Answer
There is a total of $1.56 \times 10^{24}$ atoms of $H$ in 0.185 mole of $C_6H_{14}O$.
Work Step by Step
1. Identify the conversion factors for $C_6H_{14}O$ moles to $H$ atoms.
- Each $C_6H_{14}O$ molecule has 14 $H$ atoms. Therefore:
$\frac{1mole(C_6H_{14}O)}{14moles(H)}$ and $\frac{14moles(H)}{1mole(C_6H_{14}O)}$
- Using Avogadro's number:
1 mole $(X)$ $= 6.02 \times 10^{23}$ $(X)$ $atoms$
$\frac{1 mole (X)}{6.02 \times 10^{23} atoms (X)}$ and $\frac{6.02 \times 10^{23} atoms (X)}{1 mole (X)}$
2. Calculate the amount of $H$ atoms in $0.185$ mole of $C_6H_{14}O$:
$0.185 mole(C_6H_{14}O) \times \frac{14moles(H)}{1mole(C_6H_{14}O)} \times \frac{6.02 \times 10^{23} atoms (H)}{1 mole (H)}= 1.56 \times 10^{24} $ atoms $ (H)$
