A 46.6-mgmg sample of boron reacts with oxygen to form 150 mgmg of the compound boron oxide. Part A What is the empirical formula of boron oxide

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.

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