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Sladkaya [172]
3 years ago
14

You plan to separate a polar and a non-polar compound using normal phase column chromatography technique. Methylene chloride (CH

2Cl2) and hexanes (C6H14)are the solvents to be used to elute the column. Which solvent should be used first
Chemistry
1 answer:
marissa [1.9K]3 years ago
3 0

Answer:

Hexane should be used first.

Explanation:

Chromatography is a method of separating the constituents of a mixture by taking advantage of their different rates of movement in a solvent over an adsorbent medium. Chromatography is a means of separation and analysis that utilises fractional separation. It is based on the principle  that if a fluid containing a number of substances is allowed to pass though an adsorbent medium, the different substances in the fluid may travel at different rates and be separated.

The rate of movement depends on the relative affinities of the constituents for the solvent and adsorbent medium. i.e solutes which are weakly adsorbed by the adsorbent medium are easily redissolved by the ascending solvent and quickly travel up the adsorbent medium. In addition to that , solutes which are very soluble in the solvent move up at a faster rate than those which  are not soluble.

In column Chromatography;

A non-polar solvent should be initiated and applied first. This is because , in a column chromatography, a non-polar compound will be removed at first then later polar compound.

Assuming a polar compound is used first, the polar compounds will be removed alongside with all the non-polar compounds.

From the two Compounds given;

We know that :

Hexane is a non-polar compound and Methylene chloride  is a polar compound. As such, Hexane should be used first.

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Gas A is Carbon Dioxide
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Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n
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The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide (CO_2), nitrogen dioxide (NO_2), and no other substances. A small sample gives 2.389 g CaO, 1.876 g CO_2, and 3.921 g NO_2 Determine the empirical formula of the compound.

<u>Answer:</u> The empirical formula for the given compound is CaCN_2

<u>Explanation:</u>

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of CO_2=1.876g

Mass of NO_2=3.921g

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, \frac{12}{44}\times 1.876=0.5116g of carbon will be contained.

<u>For calculating the mass of nitrogen:</u>

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, \frac{14}{46}\times 3.921=1.193g of nitrogen will be contained.

<u>For calculating the mass of calcium:</u>

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, \frac{40}{56}\times 2.389=1.706g of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Calcium =\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles

Moles of Nitrogen = \frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = \frac{0.0426}{0.0426}=1

For Carbon = \frac{0.0426}{0.0426}=1

For Nitrogen = \frac{0.0852}{0.0426}=2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is CaCN_2

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Explanation:

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