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N76 [4]
3 years ago
9

Consider the following four pairs of molecules. You may use 1H-NMR, 13C-NMR, or IRspectroscopyto differentiate the structures, b

ut you may use a technique only once on this page(i.e. each pair mustuse a different technique).For each pair, pick one of the above forms of spectroscopy and describe the single clearest difference between the two compounds by that analytical technique.

Chemistry
1 answer:
GREYUIT [131]3 years ago
4 0

Answer:

The question is incomplete. The structures were not added to the question. Find attached of the structure and the given answer.

Explanation:

See the attached file for explanation

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An unknown liquid is composed of 34.31% c, 5.28% h, and 60.41% i. The molecular weight is 210.06 amu. What is the molecular form
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In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.

Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.

To calculate molecular formula of compound, convert mass into number of moles as follows:

n=\frac{m}{M}

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.

Taking the ratio:

C:H:I=n_{C}:n_{H}:n_{I}

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C:H:I=\frac{34.31 g}{12 g/mol}:\frac{5.28 g}{1 g/mol}:\frac{60.41 g}{126.90 g/mol}=6:11:1

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