How do u calculate the empirical formulae of an unknown compound a if it contains 46.6% of carbon, 13.6% of nitrogen, 8.47% of hydrogen and ? % of nb: ram of c= 12, h= 1, n= 14 and o= 16

C₄NH₉O₂

Explanation:

The question requires we determine the empirical for an unknown compound A.

Carbon-46.6%

Nitrogen - 13.6%

Hydrogen - 8.47%

Oxygen = 100% - (46.65 + 13.6% + 8.47%)

             = 31.28%

Step 1: Determine the moles of each element

Moles of Carbon = 46.6 ÷ 12

                            = 3.88 moles

Moles of Nitrogen = 13.6 ÷ 14

                              = 0.971 moles

Moles of Hydrogen = 8.47 ÷ 1

                                 = 8.47 moles

Moles of Oxygen = 31.28 ÷ 16

                             = 1.955 moles

Step 2: Determine the mole ratio of the elements

Mole ratio = C :  N : H : O

                = 3.88 moles : 0.971 moles : 8.47 moles : 1.955 moles

We divide by the smallest (0.971 moles )

               = 3.99 : 1 : 8.72 : 2.01

                = 4 : 1 : 9 : 2

     E.F      = C₄NH₉O₂

Therefore, the empirical formula of compound A is C₄NH₉O₂

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