A 3.8mg sample of an organic compound gave 5.80mg CO2 and 1.58mg H20 on combustion, produce the empirical formula of the compound.
Calculate the energy, frequency and wavelength (in nm), associated with a transition of electron from n = 3 to n = 1
To determine the empirical formula of the organic compound, we need to calculate the number of moles of carbon and hydrogen in the given sample.
First, let's calculate the number of moles of CO2 and H2O produced during combustion.
Molar mass of CO2 = Molar mass of C + 2 * Molar mass of O
= 12.01 g/mol + 2 * 16.00 g/mol
= 44.01 g/mol
Number of moles of CO2 = Mass of CO2 / Molar mass of CO2
= 5.80 mg / 44.01 g/mol
= 0.1319 mol
Molar mass of H2O = 2 * Molar mass of H + Molar mass of O
= 2 * 1.01 g/mol + 16.00 g/mol
= 18.02 g/mol
Number of moles of H2O = Mass of H2O / Molar mass of H2O
= 1.58 mg / 18.02 g/mol
= 0.0876 mol
Now, we need to find the ratio of moles of carbon to moles of hydrogen. Divide both values by the smaller mole value (in this case, moles of H2O).
Carbon-to-hydrogen ratio = (0.1319 mol / 0.0876 mol) ≈ 1.50
Since the ratio is approximately 1.5, we can assume that the empirical formula of the compound is C1.5Hx.
To get the whole-number ratio, we multiply both subscripts by a common factor that will make the ratio whole-numbered.
In this case, we can multiply both subscripts by 2 to get the simplest whole-number ratio.
C1.5Hx * 2 = C3H2x
Therefore, the empirical formula of the compound is C3H2x.
Moving on to the second question,
To calculate the energy associated with the transition of an electron from n = 3 to n = 1, we can use the Rydberg formula:
E = -R_H *(1/n1^2 - 1/n2^2)
Here, E represents the energy, R_H is the Rydberg constant (2.18 × 10^(-18) J), and n1 and n2 are the initial and final energy levels, respectively.
Plugging in the values:
E = -2.18 × 10^(-18) J * (1/1^2 - 1/3^2)
= -2.18 × 10^(-18) J * (1/1 - 1/9)
= -2.18 × 10^(-18) J * (8/9)
≈ -1.95 × 10^(-18) J
Next, we can calculate the frequency (v) using the equation:
E = h * v
Here, E represents the energy, h is Planck's constant (6.63 × 10^(-34) J·s), and v is the frequency.
Solving for v:
v = E / h
= (-1.95 × 10^(-18) J) / (6.63 × 10^(-34) J·s)
≈ -2.95 × 10^15 Hz
Finally, to find the wavelength (λ), we can use the equation:
v = c / λ
Here, v is the frequency and c is the speed of light (3 × 10^8 m/s).
Rearranging the equation to solve for λ:
λ = c / v
= (3 × 10^8 m/s) / (-2.95 × 10^15 Hz)
≈ -1.02 × 10^(-7) m
To convert this to nm, we multiply by 10^9:
λ (in nm) ≈ -1.02 × 10^(-7) m * 10^9 nm/m
≈ -1.02 × 10^2 nm
Since wavelength cannot be negative, the absolute value is taken.
Therefore, the wavelength associated with the electron transition is approximately 102 nm.