Answer to Question #87383 in General Chemistry for Nwadike chidinma

Question

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

Answer 1

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)

$\nu_{n_1, n_2} = R \cdot \lparen \frac{1}{n_2^2} - \frac{1}{n_1^2} \rparen$

The energy of such a photon

$E=h\nu_{n_1 , n_2}$

Corresponding wavelength

$\lambda = \frac c \nu$

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)

$\nu_{3, 1} = 3,29 \cdot 10^{15} \cdot \lparen \frac{1}{1^2} - \frac{1}{3^2} \rparen = 2,92 \cdot 10^{15} s^{-1}$

$E=h\nu_{n_1 , n_2} = 6,63 \cdot 10^{-34} \cdot 2,92 \cdot 10^{15} = 1,94 \cdot 10^{-18} J$

or for energy in eV

$E=h\nu_{n_1 , n_2} = 4,136 \cdot 10^{-15} \cdot 2,92 \cdot 10^{15} = 12,08 \, eV$

Wavelength

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