Answer:
5.68%
Step-by-step explanation:
Percent Composition of Boron = Atomic Mass of Boron / Molar mass of Sr3(BO3)2 * 100%
Atomic mass of B = 10.811 g/mol
Molar mass of Sr3(BO3)2 = 380.4784 g/mol
Percent composition = 2 (10.811) / 380.4784 * 100
Percent composition = 21.622 / 380.4784 * 100
Percent composition = 5.68%