# Question #9b11e

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

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)|)))#

**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.*

For part **(c)**, you know that zinc oxide has a *surface area per gram* equal to

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

#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, **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"#