Question

6g of Oxyfluoride was prepared by mixing oxygen and fluorine in the proper ratio at 60K. This compound contains 1.926g fluorine and 4.079g oxygen. What is the empirical formula of the compound? Calculate the percentage composition of Fluorine and Oxygen in the compound​

Answer 1

Dear Student


Mass of Fluorine, F = 1.926 g

Mass of Oxygen, O = 4.079 g

Total Mass of Compound = (1.926 + 4.079) g = 6.005 g

Given Mass of Oxyfluoride = 6 g


We know,

No. of moles = $\frac{Mass}{MolarMass}$

Therefore, No. of moles of Fluorine, F = $\frac{1.926g}{19g/mol}$ = 0.1013 mol

No. of moles of Oxygen, O = $\frac{4.079g}{16g/mol}$ = 0.2549 mol


The Simple Whole Ratio of -

Fluorine, F = $\frac{0.1013}{0.1013}$ = 1

Oxygen, O = $\frac{0.2549}{0.1013}$ = 2.5

So, the molecular ratio is 1 : 2.5

Multiplying both the side by 2, to get a whole number, we get

1 x 2 : 2.5 x 2

2 : 5

So, the Empirical Formula will be F₂O₅


Now, the percentage combination of Fluorine and Oxygen in the compound ???

So,

the Percentage Composition of Fluorine in F₂O₅ = $\frac{MassofF}{TotalMassof{F}_2{O}_5}x100\%$

= $\frac{1.926g}{6.005g}x100\%$ = 0.3207 x 100% = 32.07%


the Percentage Composition of Oxygen in F₂O₅​ = ​$\frac{MassofO}{TotalMassof{F}_2{O}_5}x100\%$

= ​$\frac{4.079g}{6.005g}x100\%$ = 0.6792 x 100% = 67.93%



Regards