Chem 103 Lab Exam 3 Review

Chem 103 Lab Exam 3 Review Question 1 10 / 10 pts Answer the following two questions: 1. Why was the paper chromatogram sprayed with ninhydrin solution in the chromatogram experiment? 2. Why was no peak for Hydrogen present in the SEM spectrum for aspirin (C9H8O9)? Your Answer: paper chromatogram sprayed with ninhydrin solution because it reacts with amino acids because amino acids are colorless and color compounds are formed allowing us to visualize the spots. 2. because H cannot be detected by sem 1. The paper chromatogram was sprayed with ninhydrin solution to make the amino acids visible. 2. No peak for Hydrogen was present in the SEM spectrum for aspirin (C9H8O9) because H cannot be detected by SEM. Question 2 10 / 10 pts Answer the following question. A 1.1000 gram hydrate sample chosen from Na2CO3∙10H2O, AlCl3∙6H2O, MgCl2∙6H2O and BaCl2∙2H2O was heated and found to lose 0.6920 gram of H2O. (1) Show the calculation of the % H2O in the unknown hydrate sample. (2) Show the calculation of the % H2O in each of the hydrate compounds and identify the unknown hydrate from the list. Atomic weights: H = 1.008, O = 16.00. MWs: Na2CO3∙10H2O = 286.15, AlCl3∙6H2O = 241.43, MgCl2∙6H2O = 203.301 and BaCl2∙2H2O = 244.462 Your Answer: 1)The hydrate sample (2)= (2) / (ℎ) × 100% = 0.6920 / 1.1000 × 100% = .% 2)Na2CO3 x 10H2O (2) = (2) x (2) / (23) × 100%, N(H2O) – number of molecules H2O in hydrate compound (2) = 2 × 1.008 + 16.00 = 18.02 (2)=10∙18.02286.15×100% = .% 3) AlCl3 x 6H2O (2) = 6 x 18.02 / 241.43 × 100% = 44.78% 4) MgCl2 x 6H2O (2) = 6 x 18.02 / 203.301 × 100% = 53.18% 5) BaCl2 x 2H2O (2) = 2 x 18.02 / 244.462 × 100%=14.74% The unknown hydrate is Na2CO3 x 10H2O. % H2O in unknown = (0.6920 / 1.1000) x 100 = 62.91% % H2O in Na2CO3∙10H2O = 180.16 / 286.15 x 100 = 62.96% unknown is Na2CO3∙10H2O Question 3 9 / 10 pts Answer the following two questions: 1. A paper chromatography experiment was conducted on a mixture of amino acids including Aspartic acid (Asp), Glutamic acid (Glu), Histidine (His), Leucine (Leu), Phenylalanine (Phe) and Proline (Pro). The following Rf values were determined: Asp (0.245), Glutamic (0.275), His (0.295), Leu (0.745), Phe (0.685) and Pro (0.445). In the experiment, a colored spot for unknown amino acid "X" was found centered at a distance 61 mm from the start line when the solvent line had travelled 82 mm. (1) Show the calculation for the Rf value for amino acid "X" and (2) identity amino acid "X". 2. Which of the following is the formula of unknown #1 which gives the following SEM spectrum. Explain your answer. BS2O3 SO3Br Ba(HSO4)2 Your Answer: 1) 61mm / 82mm = 0.744 X = Leu 2) It is Ba(HSO4)2 since it is the only formula that has Ba and H does not show on the SEM. 1. Rf for X = 61/82 = 0.744 matches best with 0.745 (X = Leucine) 2. Ba(HSO4)2 is unknown #1, H cannot be detected by SEM. The name of the amino acid is Leucine. Quiz Score: 29 out of 30