1. The problem statement, all variables and given/known data 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 dont get is how the prof. found the number of moles Ti in TiO2 Any help appericiated 3. 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.