Problem 3-95

This question has two problems based on the same reaction:

Hg_xO_y(s) → xHg(ℓ) + y(oxygen)

First problem:

Initial mass of mercury oxide = 0.6498 g
Remaining mercury (Hg residue) after heating = 0.6018 g.

Therefore Mass of Hg = 0.6018 g
mass of oxygen = 0.6498 - 0.6018 = 0.0480 g

mole of Hg = 0.6018 g / 200.6 g.mol-1 = 0.00300 mol
mole of O = 0.0480 g / 16.00 g.mol-1 = 0.00300 mol
(molar mass of oxygen = 16.00 g/mol since it is not a gas (O₂) when combined with mercury.)

Dividing both element by the smallest number of mole (0.0030) gives the empirical formula: HgO

Second problem:

Hg_xO_y(s) = 0.4172 g and the oxygen lost = 0.016 g.
Again, this problem is solved the same way however here, it is the mass of Hg(ℓ) that is found by difference.
The answer is Hg₂O