Question

A mixture consisting of only rubidium chloride (RbCl, 120.92 g/mol) and sodium chloride (NaCl, 58.44 g/mol) weighs 1.0333 g. When the mixture is dissolved in water and an excess of silver nitrate is added, all the chloride ions associated with the original mixture are precipitated as insoluble silver chloride (AgCl, 143.32 g/mol). The mass of the silver chloride is found to be 1.6048 g. Calculate the mass percentage of rubidium chloride in the original mixture.

Answer 1

mass of NaCl + mass of RbCl = 1.0333 g

Let mass of NaCl = X
mass of the RbCl = Y

X + Y = 1.0333 g

mass of AgCl = 1.6048 g

Moles of AgCl = mass/molecular weight

= 1.6048g / 143.32 g/mol

= 0.011197 mol

1 mol of AgCl requires 1 mole of either NaCl or RbCl

Moles of NaCl + moles RbCl = 0.011197 moles

[X / 58.44 g/mol] + [Y / 120.92 g/mol] = 0.011197 moles

but Y = 1.0333 – X

[X / 58.44 g/mol] + [(1.0333 – X) / 120.92 g/mol] = 0.011197 moles

[X * 120.92] + [(1.0333 – X) * 58.44] = 0.011197 * 120.92 * 58.44

120.92 X – 58.44 X + (1.0333 * 58.44) = 79.124

62.48 X = 18.737

X = 0.2999

Y = 1.0333 g – X = 1.0333 – 0.2999 = 0.73342 g

% NaCl = (0.2999 g / 1.0333 g) * 100 = 29.02 %
RbCl = (0.73342 g / 1.0333 g) * 100 = 70.97 %