# What is the mass (in grams) of 8.34 moles calcium acetate?

One mole of $C a {\left\{O \left(O =\right) C - C {H}_{3}\right\}}_{2}$ has a mass of $158.17$ $g$.
So, if there are $8.34$ mol calcium acetate, there are $8.34 \cdot \cancel{m o l} \times 158.17 \cdot g \cdot \cancel{m o {l}^{-} 1}$ $\cong$ $1300$ $g$.