Question

An element X forms both a dichloride (XCl₂) and a tetrachloride (XCl₄). Treatment of 13.62 g XCl₂ with excess chlorine forms 17.09 g XCl₄.

1. Calculate the atomic mass of X.
   in g/mol
2. What is the name of X?

Answer 1

A)

The reaction is

XCl2 + Cl2 = XCl4

Mass of Cl2 reacting = Mass of XCl4 - Mass of XCl2

= 17.09 g - 13.62 g

= 3.47 g

Molar mass of Cl2 = 70.9 g/mol

mass(Cl2)= 3.47 g

use:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(3.47 g)/(70.9 g/mol)

= 4.894*10^-2 mol

From reaction,

Mol of XCl2 reacting = mol of Cl2

= 4.894*10^-2 mol

Now use:

Molar mass of XCl2 = mass of XCl2 / Mol of XCl2

= 13.62 g / 4.894*10^-2 mol

= 278.3 g/mol

Molar mass of XCl2 = MM(X) + 2*MM(Cl)

278.3 = MM(X) + 2*35.45

MM(X) = 207.4 g/mol

Answer: 207.4 g/mol

B)

This is atomic mass of lead

Answer: Lead