Question

Question #f5bf1

Answer 1

`"KCl"`

Explanation:

The cool thing about this problem is that you can give a very quick answer by looking in the periodic table.

More specifically, take a look at the molar masses of chlorine, `"Cl"`, and potassium, `"K"`.

`"For Cl: " "35.4527 g mol"^(-1)`

`"For K: " "39.0983 g mol"^(-1)`

This tells you that every mole of chlorine will have mass of `"35.4527 g"` and every mole of potassium will have a mass of `"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 `"100 g"` of compound. If you have exactly the same amount of chlorine and potassium per `"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

`~~"39 g for K "` versus `" "~~ "35.5 g for Cl"`

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

`"52.7% for K" "` versus `" ""47.3% for Cl"`

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

`color(green)(|bar(ul(color(white)(a/a)"K"_1"Cl"_1 implies "KCl"color(white)(a/a)|)))`

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

`"For K: " "52.7 g"`

`"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

`"For K: " 52.7 color(red)(cancel(color(black)("g"))) * "1 mole K"/(39.0983color(red)(cancel(color(black)("g")))) = "1.348 moles K"`

`"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

`color(green)(|bar(ul(color(white)(a/a)"K"_1"Cl"_1 implies "KCl"color(white)(a/a)|)))`