This question concerns collecting a gas over water.

Question

A sample of mercury oxide, HgO, is decomposed by heating from pure mercury metal and oxygen gas. A volume of oxygen equaling 355 mL is collected over water at 18*C and 725 Torr total pressure. Determine how many grams of mercury oxide are decomposed.

Answer 1

Whenever you collect a sample of gas over water, you MUST subtract the vapor pressure of the water from the total vapor pressure to obtain the pressure of the gas.

2HgO ==> 2Hg(s) + O₂(g) .. balanced equation

Volume of O₂ = 355 ml = 0.355 L

Temperature = 18ºC + 273 = 291K

Total pressure = 725 torr

Looking up the vapor pressure of water @ 18ºC, we find it to be 15.45 torr (your value may differ depending on the source you use).

Pressure of O₂ gas = 725 torr - 15.45 torr = 709.55 torr

Use the ideal gas law to find the moles of O₂ gas

PV = nRT

n = PV/RT = (709.55 torr)(0.355 L) / (0.08314 Ltorr/Kmol)(291K)

n = 10.41 moles O₂ gas

Converting this to grams of O₂ and correcting to 2 sig.figs. (based on 18º) we get ..

10.41 mols O₂ x 32.0 g / mol = 333.12 g = 330 g O₂