## Chemistry: The Central Science (13th Edition)

5.00 mg of that substance contains $3.08 \times 10^{-5} \space mole \space of \space allicin$.
1. Identify the conversion factors: - $mg$ to grams : $\frac{1 \space g}{1000 \space mg}$ - Grams (allicin) to moles (allicin): $\frac{1 \space mole \space (allicin)}{162.28 \space g \space (allicin)**}$ ** This was calculated in 3.39a 2. Calculate the number of moles in 5.00 mg of that substance. $5.00 \space mg \space (allicin) \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mole \space (allicin)}{162.28 \space g \space (allicin) }=$ $3.08 \times 10^{-5} \space mole \space (allicin)$