# What is the number of H atoms in 0.183 mole C_6H_14O?

Nov 16, 2015

There are $1.54 \times {10}^{24} \text{H atoms}$ in $\text{0.183 mol C"_6"H"_14"O}$.

#### Explanation:

One mole of $\text{C"_6"H"_14"O}$ contains 14 moles of hydrogen atoms. This can be written as $\left(14 \text{mol H atoms")/(1"mol C"_6"H"_14"O}\right)$.

To determine how many moles of hydrogen are contained in $\text{0.183 mol C"_6"H"_14"O}$, multiply $\text{0.183 mol C"_6"H"_13"O}$ times 14 moles of H atoms per 1 mole of $\text{C"_6"H"_14"O}$. Then multiply times $6.022 \times {10}^{23} \text{atoms H}$.

0.183cancel("mol C"_6"H"_14"O")xx(14cancel("mol H" "atoms"))/(1cancel("mol C"_6"H"_14"O"))xx(6.022xx10^23"H atoms")/(1cancel("mol H atoms"))=1.54xx10^24 "H atoms"