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Homework Statement
A sample of titanium dioxide (TiO2) weighing 1.598 g is heated in H2 gas to give
water vapour and 1.438 g of another titanium oxide. What is the formula of the
oxide produced?
TiO2 (s) + H2 (g) TixOy (s) + H2O (g) (unbalanced)
The solution posted is as follows but the main part i don't get is how the prof. found the number of moles Ti in TiO2
Any help appericiated
The Attempt at a Solution
1.598 g TiO2 × 1 1 mol TiO2 × 1 mol Ti = 0.02001 mol Ti
79.878 g TiO2 1 mol TiO2
mol Ti in TiO2 = mol Ti in TixOy
∴mol Ti in TixOy = 0.02001
mass of Ti in TixOy = 0.02001 mol Ti × 47.88 g Ti = 0.9579 g Ti
1 mol Ti
mass of O in TixOy = mass of sample – mass Ti
mass of O in TixOy = 1.438 g – 0.9579 g = 0.4801 g O
mol of O in TixOy = 0.4801 g O × 1 mol O = 0.03001 mol
15.999 g O
Thus we have a molar ratio of: Ti0.02001O0.03001
Divide each number by the smallest value:
Ti0.02001/0/02001O0.03001/0/02001 = Ti1O1.5 or, in whole numbers, Ti2O3.