Agaseous compound is 78.14 percent boron and 21.86 percent hydrogen. at 27°c, 74.3 ml of the gas exerted a pressure of 1.12 atm. if the mass of the gas was 0.0934 g, what is its molecular formula?

B₂H₆

Explanation:

Given parameters:

Mass percentage of hydrogen = 21.86%

Mass percentage of boron = 78.14%

Temperature of the compound = 27°C

Pressure exerted by the gas = 1.12atm

Volume of the gas = 74.3mL = 0.0743L or 0.0743dm³

Mass of the gas = 0.0934g

Unknown:

Molecular formula of the gas = ?

We have to convert the temperature to a unit we can work with which is Kelvin:

      27°C to K gives 273 + 27, which is 300K

Solution

      The empirical formula is the simplest formula of a compound.

The procedure is shown below:

                                                      B                                      H

Mass precentage                       78.14                                21.86

Number of moles                   78.14/10.8                           21.86/1

                                                    7.24                                  21.86

Dividing by the

smallest                                    7.24/7.24                          21.86/7.24

                                                        1                                         3

The empirical formula of the compound is BH₃

To derive the molecular formula:

               M.F = (E.F) x n

n is the number of repeating units of the empirical formula present in one mole.

          ideal gas law:  PV = nRT

     The unknown is n, and we make it the subject of the expression:

                            n =

Where P is pressure of the gas

            V is the volume of the gas

            R is the gas constant which is given as 0.082atmdm³mol⁻¹K⁻¹

            T is the temperature

         n =

         n = 0.0034mole

  From the expression:

                 number of moles =

                 molar mass =

                 molar mass =

                 molar mass =  27.5g/mol of the molecular formula

From M.F = (E.F) x n we need to solve for n:

    E.F is BH₃ this gives a mass of [10.8 + 3(1)] = 13.8g/mol

     n =

     n = = 2

Molecular formula of the compound is then (BH₃)₂ which gives B₂H₆

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