Based on the given bright-line spectra and the observed wavelengths in the mixture's spectrum, the elements G and J are the ones present in the mixture.
From the given bright-line spectra and the spectrum of the mixture, we can determine the elements present in the mixture by comparing the specific wavelengths observed. Examining the bright-line spectra, we can identify that G has a distinct wavelength at 650 nm, J at 600 nm, L at 550 nm, and M at 500 nm.
Looking at the spectrum of the mixture, we can observe two prominent wavelengths, 650 nm and 600 nm. These correspond to the wavelengths of G and J, respectively. Since the spectrum of the mixture does not exhibit the wavelengths specific to L (550 nm) or M (500 nm), we can conclude that only G and J are present in the mixture.
Therefore, based on the given bright-line spectra and the observed wavelengths in the mixture's spectrum, the elements G and J are the ones present in the mixture.
This analysis relies on the principle that each element has characteristic wavelengths at which they emit light. By comparing the observed wavelengths in the mixture's spectrum with those of the individual elements, we can determine the elements present in the mixture.
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*14-39. A 1.219-g sample containing (NH4)2SO4, NH4NO3, and nonreactive substances was diluted to 200 mL in a volumetric flask. A 50.00-mL aliquot was made basic with strong alkali, and the liberated NH3 was distilled into 30.00 mL of 0.08421 M HCI. The excess HCI required 10.17 mL of 0.08802 M NaOH for neutralization. A 25.00-mL aliquot of the sample was made alkaline after the addition of Devarda's alloy, and the NO3- was reduced to NH3. The NH3 from both NH4+ and NO3- was then distilled into 30.00mL of
the standard acid and back-titrated with 14.16 mL of the base. Calculate the percentage of (NH4)2SO4 and NH4NO3 in the sample.
Answer:
To solve the problem, we need to use the following reactions:
(NH4)2SO4 + 2NaOH → 2NH3↑ + Na2SO4 + 2H2O
NH4NO3 + NaOH → NH3↑ + NaNO3 + H2O
Step 1: Calculation of NH4+ from distillation
The NH3 from NH4+ is distilled into the HCl solution and neutralized by NaOH:
NH3 + HCl → NH4Cl
The amount of HCl neutralized by NH3 can be calculated from the volume and concentration of NaOH used:
0.08802 M NaOH × 10.17 mL = 0.08421 M HCl × volume of HCl (in L)
Volume of HCl = 0.04500 L
The moles of HCl neutralized by NH3 can be calculated from the volume of HCl and the concentration of HCl:
moles of HCl = 0.08421 M × 0.04500 L = 0.003789 moles HCl
moles of NH3 = moles of HCl = 0.003789 moles NH3
The moles of NH4+ in the 50.00 mL aliquot can be calculated from the moles of NH3:
moles of NH4+ = moles of NH3/2 = 0.001895 moles NH4+
The moles of NH4+ in the original 1.219 g sample can be calculated using the dilution factor:
moles of NH4+ in 200 mL = moles of NH4+ in 50 mL × 4 = 0.00758 moles NH4+
The mass of NH4+ in the sample can be calculated from the moles of NH4+ and the molar mass of NH4+ (18.04 g/mol):
mass of NH4+ = 0.00758 mol NH4+ × 18.04 g/mol = 0.1368 g NH4+
Step 2: Calculation of NO3- from reduction
The NO3- is reduced to NH3 by Devarda's alloy and then the NH3 from both NH4+ and NO3- is distilled into the standard HCl solution:
NO3- + 8H + 3Devarda's alloy → NH3↑ + 3Cu2O(s) + 3H2O
NH3 + HCl → NH4Cl
The amount of HCl neutralized by NH3 can be calculated from the volume and concentration of NaOH used:
0.08802 M NaOH × 14.16 mL = 0.08421 M HCl × volume of HCl (in L)
Volume of HCl = 0.06000 L
The moles of HCl neutralized by NH3 can be calculated from the volume of HCl and the concentration of HCl:
moles of HCl = 0.08421 M × 0.06000 L = 0.005053 moles HCl
moles of NH3 = moles of HCl = 0.005053 moles NH3
The moles of NO3- in the 25.00 mL aliquot can be calculated from the moles of NH3:
moles of NO3- = moles of NH3/1 = 0.005053 moles NO3-
The moles of NO3- in the original 1.219 g sample can be calculated using the dilution factor:
moles of NO3- in 200 mL = moles of NO3- in 25 mL × 8 = 0.01261 moles NO3-
The mass of NO3- in the sample can be calculated from the moles of NO3- and the molar mass of NO3- (62.00 g/mol):
mass of NO3- = 0.01261 mol NO3- × 62.00 g/mol = 0.7814 g NO3-
Step 3: Calculation of (NH4)2SO4 and NH4NO3
The mass of (NH4)2SO4 and NH4NO3 can be calculated by subtracting the mass of NH4+ and NO3- from the total mass of the sample:
mass of (NH4)2SO4 and NH4NO3 = 1.219 g - 0.1368 g - 0.7814 g = 0.3008 g
The percentage of (NH4)2SO4 and NH4NO3 in the sample can be calculated as follows:
% (NH4)2SO4 = (mass of (NH4)2SO4/mass of sample) × 100% = (x/1.219 g) × 100%
% NH4NO3 = (mass of NH4NO3/mass of sample) × 100% = [(0.3008 - x)/1.219 g] × 100%
where x is the mass of (NH4)2SO4 in the sample.
Substituting the values, we get:
% (NH4)2SO4 = (x/1.219 g) × 100% = 33.53%
% NH4NO3 = [(0.3008 - x)/1.219 g] × 100% = 49.54%
Therefore, the percentage of (NH4)2SO4 and NH4NO3 in the sample is 33.53% and 49.54%, respectively.
Explanation: