Question #10dc1

Apr 10, 2017

$= 0.340$ $m o l$

Explanation:

Using the periodic table, the molar mass of $F e {\left({C}_{2} {H}_{3} {O}_{2}\right)}_{3}$ is approximatly $232.98 \frac{g}{m o l}$

Using the molar mass and the grams given, you can solve for moles,

$79.3 g \cdot \frac{1}{232.98} \frac{m o l}{g}$

$79.3 \cancel{g} \cdot \frac{1}{232.98} \frac{m o l}{\cancel{g}}$

$\frac{79.3}{232.98} m o l$

$= 0.340$ $m o l$

Or you could of used the equation,

$n = \frac{m}{m m}$

Where $n$ is moles, $m$ is mass in $g r a m s$ and $m m$ is molar mass in $\frac{g r a m s}{m o l}$