The empirical formula of a compound containing 0.067mol Cu and 0.066mol O is [Cu = 63.5, O = 16]

To determine the empirical formula of the compound, we need to find the simplest whole number ratio of the atoms present in the compound. Given that the compound contains 0.067 mol Cu and 0.066 mol O, we need to first convert the moles to whole numbers. Dividing both values by the smaller value (0.066 mol), we get: Cu: 0.067 mol ÷ 0.066 mol ≈ 1.02 ≈ 1 O: 0.066 mol ÷ 0.066 mol = 1 So, the ratio of Cu to O atoms is 1:1. Therefore, the empirical formula of the compound is CuO. We can check that the formula is correct by calculating the molar mass of CuO and comparing it to the experimental data. Molar mass of CuO = 63.5 g/mol + 16 g/mol = 79.5 g/mol The experimental data gives us: (0.067 mol)(63.5 g/mol) + (0.066 mol)(16 g/mol) = 4.2495 g The molar mass calculated from the experimental data is: (4.2495 g) / (0.133 mol) ≈ 31.97 g/mol This is close to the molar mass of CuO (79.5 g/mol), which confirms that the empirical formula is CuO.