$ $g F_2 = (3.01 * 10^(23) mol) * (37.996 g/[mol] )$

$g F_{2} = 1.1436796 \cdot 10^{23}$ $g F_2 = 1.1436796 * 10^(23)$

Using correct number of significant figures,
$g F_{2} = 1.14 \cdot 10^{25} g r a m s$ $g F_2 = 1.14 * 10^25 grams$

note:
molar mass of $F_{2}$ $F_2$ means you have to multiply Fluorine's molar mass by 2.

molar mass of $F_{2} = 2 \cdot (18.992 \frac{g}{m o l}) = 37.996 \frac{g}{m o l}$ $F_2 = 2 * ( 18.992 g/[mol]) = 37.996 g/[mol]$

What is the mass in 3.01⋅10233.01 * 10^23 molecules of F2F_2?