Question

Analysis of a compound containing lead and chlorine reveals that the compound is 59.37% lead. The molar mass of the compound is 349.0 g/n. What is the empirical formula and the molecular formula of this compound?

Answer 1

The empirical formula and molecular formula of the compound are:

• Empirical formula =
  `PbCl_4`
• Molecular formula =
  `PbCl_4`

How to calculate the empirical and molecular formula?

The empirical formula can be calculated as follow:

• Lead (Pb) = 59.37%
• Chlorine (Cl) = 100 - 59.37 = 40.63%
• Empirical formula =?

Divide each by their molar mass

Pb = 59.37 / 207 = 0.287

Cl = 40.63 / 35.5 = 1.145

Divide by the smallest

Pb = 0.287 / 0.287 = 1

Cl = 1.145 / 0.287 = 4

Thus, the empirical formula is
`PbCl_4`

The molecular formula of the compound can be calculated as follow:

• Empirical formula =
  `PbCl_4`
• Molar mass of compound = 349.0 g/mol
• Molecular formula = ?

Molecular formula = Empirical formula × n

But,

Molecular formula = Molar mass

Thus, we have

Empirical formula × n = Molar mass

[207 + (35.5 × 4)]n = 349

349n = 349

n = 349 / 349

n = 1

Molecular formula of compound = Empirical formula × n

=
`PbCl_4` × 1

=
`PbCl_4`