Question #b85e8

1 Answer
Jun 2, 2016

Answer:

#"KMnO"_4#

Explanation:

Your goal here is to find the smallest whole number ratio that exists between the elements that make up this compound.

The problem provides you with the compound's percent composition, so your strategy here will be to go from percent by mass to grams, and then from grams to moles

#"percent by mass " stackrel(color(white)(acolor(red)(1)aaa))(->) " grams " stackrel(color(white)(acolor(blue)(2)aaa))(->) "moles " stackrel(color(white)(acolor(green)(3)aaa))(->) " mole ratios"#

#color(white)()#

#color(red)(1)#. Percent by mass to grams

To keep the calculations simple, pick a #"100-g"# sample of this unknown compound. According to the given percent composition, this sample will contain

  • #24.56% -> "24.56 g"# of potassium, #"K"#
  • #34.81% -> "34.81 g"# of manganese, #"Mn"#
  • #40.50% -> "40.50 g"# of oxygen, #"O"#

#color(white)()#

#color(blue)(2)#. Grams to moles

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

#"For K: " 24.56 color(blue)(cancel(color(black)("g"))) * "1 mole K"/(39.0983color(blue)(cancel(color(black)("g")))) = "0.6282 moles K"#

#"For Mn: " 34.81 color(blue)(cancel(color(black)("g"))) * "1 mole Mn"/(54.938color(blue)(cancel(color(black)("g")))) = "0.6336 moles Mn"#

#"For O: " 40.50 color(blue)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(blue)(cancel(color(black)("g")))) = "2.5313 moles O"#

#color(white)()#

#color(green)(3)#. Moles to mole ratios

To determine the mole ratios that exist between the three elements in the compound, divide all values by the smallest one

#"For K: " (0.6282 color(green)(cancel(color(black)("moles"))))/(0.6282color(green)(cancel(color(black)("moles")))) = 1#

#"For Mn: " (0.6336color(green)(cancel(color(black)("moles"))))/(0.6282color(green)(cancel(color(black)("moles")))) = 1.009 ~~ 1#

#"For O: " (2.5313 color(green)(cancel(color(black)("moles"))))/(0.6282color(green)(cancel(color(black)("moles")))) = 4.03 ~~ 4#

Since #1:1:4# is already the smallest whole number ratio that can exist in this case, the empirical formula of the compound will be

#"K"_1"Mn"_1"O"_4 implies color(green)(|bar(ul(color(white)(a/a)color(black)("KMnO"_4)color(white)(a/a)|)))#