Answer
There is a total of $6.68 \times 10^{23}$ atoms of $C$ 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 $C$ atoms.
- Each $C_6H_{14}O$ molecule has 6 $C$ atoms. Therefore:
$\frac{1mole(C_6H_{14}O)}{6moles(C)}$ and $\frac{6moles(C)}{1mole(C_6H_{14}O)}$
- 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{6moles(C)}{1mole(C_6H_{14}O)} \times \frac{6.02 \times 10^{23} atoms (C)}{1 mole (C)}= 6.68 \times 10^{23} $ atoms $ (C)$
