Question

What mass of xenon tetrafluoride, xef 4 , has the same number of fluorine atoms as 25.0 g of oxygen difluoride, of2?

Answer 1

Answer is: mass of xenon tetrafluoride is 47.98 grams.

1) m(OF₂) = 25.0 g; mass of oxygen difluoride.

n(OF₂) = m(OF₂) ÷ M(OF₂).

n(OF₂) = 25 g ÷ 54 g/mol.

n(OF₂) = 0.462 mol; amount of substance.

In one oxygen difluoride (OF₂) there are two fluorine atoms: n(F) : n(OF₂) = 2 : 1.

n(F) = 0.462 mol · 2.

n(F) = 0.926 mol.

N(F) = n(F) · Na(Avogadro constant).

N(F) = 0.926 mol · 6.022·10²³ 1/mol.

N(F) = 5.576·10²³; number of fluorine atoms in oxygen difluoride.

2) In one molecule of xenon tetrafluoride there are four fluorine atoms:

n(XeF₄) : n(F) = 1 :4.

n(XeF₄) = 0.926 mol ÷ 4.

n(XeF₄) = 0.2315 mol; amount of xenon tetrafluoride.

m(XeF₄) = 0.2315 mol · 207.28 g/mol.

m(XeF₄) = 47.98 g.

Answer 2

To determine the mass of xenon tetrafluoride, we need to know first the number of fluorine atoms present in oxygen difluoride. We need to convert first the mass into moles then make use of the relation of the elements from the chemical formula. Then, use the avogadro's number to convert it to number of atoms. Then, we do the reverse of the steps above but this time for xenon tetrafluoride.

25.0 g OF2 ( 1 mol / 54 g ) ( 2 mol F / 1 mol OF2 ) ( 6.022 x10^23 atoms F / 1 mol F ) ( 1 mol / 6.022x10^23 atoms) ( 1 mol XeF4 / 4 mol F ) (207.3 g / 1 mol XeF4) = 47.99 g XeF4