# Question 5956f

Apr 17, 2016

$2.3 \cdot {10}^{22} \text{atoms}$

#### Explanation:

The first thing to do here is figure out how much pure gold you have in your $\text{10-g}$ sample of $18$ carat gold.

The carat is simply a measure of the purity of a given metal alloy based on fractions of $24$. In your case, $18$ carat gold will contain $18$ parts gold for every $24$ parts of alloy by mass, which means that it has a percent purity of

$\text{% gold" = (18 color(red)(cancel(color(black)("parts gold"))))/(24color(red)(cancel(color(black)("parts alloy")))) xx 100 = "75% Au}$

So, if $18$ carat gold has a purity of 75%, it follows that your sample will contain

10 color(red)(cancel(color(black)("g 18 carat gold"))) * overbrace("75 g pure gold"/(100color(red)(cancel(color(black)("g 18 carat gold")))))^(color(purple)("= 75% Au")) = "7.5 g pure gold"

In order to determine how many atoms you have in $\text{7.5 g}$ of pure gold, convert the mass to moles by using the element's molar mass.

Gold has a molar mass of ${\text{197 g mol}}^{- 1}$, which means that one mole of gold will have a mass of $\text{197 g}$. This means that your sample will contain

7.5 color(red)(cancel(color(black)("g"))) * overbrace("1 mole Au"/(197color(red)(cancel(color(black)("g")))))^(color(green)("the molar mass of Au")) = "0.0381 moles Au"#

Now that you know how many moles of gold you have, use Avogadro's number to convert them to atoms of gold

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"atoms} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's number

So, if one mole of gold contains $6.022 \cdot {10}^{23}$ atoms, it follows that $0.0381$ moles will contain

$0.0381 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Au"))) * (6.022 * 10^(23)"atoms")/(1color(red)(cancel(color(black)("mole Au")))) = color(green)(|bar(ul(color(white)(a/a)2.3 * 10^(22)"atoms} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs.