Sample of titanium dioxide (TiO2)

[a.a]

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.

 

[a.a]

oh wow.. i figured it out a different way.. wow.. soo suprised at my self

what i did was first find the mass of Ti in the given mass if TiO2 rather than finding the monber of moles first.. i have a tendency of liking to stay in the quantity given to me.. it actually came inhandy here.. :)

 

[Borek]

1 mole of TiO₂ is 79.87 g
x moles is 1.598 g

x = 1.598/79.87