# Question 1ef73

May 27, 2017

$1.5 \times {10}^{22} \text{atoms Au}$

#### Explanation:

What we'll need to do to solve this problem is

a. find the volume of the gold foil sheet

b. use this volume and the given density for gold to calculate the mass of gold in the sample

c. use the molar mass of gold to calculate the number of moles of gold in the sample

d. use Avogadro's number to calculate the number of atoms of gold in the sample

a.

First, let's find the volume of the gold sheet. Since the density of gold is given in grams per cubic centimeter, let's convert our given measurements to centimeters for convenience. We then have

$2.50 {\text{cm" xx 4.00"cm" xx 0.025"cm" = color(red)(0.25 "cm}}^{3}$

b.

Next, let's use the given density of gold, 19.32 "g"/("cm"^3), and this measured volume to calculate the mass in grams of the gold sheet, using dimensional analysis:

color(red)(0.25cancel("cm"^3))((19.32"g Au")/(1cancel("cm"^3))) = color(darkorange)(4.83 "g Au")

c.

Now, let's use the molar mass of gold (equivalent to its relative atomic mass found on a periodic table), $196.97 \text{g"/"mol}$, to calculate the number of moles of gold in the sample:

color(darkorange)(4.83 cancel("g Au"))((1 "mol Au")/(196.97 cancel("g Au"))) = color(green)(0.0245 "mol Au"

d.

Lastly, let's use this number and Avogadro's number, $6.022 \times {10}^{23} \text{particles (atoms)"/"mol}$, to calculate the total number of gold atoms in the sheet:

color(green)(0.0245 cancel("mol Au"))((6.022 xx 10^23 "atoms Au")/(1 cancel("mol Au"))) = color(blue)(1.5 xx 10^22 "atoms Au"

rounded to $2$ significant figures.

May 27, 2017

$1.5 \cdot {10}^{22}$

#### Explanation:

The first thing to do here is to figure out the volume of the foil. Since you're dealing with a rectangular prism, you can say that the volume of the foil will be equal to

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{V = L \cdot l \cdot h}}}$

Here

• $L$ is the length of the foil
• $l$ is its width
• $h$ is its height, or thickness

Now, notice that the density of gold is given to you in grams per cubic centimeter, ${\text{g cm}}^{- 3}$.

This tells you that you need to convert the dimensions of the foil from millimiters to centimeters by using the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 cm = 10 mm}}}}$

You can thus say that the volume of the foil will be equal to

V = (40.0 color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm"))))) * (25.0 color(red)(cancel(color(black)("mm"))) * "1 cm"/(10 color(red)(cancel(color(black)("mm"))))) * (0.25 color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))))

$V = {\text{0.25 cm}}^{3}$

Next, use the desnity of gold to find the mass of the foil.

0.25 color(red)(cancel(color(black)("cm"^3))) * "19.32 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "4.83 g"

Now, in order to find the number of atoms of gold present in this sample, you must convert the number of grams to moles by using the element's molar mass

4.83 color(red)(cancel(color(black)("g"))) * "1 mole Au"/(196.97 color(red)(cancel(color(black)("g")))) = "0.02452 moles Au"

Finally, use Avogadro's constant, i.e. the definition of a mole, to convert the number of moles to number of atoms

0.02452 color(red)(cancel(color(black)("moles Au"))) * overbrace( (6.022 * 10^(23)color(white)(.)"atoms of Au")/(1color(red)(cancel(color(black)("mole Au")))))^(color(blue)("Avogadro's constant"))#

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.5 \cdot {10}^{22} \textcolor{w h i t e}{.} \text{atoms of Au}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the thickness of the foil.