# Question f5bf1

Mar 8, 2016

$\text{KCl}$

#### Explanation:

More specifically, take a look at the molar masses of chlorine, $\text{Cl}$, and potassium, $\text{K}$.

${\text{For Cl: " "35.4527 g mol}}^{- 1}$

${\text{For K: " "39.0983 g mol}}^{- 1}$

This tells you that every mole of chlorine will have mass of $\text{35.4527 g}$ and every mole of potassium will have a mass of $\text{39.0983 g}$.

Now, take a loo at the compound's percent composition.

Notice that you have approximately equal amounts of chlorine and potassium per $\text{100 g}$ of compound. If you have exactly the same amount of chlorine and potassium per $\text{100 g}$ of compound, you also have a 50% percent composition for both elements.

However, since you get slightly more potassium per mole than chlorine

$\approx \text{39 g for K }$ versus $\text{ "~~ "35.5 g for Cl}$

it makes sense to have a percent composition of potassium that's slightly higher than that of chlorine

$\text{52.7% for K" }$ versus $\text{ ""47.3% for Cl}$

This means that the empirical formula of the compound has to be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{K"_1"Cl"_1 implies "KCl} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, here's how you can prove this by doing some calculations. Pick a $\text{100-g}$ sample of this compound. According to the given percent composition, this sample will contain

$\text{For K: " "52.7 g}$

$\text{For Cl: " "47.3 g}$

Use the molar masses of the two elements to determine how many moles of each you get in this sample

$\text{For K: " 52.7 color(red)(cancel(color(black)("g"))) * "1 mole K"/(39.0983color(red)(cancel(color(black)("g")))) = "1.348 moles K}$

$\text{For Cl: " 47.3color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.4527color(red)(cancel(color(black)("g")))) = "1.334 moles Cl}$

The empirical formula of a compound tells you the smallest whole number ratio that exists between its constituent elements. To find that ratio, divide both values by the smallest one

"For K: " (1.348color(red)(cancel(color(black)("moles"))))/(1.344color(red)(cancel(color(black)("moles")))) = 1.003 ~~ 1

"For Cl: " (1.344color(red)(cancel(color(black)("moles"))))/(1.344 color(red)(cancel(color(black)("moles")))) = 1#

Once again, the empirical formula comes out to be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{K"_1"Cl"_1 implies "KCl} \textcolor{w h i t e}{\frac{a}{a}} |}}}$