Question

Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers: (a) Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O (b) Saran; 24.8% C, 2.0% H, 73.1% Cl (c) polyethylene; 86% C, 14% H (d) polystyrene; 92.3% C, 7.7% H (e) Orlon; 67.9% C, 5.70% H, 26.4% N

Answer 1

Answer:

(a) $C_5H_8O_2$

(b) $CHCl$

(c) $CH_2$

(d) $CH$

(e) $C_3H_3N$

Explanation:

Hello,

(a) For the lucite, one computes the moles of C, H and O that are present:

$n_C=0.599gC*\frac{1molC}{12gC}=0.05molC\\n_H=0.0806gH*\frac{1molH}{1gH}=0.0806molH\\n_O=0.32gO*\frac{1molO}{16gO}=0.02molO$

Now, dividing each moles by the smallest moles (oxygen's moles), one obtains:

$C=\frac{0.05}{0.02} =2.5;H=\frac{0.0806}{0.02} =4;O=\frac{0.02}{0.02} =1$

Finally, we look for the smallest whole number subscript by multiplying by 2, so the empirical formula turns out into:

$C_5H_8O_2$

(b) For the Saran, one computes the moles of C, H and Cl that are present:

$n_C=0.248gC*\frac{1molC}{12gC}=0.021molC\\n_H=0.02gH*\frac{1molH}{1gH}=0.02molH\\n_{Cl}=0.731gCl*\frac{1molCl}{35.45gCl}=0.021molCl$

Now, dividing each moles by the smallest moles (hydrogen's moles), one obtains:

$C=\frac{0.021}{0.02} =1;H=\frac{0.02}{0.02} =1;Cl=\frac{0.021}{0.02} =1$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$CHCl$

(c) For the polyethylene, one computes the moles of C and H that are present:

$n_C=0.86*\frac{1molC}{12gC}=0.072molC\\n_H=0.14gH*\frac{1molH}{1gH}=0.14molH$

Now, dividing each moles by the smallest moles (carbon's moles), one obtains:

$C=\frac{0.072}{0.072} =1;H=\frac{0.14}{0.072} =2$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$CH_2$

(d) For the polystyrene, one computes the moles of C and H that are present:

$n_C=0.923*\frac{1molC}{12gC}=0.077molC\\n_H=0.077gH*\frac{1molH}{1gH}=0.077molH$

Now, dividing each moles by the smallest moles (either carbon's or hydrogen's moles), one obtains:

$C=\frac{0.077}{0.077} =1;H=\frac{0.077}{0.077} =1$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$CH$

(e) For the orlon, one computes the moles of C, H and N that are present:

$n_C=0.679*\frac{1molC}{12gC}=0.057molC\\n_H=0.057gH*\frac{1molH}{1gH}=0.057molH\\n_N=0.264gN*\frac{1molN}{14gN}=0.019molN$

Now, dividing each moles by the smallest moles (nitrogen's moles), one obtains:

$C=\frac{0.057}{0.019} =3;H=\frac{0.057}{0.019} =3;N=\frac{0.019}{0.019} =1$

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

$C_3H_3N$

Best regards.