Question

What is the percent of Boron in Sr3(BO3)2

Answer 1

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%