24%), N 10 Elemental analysis of an unknown organic compound provided the follow

Question

24%), N 10 Elemental analysis of an unknown organic compound provided the following mass percentages: C 53.09%), H (6 (12.38%), O (28.29%). The molecular weight of the unknown was determined to be 1 13.1 1 gmol. a. (2 pts) Determine the molecular formula. Show all work. b. (2 pts) Calculate the index of hydrogen deficiency from the molecular formula. Show all work. c. (2 pts) The IR for the unknown is shown below. Draw a structure for this compound using all of this information. 0.8 0.6 0.4 0.2 3000 2000 1000 Wavenumber (cm-1)

Explanation / Answer

a. From Mass percentage, number of carbon atom=5,Number of H atom=7, no of N atom=1, number of O atom =2. So molecular formula will be C5H7NO2.

b. We know IHD =(2C+2+N-H-X)/2, where, C=#carbons, H=#hydrogens, N=# nitrogens, X=# halogens. So from this formula,we can get IHD=(10+2+1-7)/2 =3.

c. In IR spectrum,peak near 3000 cm-1 is for Sp3 C-H,peak near 2240 cm-1 suggests cyanide group (-CN) present,peak near 1770 cm-1 suggests ester group present. So the compound structure is CH3-CH2-O-(CO)-CH2-CN( Ethyl cyano acetate). IHD also matches, i e; 3.

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