What is the empirical formula of a compound whose percentage composition is found to be 62.60% lead, 8.40% nitrogen, and 29.0% oxygen?

The empirical formula of the compound containing 62.60% lead, 8.40% nitrogen, and 29.0% oxygen is PbN₂O₆ or Pb(NO₃)₂

From the question given above, the following data were obtained:

Lead (Pb) = 62.60%

Nitrogen (N) = 8.40%

Oxygen (O) = 29.0%

The empirical formula of the compound can be obtained as illustrated below:

Divide by their molar mass

Pb = 62.60 / 207 = 0.302

Nitrogen (N) = 8.40 / 14 = 0.6

Oxygen (O) = 29.0 / 16 = 1.8125

Divide by the smallest

Pb = 0.302 / 0.302 = 1

Nitrogen (N) = 0.6 / 0.302 = 2

Oxygen (O) = 1.8125 / 0.302 = 6

Thus, the empirical formula of the compound is PbN₂O₆ or Pb(NO₃)₂

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PbN₂O₆

Explanation:

Empirical formula is the simplest formula of a compound.

To find the empirical formula, we use the system below:

                                  Lead                      Nitrogen                Oxygen

Percentage

Composition              62.6                           8.4                          29

Molar mass                  207.2                         14                           16

Number of

Moles                        62.6/207.2                 8.4/14                     29/16

                                      0.302                        0.6                        1.813

Divide by the

smallest                     0.30/0.302                   0.6/0.302          1.813/0.302

                                            1                                  2                           6

Empirical formula is PbN₂O₆