Answer:
1. EF = PSCl₃; 2. MF = PSCl₃
Explanation:
1. Empirical formula
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our first job is to calculate the molar ratio of P:S:Cl.
Assume 100 g of the compound.
(a) Calculate the mass of each element.
Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.
(b) Calculate the moles of each element
[tex]\text{Moles of P} = \text{18.28 g C} \times \dfrac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\\\\\text{Moles of S} = \text{18.93 g S} \times \dfrac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\\\\\text{Moles of Cl} = \text{62.78 g Cl} \times \dfrac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}[/tex]
(c) Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3
(d) Write the empirical formula
EF = PSCl₃
The empirical formula for this compound is PSCl₃.
2. Molecular formula
(a) Calculate the ratio of the molecular and empirical formula masses
n = (169.4 u)/(169.40 u) = 1.000 ≈ 1
(b) Calculate the molecular formula
MF = (EF)ₙ = (EF)₁ = PSCl₃
The molecular formula for this compound is PSCl₃.
