# When 1.50 g of copper is heated in air, it reacts with oxygen to achieve a final mass of 2.14 g. How do you calculate the empirical formula of this copper oxide?

Oct 4, 2016

See below.

#### Explanation:

Recall that by using the molar mass of an element and the mass of the element in the sample, you can find the amount of moles in the sample. We're gonna find the moles of copper and oxygen within the sample of copper oxide.

1. Calculate the moles of copper given
Remember that moles$= \frac{m a s s}{m o l a r m a s s}$
The molar mass of copper is $63.55 g m o {l}^{-} 1$, as the atomic mass of an element corresponds to its molar mass in grams.
$n \left(C u\right) = \frac{1.50 g}{63.55 g m o {l}^{-} 1} = 0.0236 \ldots m o l C u$
2. Calculate the moles of oxygen in the final sample of copper oxide.
Mass of oxygen = mass of copper oxide - initial mass of copper
$m \left(O\right) = 2.14 g - 1.50 g = 0.64 g$
Molar mass of oxygen = $16.00 g m o {l}^{-} 1$
Now that we have the mass and the molar mass, we can work out the moles.
$n \left(O\right) = \frac{0.64 g}{16.00 g m o {l}^{-} 1} = 0.04 m o l O$

Now that we have the moles of copper and oxygen in the sample, compare the mole ratio.
$C u : O = 0.0236 : 0.04 \approx 3 : 5$
Therefore, the empirical formula is $C {u}_{3} {O}_{5}$ based on the data from the experiment.