Answer to Question #342629 in Organic Chemistry for Koketso

Question

the reaction of diatomic fluorine gas with compound X yields a single product, compound Y. The mass percent composition of the product, compound Y is 61.7%F2 and 38.3%CL. What is the empirical formula for compound x?

Answer 1

Solution:

Schematic chemical equation:

F₂ + X → Y

Compound Y: w(F) + w(Cl) = 61.7% + 38.3% = 100%

Thus, the compound Y consists of only fluorine and chlorine (Y = Cl_xF_y)

Assume we were given 100 g of compound Y.

Convert %values to grams:

Mass of F = w(F) × Mass of compound Y = 0.617 × 100 g = 61.7 g F

Mass of Cl = w(Cl) × Mass of compound Y = 0.383 × 100 g = 38.3 g Cl

Convert grams to moles:

The molar mass of F is 18.998 g/mol

The molar mass of Cl is 35.453 g/mol

Therefore,

Moles of F = (61.7 g F) × (1 mol F / 18.998 g F) = 3.2477 mol F

Moles of Cl = (38.3 g Cl) × (1 mol Cl / 35.453 g Cl) = 1.0803 mol Cl

Divide moles by the smallest of the results:

F : 3.2477 / 1.0803 = 3.006

Cl : 1.0803 / 1.0803 = 1.000

Thus, the empirical formula for compound Y is ClF₃

F₂ + X → ClF₃

So, the empirical formula for compound X is ClF

Balanced chemical equation:

F₂(g) + ClF(g) → ClF₃(g)

Answer: The empirical formula for compound X is ClF