A1.32-g sample of aluminum reacted with 1.174 g oxygen to form aluminum oxide. determine the empirical formula of the aluminum oxide.

Al2O3

Explanation:

Moles of Aluminium = 1.32 g/27 g/mole

                                 = 0.0489 moles

Moles of oxygen = 1.174 g /16 g/mole

                            = 0.0734 moles

Mole Ratio = 1 : 0.0734/0.0489

                  = 1 : 1.5

We multiply by 2 to make them whole numbers;

                  = (1 : 1.5) 2

                  = 2 : 3

Therefore;

The empirical formula is Al2O3

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b