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

(a)
`C_5H_8O_2`

(b)
`CHCl`

(c)
`CH_2`

(d)
`CH`

(e)
`C_3H_3N`

Step-by-step explanation:

Hello,

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

`n_C=0.599gC*(1molC)/(12gC)=0.05molC\\n_H=0.0806gH*(1molH)/(1gH)=0.0806molH\\n_O=0.32gO*(1molO)/(16gO)=0.02molO\\`

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

`C=(0.05)/(0.02) =2.5;H=(0.0806)/(0.02) =4;O=(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*(1molC)/(12gC)=0.021molC\\n_H=0.02gH*(1molH)/(1gH)=0.02molH\\n_(Cl)=0.731gCl*(1molCl)/(35.45gCl)=0.021molCl\\`

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

`C=(0.021)/(0.02) =1;H=(0.02)/(0.02) =1;Cl=(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*(1molC)/(12gC)=0.072molC\\n_H=0.14gH*(1molH)/(1gH)=0.14molH`

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

`C=(0.072)/(0.072) =1;H=(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*(1molC)/(12gC)=0.077molC\\n_H=0.077gH*(1molH)/(1gH)=0.077molH`

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

`C=(0.077)/(0.077) =1;H=(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*(1molC)/(12gC)=0.057molC\\n_H=0.057gH*(1molH)/(1gH)=0.057molH\\n_N=0.264gN*(1molN)/(14gN)=0.019molN`

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

`C=(0.057)/(0.019) =3;H=(0.057)/(0.019) =3;N=(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.