Question

Xenon (xe) and fluorine (f) can combine to form several different compounds, including xef2, which contains 0.2894 g fluorine for every gram of xenon. use the law of multiple proportions to determine, n, which represents the number of f atoms in another compound, xefn, given that it contains 0.8682 g of f for every gram of xe.

Answer 1

We can say that for every 2 atoms of F, there is a mass of 0.2894 g as evident from the compound XeF2. Now to find for f, we can say that there are n atoms of F for every 0.8682 g of F. Taking the proportion:

2 atoms / 0.2894 g = n / 0.8682 g

Calculating for n:

n = 2 * 0.8682 / 0.2894

n = 6

Answer 2

Answer is: n (number of fluorine atoms) is 6, formula of the compound is XeF₆.

m(F) = 0.8682 g; mass of fluorine.

n(F) = m(F) ÷ M(F).

n(F) = 0.8682 g ÷ 19 g/mol.

n(F) = 0.0457 mol; amount of substance.

m(Xe) = 1.00 g.

n(Xe) = 1.00 g ÷ 131.293 g/mol.

n(Xe) = 0.00761 mol.

n(Xe) : n(F) = 0.00761 mol : 0.0457 mol.

n(Xe) : n(F) = 1 mol : 6 mol.