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

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

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