# How many moles are in 68 grams of copper (II) hydroxide, Cu(OH)2?

Nov 28, 2015

There are $0.70$ moles of $C u {\left(O H\right)}_{\text{2}}$.

#### Explanation:

We can use the following formula to determine the number of moles:

$m = n M$

where:
m = mass (grams)
n = number of moles
M = molar mass

To find the number of moles, we first have to find the molar mass of $C u {\left(O H\right)}_{\text{2}}$. We can do this by adding the molar masses of copper, oxygen, and hydrogen together, accounting for the number of atoms for each element.

${M}_{\textcolor{b r o w n}{\text{copper}}} = \textcolor{b r o w n}{\frac{63.55 g}{m o l}}$

${M}_{\textcolor{b l u e}{\text{hydrogen}}} = \frac{1.01 g}{m o l} \cdot 2 = \textcolor{b l u e}{\frac{2.02 g}{m o l}}$

${M}_{\textcolor{g r e e n}{\text{oxygen}}} = \frac{16.00 g}{m o l} \cdot 2 = \textcolor{g r e e n}{\frac{32.00 g}{m o l}}$

${M}_{\textcolor{p u r p \le}{\text{copper (II) hydroxide}}} = \textcolor{b r o w n}{\frac{63.55 g}{m o l}} + \textcolor{b l u e}{\frac{2.02 g}{m o l}} + \textcolor{g r e e n}{\frac{32.00 g}{m o l}} = \textcolor{p u r p \le}{\frac{97.57 g}{m o l}}$

Now we can substitute our known values into the formula to solve for number of moles:

$m = n M$

$n = \frac{m}{M}$

$n = 68 g \div \textcolor{p u r p \le}{\frac{97.57 g}{m o l}}$

$n = 68 g \cdot \frac{m o l}{97.57 g}$

$n = 68 \textcolor{red}{\cancel{\textcolor{b l a c k}{g}}} \cdot \frac{m o l}{97.57 \textcolor{red}{\cancel{\textcolor{b l a c k}{g}}}}$

$n = \frac{68 m o l}{97.57}$

$n = 0.6969355335 m o l$

$n = 0.70 m o l$

Note:
The final answer should always have the correct number of significant figures.

$\therefore$, there are $0.70$ moles of $C u {\left(O H\right)}_{\text{2}}$.