Polymers are large molecules composed of simple units repeated many times. they often have relatively simple
empirical formulas. calculate the empirical formulas of the following polymers.
(a)lucite (p; exigals); 59.9% c, 8.06% h, 32.0% o
(b)saran; 24.8% c, 2.00% h, 73.1% cl
(c)plolyethylene; 86 % c, 14%h
(d)polystyrene; 92.3% c, 7.7 % h
(e)orlon; 67.9% c, 5.70% h, 26.4% n

Empirical formula:

(a) Lucite: C₅H₈O₂

(b) Saran: CHCl

(c) Polyehtylene: CH₂

(d) Polystyrene: CH

(e) Orlon: C₃H₃N

Explanation:

You must divide this percentages of mass in the molecular weight of each atom to see how many moles do you have

(a) Lucite:

Carbon:

59,9% × (1mol C/ 12 g) = 5 mol

Hydrogen:

8,06% × (1mol H/ 1g) = 8 mol

Oxygen:

32% × (1mol O/ 16g) = 2 mol

Thus, empirical formula is:

C₅H₈O₂

(b) Saran:

Carbon:

24,8% × (1mol C/ 12 g) = 2 mol

Hydrogen:

2,00% × (1mol H/ 1g) = 2 mol

Chlorine:

73,1% × (1mol Cl/ 35,45g) = 2 mol

Empirical formula is the simplest way to express a formula, dividing each mole in two, the empirical formula is:

CHCl

(c) Polyehtylene:

Carbon:

86% × (1mol C/ 12 g) = 7 mol

Hydrogen:

14% × (1mol H/ 1g) = 14 mol

dividing in the least common multiple (7), the empirical formula obtained is:

CH₂

(d) Polystyrene:

Carbon:

92,3% × (1mol C/ 12 g) = 7,7 mol

Hydrogen:

7,7% × (1mol H/ 1g) = 7,7 mol

Dividing in 7,7 each mole the obtained formula is:

CH

(e) Orlon:

Carbon:

67,9% × (1mol C/ 12 g) = 5,7 mol

Hydrogen:

5,70% × (1mol H/ 1g) = 5,7 mol

Nitrogen:

26,4% × (1mol N/ 14g) = 1,9 mol

Dividing each number in 1,9 the obtained empirical formula is:

C₃H₃N

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