When hydrogen gas was passed over 1.59 g of copper oxide, 1.27 g of metallic copper were produced - Leaving Cert Chemistry - Question e - 2013

To find the mass of oxygen, we subtract the mass of copper from the original mass of copper oxide:

$\text{Mass of oxygen} = 1.59 ext{ g (copper oxide)} - 1.27 ext{ g (copper)} = 0.32 ext{ g}$

Next, we need to calculate the number of moles of copper using its molar mass (approximately 63.5 g/mol):

$\text{Moles of copper} = \frac{1.27 ext{ g}}{63.5 ext{ g/mol}} \approx 0.020 ext{ mol}$

Now, calculate the number of moles of oxygen using its molar mass (approximately 16 g/mol):

$\text{Moles of oxygen} = \frac{0.32 ext{ g}}{16 ext{ g/mol}} = 0.02 ext{ mol}$

Now we find the simplest whole number ratio between the moles of copper and oxygen.

For copper: (0.020 \text{ mol} )\ For oxygen: (0.020 \text{ mol} )

This gives us a ratio of 1:1, leading to the empirical formula:

$\text{Empirical formula} = \text{CuO}$