Question

Ferrophosphorus (Fe2P) reacts with pyrite (FeS2), producing iron (II) sulfide and a compound that is 27.87% P and 72.13% S by mass and has a molar mass of 444.56 g/mol. Determine the empirical and molecular formulas of the compound.

Answer 1

Answer: The empirical and molecular formula for the formed compound is
`P_2S_5` and
`P_4S_(10)` respectively.

Step-by-step explanation:

We are given:

Percentage of P = 27.87 %

Percentage of S = 72.13 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 27.87 g

Mass of S = 72.13 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Phosphorus =
`\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=(27.87g)/(31g/mole)=0.899moles`

Moles of Sulfur =
`\frac{\text{Given mass of Sulfur}}{\text{Molar mass of Sulfur}}=(72.13g)/(32g/mole)=2.25moles`

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.899 moles.

For Phosphorus =
`(0.899)/(0.899)=1`

For Sulfur =
`(2.25)/(0.899)=2.5`

• Step 3: Taking the mole ratio as their subscripts.

The ratio of P : S = 1 : 2.5

Converting the mole ratio into whole number by multiplying it by '2'

Now, the ratio becomes, P : S = 2 : 5

The empirical formula of the compound is
`P_2S_5`

• For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

`n=\frac{\text{Molecular mass}}{\text{Empirical mass}}`

We are given:

Mass of molecular formula = 444.56 g/mol

Mass of empirical formula = 222 g/mol

Putting values in above equation, we get:

`n=(444.56g/mol)/(222g/mol)=2`

Multiplying this valency by the subscript of every element of empirical formula, we get:

`P_((2* 2))S_((5* 2))=P_4S_(10)`

Hence, the empirical and molecular formula for the formed compound is
`P_2S_5` and
`P_4S_(10)` respectively.