Answer to Question #132607 in General Chemistry for Om Barvaliya

Question

John wants to buy a ring with sapphire( Al2O3 in its purest form) that contains 2.22 x 10^22 molecules of Oxygen. What volume of Al2O3 would he need in cm^3? The molar mass of Al2O3 is 101.96 g/mol and the density is 3.95 g/cm^3.

Answer 1

From the number of Oxygen particles the number of moles may be found:

n=N/N_a, where N_a is Avogadro's constant.

n=2.22×10²²/6.02×10²³=0.0369 (moles).

There are 3 atoms of Oxygen in each molecule of Al₂O₃, so it's number of moles is 3 times lower:

n(Al₂O₃)=0.0369/3=0.0123(moles).

From molar mass mass of sapphire may be found:

m=n×M.

m=0.0123×101.96=1.2541(g).

From density the needed volume may be found:

V=m/d.

V=1.2541/3.95=0.3175(cm³).

Answer: 0.3175 cm³.