The empirical formula of a compound of uranium and fluorine that is composed of 67.6% uranium and 32.4% fluorine is A) U₂F. B) U₃F₄. C) UF₄. D) UF₆. E) UF₈.

Question

Quizplus · Accepted Answer

To find the empirical formula, we calculate the moles of each element based on their given percentages and atomic masses. Uranium (U) has an atomic mass of approximately 238 g/mol, and fluorine (F) has an atomic mass of approximately 19 g/mol. For uranium: $67.6 \, \text{g} \, \text{U} \times \frac{1 \, \text{mol} \, \text{U}}{238 \, \text{g}} \approx 0.284 \, \text{mol} \, \text{U}$. For fluorine: $32.4 \, \text{g} \, \text{F} \times \frac{1 \, \text{mol} \, \text{F}}{19 \, \text{g}} \approx 1.705 \, \text{mol} \, \text{F}$. The ratio of U to F is approximately 1:6 when simplified, leading to the empirical formula UF6.

The Empirical Formula of a Compound of Uranium and Fluorine