Answer:
B₂O₃
Explanation:
Step 1: Calculate the mass of oxygen in 150 mg of boron oxide
Of 150 mg of boron oxide, 46.6 mg belong to boron. The mass of oxygen is:
150 mg - 46.6 mg = 103.4 mg
Step 2: Calculate the percent by mass of each element
We will use the following expression.
%Element = mElement/mCompound × 100%
%B = 46.6 mg/150 mg × 100% = 31.1%
%O = 103.4 mg/150 mg × 100% = 68.9%
Step 3: Divide each percentage by the atomic mass of the element
B: 31.1/10.81 = 2.88
O: 68.9/16.00 = 4.31
Step 4: Divide both numbers by the smallest one (2.88)
B: 2.88/2.88 = 1
O: 4.31/2.88 ≈ 1.5
Step 5: Multiply both numbers by 2 so that they are integers
B: 1 × 2 = 2
O: 1.5 × 2 = 3
The empirical formula is B₂O₃.
