Chem 171 F09 Q3 Chrom Emp Key

Chem 171 F09 Q3 Chrom Emp Key - Chem 171 Quiz 3 Version A 1...

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Chem 171 Quiz 3 Version B 1. In the attached paper chromatogram, spots 1,2,3, and 4 are known compounds. What is the composition of unknown #5? The unknown # 5 is a mixture of two compounds Compound 2 Compound 3 2. Determine the empirical formula of the following compound: 0.903 g of phosphorus combined with 6.99 g of bromine. P=30.97. Br = 79.90. () 1molP 0.903 g P 30.97 g P ⎛⎞ ⎜⎟ ⎝⎠ = 0.029157 mol P 1molBr 6.99 g Br 79.90 g Br = 0.087484 mol Br 0.029157 mol P 0.029157 mol P = 1 0.087484 mol Br 0.029157 mol P = 3 The empirical formula is PBr 3 . Chem 171 Quiz 3 Version A 1 . In a paper chromatography experiment, the following values were obtained: Ring front distances for solvent 77 mm, for Fe 3+ 77 mm, Cu 2+ 43 mm, Ni 2+ 39 mm. Calculate the R f values. R f Fe3+ = 77 mm /77 mm = 1.0 R f Cu2+ = 43 mm / 77 mm = 0.56 R f Ni2 + = 39 mm / 77 mm = 0.51 2 . Cortisol (Molar mass = 362.47) is 69.6% C, 8.34% H, and 22.1 % O by mass. What is its molecular formula? C = 12.01. H =1.008. O = 16.00. Moles C = 1molC 69.6 g C 12.01 g C = 5.7952 mol C Moles H = 1molH 8.34 g H 1.008 g H = 8.2738 mol H Moles O = 1molO 22.1 g O 16.00 g O = 1.38125 mol O 5.7952 mol C 1.38125 mol O = 4.20 8.2738 mol H 1.38125 mol O = 6.00 1.38125 mol O 1.38125 mol O = 1.00 x 5 empirical formula = C 21 H 30 O 5 (12.01 g C/mol) + 30 (1.008 g H/mol) + 5 (16.00 g O/mol) = 362.45 g/mol
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This note was uploaded on 03/04/2012 for the course CHEM 171 taught by Professor Several during the Fall '08 term at Rutgers.

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Chem 171 F09 Q3 Chrom Emp Key - Chem 171 Quiz 3 Version A 1...

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