Answer:
(a) [tex]C_5H_8O_2[/tex]
(b) [tex]CHCl[/tex]
(c) [tex]CH_2[/tex]
(d) [tex]CH[/tex]
(e) [tex]C_3H_3N[/tex]
Explanation:
Hello,
(a) For the lucite, one computes the moles of C, H and O that are present:
[tex]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\\[/tex]
Now, dividing each moles by the smallest moles (oxygen's moles), one obtains:
[tex]C=\frac{0.05}{0.02} =2.5;H=\frac{0.0806}{0.02} =4;O=\frac{0.02}{0.02} =1[/tex]
Finally, we look for the smallest whole number subscript by multiplying by 2, so the empirical formula turns out into:
[tex]C_5H_8O_2[/tex]
(b) For the Saran, one computes the moles of C, H and Cl that are present:
[tex]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\\[/tex]
Now, dividing each moles by the smallest moles (hydrogen's moles), one obtains:
[tex]C=\frac{0.021}{0.02} =1;H=\frac{0.02}{0.02} =1;Cl=\frac{0.021}{0.02} =1[/tex]
Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:
[tex]CHCl[/tex]
(c) For the polyethylene, one computes the moles of C and H that are present:
[tex]n_C=0.86*\frac{1molC}{12gC}=0.072molC\\n_H=0.14gH*\frac{1molH}{1gH}=0.14molH[/tex]
Now, dividing each moles by the smallest moles (carbon's moles), one obtains:
[tex]C=\frac{0.072}{0.072} =1;H=\frac{0.14}{0.072} =2[/tex]
Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:
[tex]CH_2[/tex]
(d) For the polystyrene, one computes the moles of C and H that are present:
[tex]n_C=0.923*\frac{1molC}{12gC}=0.077molC\\n_H=0.077gH*\frac{1molH}{1gH}=0.077molH[/tex]
Now, dividing each moles by the smallest moles (either carbon's or hydrogen's moles), one obtains:
[tex]C=\frac{0.077}{0.077} =1;H=\frac{0.077}{0.077} =1[/tex]
Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:
[tex]CH[/tex]
(e) For the orlon, one computes the moles of C, H and N that are present:
[tex]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[/tex]
Now, dividing each moles by the smallest moles (nitrogen's moles), one obtains:
[tex]C=\frac{0.057}{0.019} =3;H=\frac{0.057}{0.019} =3;N=\frac{0.019}{0.019} =1[/tex]
Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:
[tex]C_3H_3N[/tex]
Best regards.
