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

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

what are the statements