Question

A 4.50 mL sample of a 0.100 M solution of an aromatic hydrocarbon dissolved in hexane is excited with a flash of light. The aromatic compound emits 19.3 J of energy at an average wavelength of 379 nm. What percentage of the aromatic compound molecules emitted a photon? Please Help, Thanks

Answer 1

molarity of hydrocarbon = 0.100M

volume of sample = 4.50mL = 4.50/1000L

moles of hydrocarbon = 0.1004.50/1000

                                = 4.5010^-4moles

no. of molecules of hydrocarbon = 4.5010^-4 6.022*10²³

                                               = 2.709910²⁰ molecules                                                  (1)

wavelength of light = 379nm = 37910^-9m

    using, E = hc/

                 = 6.62610^-34 2.99810⁸ / 37910^-9

                       = 5.24110^-19J

total amount of energy emitted = 19.3J

no. of photons emitted = 19.3 / 5.24110^-19

                                 = 3.68210¹⁹photons

percentage of molecules emitted a photon = no. of photons / no. of molecules 100

                                                              = 3.68210¹⁹ / 2.709910²⁰ 100

                                                              = 13.59%

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