Draw two repeating units of ABS assuming the three monomer units react in a 1:1:

Question

Draw two repeating units of ABS assuming the three monomer units react in a 1:1:1 mole ratio and reacts in the same order as listed above. The reaction simple addition polymerization. A sample of ABS contains 8.80% N by mass. It took 0.605 g of Br_2 to react completely with a 1.20 g sample of ABS. What is the percent by mass acrylonitrile, butadiene and styrene in this sample? ABS does not consist of a 1:1:1 mole ratio between the three monomer units. Using the results from part b, determine the relative numbers of monomer units in this sample of ABS.

Explanation / Answer

b) Acrylonitrile is C3H3N

Butadiene is C4H6

Styrene is C8H8

We have to determine the mass percent of these three in the ABS Sample

It may be assumed that the total weight of polymer is 100 gm and the mass of N is 8.80 gm

The molecular weight of Acrylonitrile or C3H3N is 53 g/mol, therefore the relative mass of the monomeric unit to N can be calculated as:

53/14 = 3.8

(We used 14 because it is the relative atomic mass of N)

Now, to find the mass or weight percent of Acrylonitrile in the ABS polymer, we will:

3.8 x 8.8 = 33.3 gm or it can also be written as, 33.3% w/w of Acrylonitrile or C3H3N

Similarly, the mass percent of Butadiene or C4H6 can be determined:

Since 0.605 gm of Br2 reacted with 1.20 gm of ABS Sample, therefore,

0.605 gm/159.8 g/mol = 0.0038 moles of Br2

(Here, 159.8 g/mol is the molecular weight of Br2)

Thus, the same number of moles must have been used by Butadience or C4H6

Hence, the share of Butadiene or C4H6 will be:

0.38 mol x 54 g/mol = 0.205 gm of C4H6

Now, 0.205 gm of C4H6 is present in 1.20 gms of ABS sample

The mass percent will be:

(0.205gm/ 1.20 gm) x 100% = 17.1% w/w Butadiene or C4H6

Next, the mass or weight percent of Styrene or C8H8 = 100 - (33.3 + 17.1) = 49.6% w/w Styrene or C8H8

c) We have already assumed that let the total weight of polymer be equal to 100 gm

Thus, the total moles in Acrylonitrile or C3H3N = 33.3 gm/ 53 g/mol = 0.63 moles

(53 gm/mol is the molecular weight of Acrylonitrile)

  for Butadiene or C4H6 = 17.1gm/ 54 gm/mol = 0.32 moles

   (54 gm/mol is the molecular weight of Butadiene)

for Styrene or C8H8 = 49.6 gm / 104 gm/mol = 0.48 moles

(104 gm/mol is the molecular weight of Styrene)

Since, 1:1:1 is not a ratio between the monomeric units, therefore, we will divide all the results of the total number of moles with the lowest value, which among the three is 0.32, hence

0.63/0.32 = 2 (Acrylonitrile or C3H3N)

0.32/0.32 = 1 (Butadiene or C4H6)

0.48/0.32 = 1.5 (Styrene or C8H8)

Thus, the ratio will become 2:1:1.5 and the round figures will provide the answer

4:2:3 (Acrylonitrile: Butadiene: Styrene)

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