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vitfil [10]
2 years ago
15

Perform an online search to find the structures of the compounds in the questions that follow. Provide a method for separating t

he mixtures of two or three compounds, dissolved in a solvent diethyl ether. In each case one of the components will be a neutral compound. You should give your answer in the form of a flow chart.
a) Benzophenone and tributylamine
b) 4-Bromoaniline, 3-nitrobenzoic acid, and benzoin
c) Fluorenone, octanoic acid, and dicyclohexylamine
d) 1-Hexanol and 4-bromoaniline
Chemistry
1 answer:
Vedmedyk [2.9K]2 years ago
5 0

The chemical separation method for two of the above named mixtures or compound indicated above are explained below.

<h3>What is a Chemical compound?</h3>

A chemical compound is a material generated by the chemical union of two or more elements in set quantities.

<h3>What are the separation methods?</h3>

<u>Benzophenone and tributylamine</u>

The combination of benzophenone and tributylamine is in diethyl ether (given in question). Tributylamine is a base that may readily be changed to a water soluble salt by adding an aqueous acid.

Add dilute aqueous acid to the solution. The amine salt will be present in the aqueous layer, whereas benzophenone will be present in the ether layer. A separatory funnel will isolate levels.

<u>Fluorenone, octanoic acid, and dicyclohexylamine</u>

Fluorenone, Octanoic acid, and dicyclohexyl amine: The combination comprises both an acid (octanoic acid) and a base (dicyclohexyl amine) (dicyclohexyl amine).

Add the diluted aqueous base to the solution (like NaCO3). Divide the two layers.

The acid is present in the aqueous layer as a salt, while the fluorenone and amine are present in the ether layer.

Add weak aqueous acid to this combination of fluorenone and dicyclohexyl amine.

Divide the two layers. The amine is present in the aqueous layer, and fluorenone is present in the ether.

Learn more about chemical separation at;
brainly.com/question/17143033
#SPJ1

<u></u>

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where,

n= moles of solute =\frac{\text{Given mass}}{\text{Molar mass}}=\frac{10.0g}{162.2g/mol}=0.0616moles

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