Question #9b11e

For part (a), use the molar mass of pyridine to calculate how many moles you have in that sample

`0.068 color(red)(cancel(color(black)("g"))) * ("1 mole C"_5"H"_5"N")/(79.10 color(red)(cancel(color(black)("g")))) = 8.597 * 10^(-4)"moles C"_5"H"_5"N"`

As you know, the number of particles needed to have one mole is given by Avogadro's number

`color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"particles"color(white)(a/a)|)))`

Your sample of pyridine will thus contain

`8.597 * 10^(-4) color(red)(cancel(color(black)("moles C"_5"H"_5"N"))) * (6.022 * 10^(23)"molec C"_5"H"_5"N")/(1color(red)(cancel(color(black)("mole C"_5"H"_5"N"))))`

`= color(green)(|bar(ul(color(white)(a/a)color(black)(5.2 * 10^(20)"molec CH"_5"H"_5"N")color(white)(a/a)|)))`

For part (b), use the same approach to calculate the number of formula units present in your sample of zinc oxide.

You will have

`5.0 color(red)(cancel(color(black)("g"))) * "1 mole ZnO"/(81.41color(red)(cancel(color(black)("g")))) = 6.142 * 10^(-2)"moles ZnO"`

and

`6.142 * 10^(-2) color(red)(cancel(color(black)("moles ZnO"))) * (6.022 * 10^(23)"f. units ZnO")/(1color(red)(cancel(color(black)("mole ZnO"))))`

`= 3.7 * 10^(22)"f. units ZnO"`

The ratio between the number of molecules of pyridine and the number of formula units of zinc oxide will thus be

`(5.2 * 10^(20))/(3.7 * 10^(22)) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.014)color(white)(a/a)|)))`

`color(white)(a)`

SIDE NOTE Notice that you can also calculate this ratio by dividing the number of moles of each compound

`(8.597 * 10^(-4) color(red)(cancel(color(black)("moles C"_5"H"_5"N"))))/(6.142 * 10^(-2)color(red)(cancel(color(black)("moles ZnO")))) = 0.014`

That is the case because a mole is simply a fixed number of molecules / formula units.

`color(white)(a)`

For part (c), you know that zinc oxide has a surface area per gram equal to `"42 m"^2"g"^(-1)`, which basically means that you can distribute `"1 g"` of zinc oxide over a surface area of `"42 m"^2`.

Now, calculate how many molecules of pyridine are adsorbed per gram of zinc oxide by using the fact that you know how many molecules are adsorbed by `"5.0 g"` of zinc oxide

`1 color(red)(cancel(color(black)("g ZnO"))) * (5.2 * 10^(20)"molec. C"_5"H"_5"N")/(5.0 color(red)(cancel(color(black)("g ZnO"))))`

`=1.04 * 10^(20)"molec C"_5"H"_5"N"`

So, `"1 g"` of zinc oxide adsorbs `1.04 * 10^(20)` molecules of pyridine, it follows that the surface area per molecule of pyridine will be

`48"m"^2 / color(red)(cancel(color(black)("1 g ZnO"))) * color(red)(cancel(color(black)("1 g ZnO")))/(1.04 * 10^(20)"molec. C"_5"H"_5"N")`

`= color(green)(|bar(ul(color(white)(a/a)color(black)(4.6 * 10^(-19)"m"^2" / molec C"_5"H"_5"N")color(white)(a/a)|)))`

The answers are rounded to two sig figs.

SIDE NOTE If you want, you can convert this value from square meters per molecule to something like square nanometers per molecule

`4.6 * 10^(-19) color(red)(cancel(color(black)("m"^2)))/("1 molec. C"_5"H"_5"N") * (10^9 * 10^9"nm"^2)/(1color(red)(cancel(color(black)("m"^2))))`

`="0.46 nm"^2" / molec. C"_5"H"_5"N"`