Question

A 2.66-g sample of a pure compound, with formula M2SO4, was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh 1.36 g. Determine the atomic mass of M, and identify M.

Answer 1

Molar mass of CaSO4,

MM = 1*MM(Ca) + 1*MM(S) + 4*MM(O)

= 1*40.08 + 1*32.07 + 4*16.0

= 136.15 g/mol

mass(CaSO4)= 1.36 g

use:

number of mol of CaSO4,

n = mass of CaSO4/molar mass of CaSO4

=(1.36 g)/(1.362*10^2 g/mol)

= 9.989*10^-3 mol

The reaction is:

M2SO4 + CaCl2 —> CaSO4 + 2 MCl

From reaction,

moles of M2SO4 reacted = mol of CaSO4 formed

= 9.989*10^-3 mol

molar mass of M2SO4 = mass / number of mol

= 2.66 g / 9.989*10^-3 mol

= 266.3 g

Let atomic mass of M be x g/mol

molar mass of M2SO4 = 2*x + 1*32 + 4*16

266.3 = 2x + 96

x = 85.2 g/mol

This is Rb

85.2 g/mol

Rb