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ABS plastic is a tough, hard plastic used in applications requiring shock resistance. The polymer consists of three monomer units: acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\right)\), butadiene \(\left(\mathrm{C}_{4} \mathrm{H}_{6}\right)\), and styrene \(\left(\mathrm{C}_{8} \mathrm{H}_{8}\right)\). a. A sample of \(\mathrm{ABS}\) plastic contains \(8.80 \% \mathrm{~N}\) by mass. It took \(0.605 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) to react completely with a \(1.20-\mathrm{g}\) sample of \(\mathrm{ABS}\) plastic. Bromine reacts \(1: 1\) (by moles) with the butadiene molecules in the polymer and nothing else. What is the percent by mass of acrylonitrile and butadiene in this polymer? b. What are the relative numbers of each of the monomer units in this polymer?

Short Answer

Expert verified
The percent by mass of acrylonitrile and butadiene in the ABS polymer sample can be calculated as follows: 1. Moles of Nitrogen = 0.00757 mol 2. Moles of acrylonitrile = 0.00757 mol 3. Mass of acrylonitrile = 0.401 g 4. Moles of Br₂ reacted = 0.00379 mol 5. Mass of butadiene = 0.205 g 6. Mass percentage of acrylonitrile = 33.4% 7. Mass percentage of butadiene = 17.1% For the relative numbers of each monomer unit: 1. Mass of styrene = 0.594 g 2. Moles of styrene = 0.00572 mol 3. The relative numbers of each monomer unit are: Acrylonitrile : Butadiene : Styrene = 1 : 1 : 1.33 So, the ABS polymer consists of 33.4% acrylonitrile, 17.1% butadiene, and the relative numbers of monomer units are in the ratio 1:1:1.33.

Step by step solution

01

Part (a): Find the mass percentages of acrylonitrile and butadiene

Step 1: Calculate the moles of Nitrogen (N) in ABS polymer sample Given that the mass of the ABS polymer is 1.20 g and it contains 8.80% Nitrogen by mass. Mass of Nitrogen in the sample = (8.80 / 100) × 1.20 g Molar mass of Nitrogen = 14 g/mol Now, we can find the moles of Nitrogen in the sample: Moles of Nitrogen = Mass of Nitrogen / Molar mass of Nitrogen Step 2: Calculate the moles of acrylonitrile Each acrylonitrile molecule contains 1 nitrogen atom, so the moles of acrylonitrile in ABS sample are equal to the moles of nitrogen atoms. Step 3: Calculate mass of acrylonitrile Molar mass of acrylonitrile, C3H3N, = (3×12) + (3×1) + 14 = 53 g/mol Mass of acrylonitrile = moles of acrylonitrile × molar mass of acrylonitrile Step 4: Calculate the moles of Bromine (Br2) that reacted with butadiene 1 mole of Bromine reacts with 1 mole of butadiene. Moles of Br2 reacted = mass of Br2 / molar mass of Br2 = 0.605 g / (2 × 80 g/mol) Step 5: Calculate the mass of butadiene in ABS polymer Molar mass of butadiene, C4H6, = (4 × 12) + (6 × 1) = 54 g/mol Mass of butadiene = moles of Br2 reacted × molar mass of butadiene Step 6: Calculate mass percentages of acrylonitrile and butadiene Mass percentage of acrylonitrile = (Mass of acrylonitrile / mass of ABS polymer) × 100% Mass percentage of butadiene = (Mass of butadiene / mass of ABS polymer) × 100%
02

Part (b): Find the relative numbers of each monomer unit

Step 1: Calculate the moles of styrene We know the total mass of ABS polymer is 1.20 g. So, to find the mass of styrene, subtract the masses of acrylonitrile and butadiene from the total mass: Mass of styrene = Mass of ABS polymer - (Mass of acrylonitrile + Mass of butadiene) Step 2: Calculate the moles of styrene Molar mass of styrene, C8H8, = (8 × 12) + (8 × 1) = 104 g/mol Moles of styrene = Mass of styrene / molar mass of styrene Step 3: Find the relative numbers of monomer units To find the relative numbers of each monomer unit, we should find the smallest molar ratio between the moles of acrylonitrile, butadiene, and styrene. Divide the moles of each monomer by the smallest value among the three. The resulting values will represent the relative numbers of each of the monomer units in the ABS polymer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polymer Chemistry
Polymers are large molecules composed of repeating structural units called monomers, which are bonded together in long chains. In the field of polymer chemistry, the synthesis, structure, and properties of these materials are studied extensively. A toolbox of reactions allows chemists to create a vast array of polymers from various monomers, each with unique characteristics suitable for different applications.

Take ABS plastic, for example, it's a widely-used polymer made from three monomers: acrylonitrile, butadiene, and styrene. Each monomer contributes to the overall properties of the plastic; acrylonitrile offers chemical and thermal stability, butadiene provides toughness and impact resistance, while styrene adds rigidity and ease of processing. By changing the proportions of these monomers, manufacturers can tweak the material properties of ABS to suit specific needs.
Monomer Units Calculation
The art of polymer chemistry involves not only knowing what monomers make up a polymer but also understanding the proportions and arrangement of these monomers within the polymer chain. Calculating these units requires a solid grasp of stoichiometry and the ability to relate mole ratios to mass ratios.

In our ABS plastic example, the percentage of nitrogen by mass can be directly tied to the amount of acrylonitrile since this monomer is the only one of the three that contains nitrogen. By calculating the moles of nitrogen in the sample, we can determine the moles -- and subsequently the mass -- of acrylonitrile in the polymer. This understanding is critical for tailoring the properties of ABS by altering the monomer ratios.
Percentage Mass Composition
Percentage mass composition is a fundamental concept in chemistry that provides insight into the quantitative aspects of chemical substances. When examining polymers, understanding the mass percentage of each constituent monomer is important for predicting the material's properties and behavior. The percentage mass composition of a polymer is calculated by dividing the mass of a particular monomer by the total mass of the polymer, then multiplying by 100%.

For example, in calculating the mass percentages of acrylonitrile and butadiene for ABS plastic, we carefully considered how bromine reacts exclusively with butadiene. By determining the moles and hence mass of butadiene that reacted with a known mass of bromine, we could calculate its mass percentage. The mass percentage of acrylonitrile was similarly found. These calculations are essential for quality control and refinement of polymer production processes.

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Most popular questions from this chapter

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 g sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(\mathrm{g}) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

A compound contains \(47.08 \%\) carbon, \(6.59 \%\) hydrogen, and \(46.33 \%\) chlorine by mass; the molar mass of the compound is \(153 \mathrm{~g} / \mathrm{mol} .\) What are the empirical and molecular formulas of the compound?

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Consider the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ Identify the limiting reagent in each of the reaction mixtures given below: a. 50 molecules of \(\mathrm{H}_{2}\) and 25 molecules of \(\mathrm{O}_{2}\) b. 100 molecules of \(\mathrm{H}_{2}\) and 40 molecules of \(\mathrm{O}_{2}\) c. 100 molecules of \(\mathrm{H}_{2}\) and 100 molecules of \(\mathrm{O}_{2}\) d. \(0.50 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}\). e. \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}_{2}\) f. \(1.0 \mathrm{~g} \mathrm{H}_{2}\) and \(0.25 \mathrm{~mol} \mathrm{O}_{2}\) g. \(5.00 \mathrm{~g} \mathrm{H}_{2}\) and \(56.00 \mathrm{~g} \mathrm{O}_{2}\)

Chloral hydrate \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\right)\) is a drug formerly used as a sedative and hypnotic. It is the compound used to make "Mickey Finns" in detective stories. a. Calculate the molar mass of chloral hydrate. b. What amount (moles) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\) molecules are in \(500.0 \mathrm{~g}\) chloral hydrate? c. What is the mass in grams of \(2.0 \times 10^{-2}\) mol chloral hydrate? d. What number of chlorine atoms are in \(5.0 \mathrm{~g}\) chloral hydrate? e. What mass of chloral hydrate would contain \(1.0 \mathrm{~g} \mathrm{Cl} ?\) \(\mathbf{f}\). What is the mass of exactly 500 molecules of chloral hydrate?

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