Question

A 11.21 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.455 g. Add subscripts below to correctly identify the empirical formula of the new oxide.

Answer 1

Answer -

Mo₂O₃ has a molar mass of 239.9 g/mole.

so 11.21 g = 11.21/239.9 = 0.04673 mole.

Moles of Mo = 2 * 0.04673 = 0.09346 moles

we added 13.455 g - 11.21 g of O = 2.245 g O = 0.1403 moles O.

Moles of O originally present in the 11.21 g = 3 * 0.04673 = 0.1403 moles.

New number of moles of O2 = 0.1403 moles + 0.1403moles (or twice the original number of moles) = 0.2806 moles

moles of O / moles of Mo = 0.2806 moles / 0.09346 moles Mo = 3 (divided by smaller no. of moles)

The empirical formula is MoO3.