The answer is c. SrCO3 Sr is in Group II of the periodic table and form Sr2+ ions (oxidation state +2) to balance the CO32- ions %Oxygen by mass = (16 x 3)/(87.6 + 12 + 16 x 3) x 100 = 32.50% Solution The answer is c. SrCO3 Sr is in Group II of the periodic table and form Sr2+ ions (oxidation state +2) to balance the CO32- ions %Oxygen by mass = (16 x 3)/(87.6 + 12 + 16 x 3) x 100 = 32.50%.