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16 June, 13:31

The empirical formula of a compound of iron (Mg) and sulfur (S) was determined using the following data. A sample of Mg was weighed into a crucible and covered with finely powdered S. The crucible was covered and heated to allow the Fe and S to react. Additional heating burned off all unreacted S. The crucible was then cooled and weighed. The following data was collected. The molecular weight of Mg and S is 24.305 g/mol and 32.065 g/mol respectively. mass of crucible and cover = 27.631 grams mass of crucible, cover and Mg = 33.709 grams mass of crucible, cover, and the compound formed 41.725 grams

Calculate the number of moles Mg and reacted and what is the empirical formula of the compound.

A. 0.25, 0.25, MgS

B. 0.20,0.30, Mg2S3

C. 0.15,0.35, Mg1S3

D. 0.45, 0.25, MgS2

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Answers (2)
  1. 16 June, 14:47
    0
    The empirical formula is MgS

    The number of moles Mg is 0.25 moles

    There will react 0.25 moles

    Option A is correct

    Explanation:

    Step 1: data given

    The molecular weight of Mg = 24.305 g/mol

    The molecular weight of S is 32.065 g/mol

    mass of crucible and cover = 27.631 grams

    mass of crucible, cover and Mg = 33.709 grams

    mass of crucible, cover, and the compound formed 41.725 grams

    Step 2: Calculate mass of Mg

    Mass of Mg = mass of crucible, cover and Mg - mass of crucible and cover

    Mass of Mg = 33.709 grams - 27.631 grams

    Mass of Mg = 6.078 grams

    Step 3: Calculate moles Mg

    Moles Mg = mass Mg / atomic mass Mg

    Moles Mg = 6.078 grams 24.305 g/mol

    Moles Mg = 0.250 moles

    Step 4: Calculate mass compound

    Mass compound = mass of crucible, cover, and the compound formed - mass of crucible and cover

    Mass compound = 41.725 grams - 27.631 grams

    Mass compound = 14.094 grams

    Step 5: Calculate mass sulfur

    Mass sulfur = 14.094 grams - 6.078 grams

    Mass sulfur = 8.016 grams

    Step 6: Calculate moles Sulfur

    Moles sulfur = 8.016 grams / 32.065 g/mol

    Moles sulfur = 0.250 moles

    Step 7: Calculate the mol ratio

    We diviide by the smallest amount of moles

    Mg: 0.250 moles / 0.250 moles = 1

    S: 0.250 moles / 0.250 moles = 1

    The empirical formula is MgS

    The number of moles Mg is 0.25 moles

    There will react 0.25 moles

    Option A is correct
  2. 16 June, 16:47
    0
    Option A. 0.25, 0.25, MgS

    Explanation:

    Step 1:

    Data obtained from the question. This includes:

    Molar Mass of Mg = 24.305 g/mol

    Molar Mass of S = 32.065 g/mol

    Mass of crucible and cover = 27.631g

    Mass of crucible, cover and Mg = 33.709 g

    mass of crucible, cover, and the compound formed = 41.725g

    Step 2:

    Determination of the mass of Mg that reacted:

    This is illustrated below:

    Mass of crucible and cover = 27.631g

    Mass of crucible, cover and Mg = 33.709 g

    Mass of Mg = (Mass of crucible, cover and Mg) - (Mass of crucible and cover)

    Mass of Mg = 33.709 - 27.631

    Mass of Mg = 6.078g

    Step 3:

    Determination of the mass of S that reacted. This is illustrated below:

    Mass of crucible and cover = 27.631g

    Mass of Mg = 6.078g

    mass of crucible, cover, and the compound formed = 41.725g

    Mass of S = (mass of crucible, cover, and the compound formed) - (Mass of crucible and cover) - (Mass of Mg)

    Mass of S = 41.725 - 27.631 - 6.078

    Mass of S = 8.016g

    Step 4:

    Determination of the number of mole of Mg that reacted. This is illustrated below:

    Molar Mass of Mg = 24.305 g/mol

    Mass of Mg = 6.078g

    Number of mole = Mass/Molar Mass

    Number of mole Mg = 6.078/24.305

    Number of mole of Mg = 0.25 mol

    Step 5:

    Determination of the number of mole of S that reacted. This is illustrated below:

    Molar Mass of S = 32.065 g/mol

    Mass of S = 8.016g

    Number of mole = Mass/Molar Mass

    Number of mole S = 8.016/32.065

    Number of mole of S = 0.25 mol

    Step 6:

    Determination of the empirical formula. This is illustrated below below:

    Mg = 6.078g

    S = 8.016g

    Divide by their molar mass

    Mg = 6.078 / 24.305 = 0.25

    S = 8.016 / 32.065 = 0.25

    Divide by the smallest

    Mg = 0.25 / 0.25 = 1

    S = 0.25 / 0.25 = 1

    Therefore, the empirical formula is MgS
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