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 o

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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

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Delwyn 4 years 2021-07-12T17:57:21+00:00 1 Answers 124 views 0

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  1. kimchi2
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    2021-07-12T17:58:51+00:00
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    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.

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