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4.5 Quantitative chemical analysis  (Page 3/14)

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A photo is shown of a flask and funnel used for filtration. The flask contains a slightly opaque liquid filtrate with a slight yellow tint. A funnel, which contains a bright yellow and orange material, sits atop the flask. The flask is held in place by a clamp and is connected to a vacuum line. The connection between the funnel and flask is sealed with a rubber bung or gasket.
Precipitate may be removed from a reaction mixture by filtration.

Gravimetric analysis

A 0.4550-g solid mixture containing MgSO 4 is dissolved in water and treated with an excess of Ba(NO 3 ) 2 , resulting in the precipitation of 0.6168 g of BaSO 4 .

MgSO 4 ( a q ) + Ba ( NO 3 ) 2 ( a q ) ⟶ BaSO 4 ( s ) + Mg ( NO 3 ) 2 ( a q )

What is the concentration (percent) of MgSO 4 in the mixture?

Solution

The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of BaSO 4 and MgSO 4 through their stoichiometric factor. Once the mass of MgSO 4 is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.

This figure shows five rectangles. The first is shaded yellow and is labeled “Mass of B a S O subscript 4.” This rectangle is followed by an arrow pointing right to a second rectangle. The arrow is labeled, “Molar mass.” The second rectangle is shaded pink and is labeled, “Moles of B a S O subscript 4.” This rectangle is followed by an arrow pointing right to a third rectangle. The arrow is labeled, “Stoichiometric factor.” This third rectangle is shaded pink and is labeled, “Moles of M g S O subscript 4.” This rectangle is followed by an arrow labeled, “Molar mass,” which points downward to a fourth rectangle. This fourth rectangle is shaded yellow and is labeled, “Mass of M g S O subscript 4.” This rectangle is followed by an arrow labeled, “Sample mass,” which points left to a fifth rectangle. This fifth rectangle is shaded lavender and is labeled, “Percent M g S O subscript 4.

The mass of MgSO 4 that would yield the provided precipitate mass is

0.6168 g BaSO 4 × 1 mol BaSO 4 233.43 g BaSO 4 × 1 mol MgSO 4 1 mol BaSO 4 × 120.37 g MgSO 4 1 mol MgSO 4 = 0.3181 g MgSO 4

The concentration of MgSO 4 in the sample mixture is then calculated to be

percent MgSO 4 = mass MgSO 4 mass sample × 100 % 0.3181 g 0.4550 g × 100 % = 69.91 %

Check your learning

What is the percent of chloride ion in a sample if 1.1324 g of the sample produces 1.0881 g of AgCl when treated with excess Ag + ?

Ag + ( a q ) + Cl − ( a q ) ⟶ AgCl ( s )

Answer:

23.76%

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The elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis    . In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product ( [link] ). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.

This diagram shows an arrow pointing from O subscript 2 into a tube that leads into a vessel containing a red material, labeled “Sample.” This vessel is inside a blue container with a red inner lining which is labeled “Furnace.” An arrow points from the tube to the right into the vessel above the red sample material. An arrow leads out of this vessel through a tube into a second vessel outside the furnace. An line points from this tube to a label above the diagram that reads “C O subscript 2, H subscript 2 O, O subscript 2, and other gases.” Many small green spheres are visible in the second vessel which is labeled below, “H subscript 2 O absorber such as M g ( C l O subscript 4 ) subscript 2.” An arrow points to the right through the vessel, and another arrow points right heading out of the vessel through a tube into a third vessel. The third vessel contains many small blue spheres. It is labeled “C O subscript 2 absorber such as N a O H.” An arrow points right through this vessel, and a final arrow points out of a tube at the right end of the vessel. Outside the end of this tube at the end of the arrow is the label, “O subscript 2 and other gases.”
This schematic diagram illustrates the basic components of a combustion analysis device for determining the carbon and hydrogen content of a sample.

Combustion analysis

Polyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a 0.00126-g sample of polyethylene yields 0.00394 g of CO 2 and 0.00161 g of H 2 O. What is the empirical formula of polyethylene?

Solution

The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:

C x H y ( s ) + excess O 2 ( g ) ⟶ x CO 2 ( g ) + y 2 H 2 O ( g )

Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x and y are needed.

First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:

This figure shows two flowcharts. The first row is a single flow chart. In this row, a rectangle at the left is shaded yellow and is labeled, “Mass of C O subscript 2.” This rectangle is followed by an arrow pointing right to a second rectangle. The arrow is labeled, “Molar mass.” The second rectangle is shaded pink and is labeled, “Moles of C O subscript 2.” This rectangle is followed by an arrow pointing right to a third rectangle. The arrow is labeled, “Stoichiometric factor.” The third rectangle is shaded pink and is labeled, “Moles of C.” This rectangle is followed by an arrow labeled “Molar mass” which points right to a fourth rectangle. The fourth rectangle is shaded yellow and is labeled “Mass of C.” Below, is a second flowchart. It begins with a yellow shaded rectangle on the left which is labeled, “Mass of H subscript 2 O.” This rectangle is followed by an arrow labeled, “Molar mass,” which points right to a second rectangle. The second rectangle is shaded pink and is labeled, “Moles of H subscript 2 O.” This rectangle is followed by an arrow pointing right to a third rectangle. The arrow is labeled, “Stoichiometric factor.” The third rectangle is shaded pink and is labeled “Moles of H.” This rectangle is followed to the right by an arrow labeled, “Molar mass,” which points to a fourth rectangle. The fourth rectangle is shaded yellow and is labeled “Mass of H.” An arrow labeled, “Sample mass” points down beneath this rectangle to a green shaded rectangle. This rectangle is labeled, “Percent composition.” An arrow extends beneath the pink rectangle labeled, “Moles of H,” to a green shaded rectangle labeled, “C to H mole ratio.” Beneath this rectangle, an arrow extends to a second green shaded rectangle which is labeled, “Empirical formula.”
mol C = 0.00394 g CO 2 × 1 mol CO 2 44.01 g/mol × 1 mol C 1 mol CO 2 = 8.95 × 10 − 5 mol C mol H = 0.00161 g H 2 O × 1 mol H 2 O 18.02 g/mol × 2 mol H 1 mol H 2 O = 1.79 × 10 − 4 mol H

The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is

mol H mol C = 1.79 × 10 −4 mol H 8.95 × 10 −5 mol C = 2 mol H 1 mol C

and the empirical formula for polyethylene is CH 2 .

Check your learning

A 0.00215-g sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of CO 2 and 0.00148 g of H 2 O in a combustion analysis. What is the empirical formula for polystyrene?

Answer:

CH

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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