Question

Fluorine reacts with uranium to produce uranium hexafluoride \[U{F_6}\] as represented by this equation: \[U(s) + 3{F_2}(g) \to U{F_6}(g)\]. How many fluorine molecules are required to produce \[7.04{\text{ mg}}\] of uranium hexafluoride from an excess of uranium? The molar mass of \[U{F_6}\] is \[352{\text{ gm/mol}}\]
A. \[3.6 \times {10^{19}}\]
B. \[3.6 \times {10^{15}}\]
C. \[4.6 \times {10^{19}}\]
D. \[3.6 \times {10^{15}}\]

Answer 1

Hint: To answer this question, you should recall the mole concept and Avogadro’s number. The concept states that a mole of any substance contains the same number of particles, it is $6.022 \times {10^{23}}$ particles per mol. It is derived from the number of atoms of the pure isotope $^{{\text{12}}}{\text{C}}$ in 12 grams of that substance and is the reciprocal of atomic mass in grams. We shall calculate the moles of uranium hexafluoride and from that, the moles of fluorine present. Then, we shall calculate the number of particles in it.

Formula Used: ${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Mass of the Substance in grams}}}}{{{\text{Molar mass of a Substance}}}} = \dfrac{{{\text{Number of Atoms or Molecules}}}}{{6.022 \times {{10}^{23}}}}$

Complete step by step answer:
According to the reaction: \[U(s) + 3{F_2}(g) \to U{F_6}(g)\].
\[7.04{\text{mg}}\] of uranium hexafluoride corresponds to \[\dfrac{{7.04}}{{1000 \times 352}} = 2 \times {10^{ - 5}}{\text{moles }} = 2 \times {10^{ - 5}} \times 6.023 \times {10^{23}} = 1.2 \times {10^{19}}{\text{molecules}}{\text{.}}\]
The number of fluorine molecules will be three times the amount for each mole of \[U{F_6}\] produced.

Hence, the required amount is \[3 \times 1.2 \times {10^{19}} = 3.6 \times {10^{19}}.\]

Note:
The mole concept is an important component in the measurement of entities in a chemical compound. The concept that a mole of any substance contains the same number of particles was formed out of research which was conducted by Italian physicist Amedeo Avogadro. Avogadro constant can be defined as the number of molecules, atoms, or ions in one mole of a substance: $6.022 \times {10^{23}}$ per mol. It is derived from the number of atoms of the pure isotope $^{{\text{12}}}{\text{C}}$in 12 grams of that substance and is the reciprocal of atomic mass in grams. Now the mole concept can be applied to ions and formula units.