Answer
There is a total of $0.185$ mole of $O$ in 0.185 mole of $C_6H_{14}O$.
Work Step by Step
1. Identify the conversion factor for $C_6H_{14}O$ moles and $O$ moles.
- Each $C_6H_{14}O$ molecule has 1 $O$ atom. Therefore:
$\frac{1mole(C_6H_{14}O)}{1mole(O)}$ and $\frac{1mole(O)}{1mole(C_6H_{14}O)}$
2. Calculate the amount of $O$ (in moles), in $0.185$ mole of $C_6H_{14}O$:
$0.185 mole(C_6H_{14}O) \times \frac{1mole(O)}{1mole(C_6H_{14}O)} = 0.185 $ mole $ (O)$
