Question #f5bf1
1 Answer
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,
#"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
Now, take a loo at the compound's percent composition.
Notice that you have approximately equal amounts of chlorine and potassium per
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
#"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)|)))#