# You have 10 grams of C10H10O2. How many moles are present? How many molecules are present?

Mar 4, 2018

Well the formula mass is $162.18 \cdot g \cdot m o {l}^{-} 1$

#### Explanation:

And so we take the quotient....

$\text{Molar quantity"="Mass"/"Molar mass} = \frac{10.0 \cdot \cancel{g}}{162.18 \cdot \cancel{g} \cdot m o {l}^{-} 1}$

$= 0.06171 \cdot \frac{1}{\frac{1}{m o l}} = 0.0617 \cdot m o l$

And we know that there are $6.022 \times {10}^{23} \cdot \text{particles} \cdot m o {l}^{-} 1$..

And so we take the product....

$6.022 \times {10}^{23} \cdot \text{particles"*mol^-1xx0.0617*mol=3.71xx10^22*"mollykewells}$

How many carbon atoms present in this quantity?

Mar 4, 2018

0.062 moles and 3.7*10^22 molecules

#### Explanation:

Molecular mass of ${C}_{10} {H}_{10} {O}_{2}$ is $162$, therefore $10$ grams are $\frac{10}{162} = 0.062$ moles. Each mole contains the Avogadro's number of molecules $0.062 \cdot 6.02 \cdot {10}^{23} = 3.7 \cdot {10}^{22}$