Question

There are two binary compounds of mercury and oxygen. Heating either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. Heating 0.6498 g of one of the compounds leaves a residue of 0.6018 g. Heating 0.4172 g of the other compound results in a mass loss of 0.016 g. Determine the empirical formula of each compound.

Short answer

The empirical formulas for the two binary compounds of mercury and oxygen are \(HgO\) for the first compound and \(Hg_2O\) for the second compound.

Step-by-step solution

Calculate moles of Mercury and Oxygen in each compound

First, we need to determine the moles of Mercury (Hg) and Oxygen (O) in each compound. To do this, we'll use the given mass data and the molar masses of Hg (200.59 g/mol) and O (16.00 g/mol). For compound 1: Initial mass = 0.6498 g Mass of residue (pure mercury) = 0.6018 g Moles of Hg in compound 1 = 0.6018 g / 200.59 g/mol ≈ 0.0030 mol Mass of oxygen in compound 1 = 0.6498 g - 0.6018 g = 0.048 g Moles of O in compound 1 = 0.048 g / 16.00 g/mol ≈ 0.0030 mol For compound 2: Initial mass = 0.4172 g Mass loss (oxygen gas) = 0.016 g Moles of O in compound 2 = 0.016 g / 16.00 g/mol ≈ 0.0010 mol Mass of mercury in compound 2 = 0.4172 g - 0.016 g = 0.4012 g Moles of Hg in compound 2 = 0.4012 g / 200.59 g/mol ≈ 0.0020 mol

Determine the simplest whole-number ratio

Now, we can find the simplest whole-number ratio to determine the empirical formulas of the compounds. For compound 1: Hg/O ratio = 0.0030 mol / 0.0030 mol = 1/1 Empirical formula: HgO For compound 2: Hg/O ratio = 0.0020 mol / 0.0010 mol = 2/1 Empirical formula: Hg2O

Conclusion

The empirical formulas for the two binary compounds of mercury and oxygen are HgO for the first compound and Hg2O for the second compound.