Answer to Question #87383 in General Chemistry for Nwadike chidinma

Question #87383
A 3.87mg of an organic compound gave 5.80mg CO2 and 1.58mg of H2O on combustion,deduce the empirical formula of the compound
B calculate the energy,frequency and wavelength (in nm)associated with a transition of electron from n=3 to n=1
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Expert's answer
2019-04-02T06:34:57-0400

A: During combustion, 5.80 / (12 + 32) = 0.1318 mmol of carbon monoxide are formed, which corresponds to 0.1318 mmol of carbon in the sample taken, and 1.58 / (2 + 16) = 0.08778 mmol of water, which corresponds to 0.08778 * 2 = 0.1756 moles of hydrogen. The ratio of the number of moles of hydrogen to carbon in the composition of the compound will be 0.1756 / 0.1318 = 1.33 = 1 + 1/3 = 4/3. Therefore, the empirical formula of the compound under consideration is C3H4.


B: In accordance with the Balmer's formula, during the transition of an electron from the n1 level to the n2 level, a photon is emitted by the frequency (for hydrogen atom)



νn1,n2=R(1n221n12)\nu_{n_1, n_2} = R \cdot \lparen \frac{1}{n_2^2} - \frac{1}{n_1^2} \rparen

The energy of such a photon


E=hνn1,n2E=h\nu_{n_1 , n_2}

Corresponding wavelength



λ=cν\lambda = \frac c \nu


For n2 = 1 and n1 = 3, Ridberg constant R = 3,29*1015 c-1 we obtain


In accordance with the Balmer's formula, during the transition of an electron from the n1 level to the n2 level, a photon is emitted by the frequency (for hydrogen atom)



ν3,1=3,291015(112132)=2,921015s1\nu_{3, 1} = 3,29 \cdot 10^{15} \cdot \lparen \frac{1}{1^2} - \frac{1}{3^2} \rparen = 2,92 \cdot 10^{15} s^{-1}

E=hνn1,n2=6,6310342,921015=1,941018JE=h\nu_{n_1 , n_2} = 6,63 \cdot 10^{-34} \cdot 2,92 \cdot 10^{15} = 1,94 \cdot 10^{-18} J

or for energy in eV


E=hνn1,n2=4,13610152,921015=12,08eVE=h\nu_{n_1 , n_2} = 4,136 \cdot 10^{-15} \cdot 2,92 \cdot 10^{15} = 12,08 \, eV

Wavelength


λ=cν=3,081082,921015=1,026107m=102,6nm\lambda = \frac c \nu = \frac {3,08 \cdot 10^8} {2,92 \cdot 10^{15} } =1,026 \cdot 10^{-7} m = 102,6 \, nm


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