Question

3. A student used the procedure described in this module to determine the empirical formula of an oxide of titanium (Ti). A crucible and crucible cover were heated to a constant mass of 34.749 g. The mass of the crucible, crucible cover, and a sample of Ti was 36.383 g. Following the reaction of the Ti metal with O2, the crucible, crucible cover, and contents were heated to a constant mass of 37 472 g (1) Determine the mass of the oxide of Ti. 31 ,472-34 .741-12-723a-ON04 (2) Determine the mass of Ti in the compound. 3 393 34 19 (3) Determine the mass of O in the compound.

Answer 1

Mass of Crucible + cover = 34.749 g

Mass of Crucible + cover + Ti = 36.383 g

Mass of Crucible + cover + heated content = 37.472

Heated content is the oxide of Ti

1. Mass of Oxide of Ti = (Mass of Crucible + cover + heated content) - Mass of Crucible + cover

= 37.472 - 34.749 g = 2.723 g

2. Mass of Ti =(Mass of Crucible + cover + Ti) -( Mass of Crucible + cover ) = 36.383 g- 34.749 g = 1.634g

3. Mass of O = (Mass of Crucible + cover + heated content)-(Mass of Crucible + cover + Ti)

= 37.472 - 36.383 g = 1.089 g

4. Molar mass of Ti = 47.9 g/mol

Moles of Ti = mass / Molar mass = 1.634g/ 47.9 g/mol = 0.034 mol

5. Molar mass of oxygen = 16g/mol

Moles of O = mass / molar mass = 1.089 g/ 16g/mol = 0.068 mol

6. Mole ratio of Ti: O is :

0.034 mol of Ti : 0.068 mol of O

Simple whole number ratio is

: $\frac{0.068}{0.034}$

null

1 : 2

6. Empirical formula of the oxide of Ti is TiO₂ as indicated by simple whole number ratio