A mixture of copper(I) oxide and copper(II) oxide was found to contain 16.3% oxygen by mass. Calculate the mass percentage of copper(I) oxide in the mixture with two significant figures. Have an exam tomorrow and can't figure this out! I found the relative quantities of the two coppers by moles, but I don't know where to go from there...
CuO (Copper(II) oxide) and Cu?O (Copper(I) oxide). This gives us a total of 2 oxygen atoms (1 in each compound) and 3 copper atoms. We are given that oxygen comprises 16.3% of the mixture's mass and since there is only one oxygen in each formula, it must constitute 8.15% of each compound's mass. Let's first assume we have 100-g of the mixture. That would give us 16.3 gram of oxygen and 100-16.3 = 83.7-g of copper. Let's start with the simpler formula CuO first. This contains equal numbers of moles of Cu and O. Since we know we have 8.15-g of O in it (8.15-g / 16.0-g/mol = 0.509 moles of oxygen) and thus 0.509 moles of Cu. 0.509 moles of Cu = 0.509 mol x 63.5-g/mol = 32.32-g of Cu in the CuO - Copper (II). Since we had a total copper of 83.7-g with 32.32-g of it in the CuO, there must be (83.7 - 32.32 = 51.38-g in the Cu?O - copper(I) . The total mass of the Copper(I) oxide is 51.38-g Cu + 8.15-g O = 59.53-g Given 100 grams of mixture and 59.53-g Cu?O we have a % mass of 59.53-g / 100-g x 100% = 59.53% = 60% to 2 sig fig.
