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 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?

Given that the sample contains 8.80% of Nitrogen by mass. We can use this information to find the moles of Nitrogen. Moles of Nitrogen = (8.80% of 1.20 g) / (Molecular weight of N) = (0.088*1.20 g) / 14.01 g/mol = 0.007556 mol

Since each acrylonitrile molecule \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) contains one Nitrogen atom, the moles of acrylonitrile are equal to the moles of Nitrogen. Moles of Acrylonitrile = Moles of Nitrogen = 0.007556 mol

We are given that Bromine reacts \(1:1\) (by moles) with butadiene molecules in the polymer. Therefore, the moles of butadiene can be found from the moles of Bromine that reacted with the butadiene. Moles of Bromine = 0.605 g / (Molecular weight of Br2) = 0.605 g / 159.80 g/mol = 0.003787 mol Since Bromine reacts with Butadiene in a 1:1 ratio, Moles of Butadiene = Moles of Bromine = 0.003787 mol

We have calculated the moles of Acrylonitrile and Butadiene in the ABS sample. Now, we can find their mass using their molecular weights. Mass of Acrylonitrile = Moles of Acrylonitrile * (Molecular weight of Acrylonitrile) = 0.007556 mol * 53.06 g/mol = 0.40096 g Mass of Butadiene = Moles of Butadiene * (Molecular weight of Butadiene) = 0.003787 mol * 54.09 g/mol = 0.20487 g

Percent by mass of a component is given by the ratio of its mass to the total mass of the sample and multiplied by 100. Percent by mass of Acrylonitrile = (Mass of Acrylonitrile / Total mass of ABS sample) * 100 = (0.40096 g / 1.20 g) * 100 = 33.42 % Percent by mass of Butadiene = (Mass of Butadiene / Total mass of ABS sample) * 100 = (0.20487g / 1.20 g) * 100 = 17.07 %

Since the total mass of the ABS sample is given, we can first find the mass of Styrene in the sample and then calculate its moles. Mass of Styrene = Total mass of ABS sample - Mass of Acrylonitrile - Mass of Butadiene = 1.20 g - 0.40096 g - 0.20487 g = 0.59417 g Moles of Styrene = Mass of Styrene / (Molecular weight of Styrene) = 0.59417g / 104.15 g/mol = 0.005707 mol

We will divide the moles of each monomer unit (Acrylonitrile, Butadiene, and Styrene) by the smallest moles value to get their relative numbers. Relative numbers of Acrylonitrile, Butadiene, and Styrene are given by: Relative number of Acrylonitrile = Moles of Acrylonitrile / Smallest moles value = 0.007556 mol / 0.003787 mol = 1.995 ≈ 2 Relative number of Butadiene = Moles of Butadiene / Smallest moles value = 0.003787 mol / 0.003787 mol = 1 Relative number of Styrene = Moles of Styrene / Smallest moles value = 0.005707 mol / 0.003787 mol = 1.506 ≈ 1.5 Thus, the relative numbers of Acrylonitrile, Butadiene, and Styrene in the ABS polymer are approximately 2:1:1.5. Answer: a. Percent by mass of Acrylonitrile in the ABS polymer is 33.42%, and Percent by mass of Butadiene is 17.07%. b. The relative numbers of the monomer units in this polymer are approximately 2:1:1.5 for Acrylonitrile, Butadiene, and Styrene, respectively.