# How many grams is .25 mol of CO_2 molecules?

Well $1$ $m o l$ carbon dioxide has a mass of $44.01$ $g$.
So $\frac{1}{4}$ $\cancel{m o l}$ $\times$ $44.01 \cdot g \cdot \cancel{m o {l}^{-} 1}$ $=$ ?? g