Question #184ab

The first thing to do here is to figure out the mass of oxygen present in the sample

Use

#m_"total" = m_"K" + m_"Cl" + m_"O"#

to get

#m_"O" = "22.35 g" - ("7.13 g" + "6.47 g")#

#m_"O" = "8.75 g"#

Now, your goal here is to find the smallest whole number ratio that exists between the three elements in this unknown compound. To do that, use their molar masses to find how many moles of each you have in this sample

#7.13 color(red)(cancel(color(black)("g"))) * "1 mole K"/(39.0983color(red)(cancel(color(black)("g")))) = "0.18236 moles K"#

#6.47 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g")))) = "0.18250 moles Cl"#

#8.75color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.54690 moles O"#

Divide each value by the smallest one to get the mole ratio that exists between the elements in the compund

#"For K: " (0.18236 color(red)(cancel(color(black)("moles"))))/(0.18236color(red)(cancel(color(black)("moles")))) = 1#

#"For O: " (0.18250color(red)(cancel(color(black)("moles"))))/(0.18236color(red)(cancel(color(black)("moles")))) = 1.001 ~~ 1#

#"For O: " (0.54690color(red)(cancel(color(black)("moles"))))/(0.18236color(red)(cancel(color(black)("moles")))) = 2.999 ~~3#

Since #1:1:3# is already the smallest whole number ratio that can exist here, you can say that the empirical formula of the compound will be

#color(green)(bar(ul(|color(white)(a/a)color(black)("K"_ 1 "Cl"_ 1"O"_ 3 implies "KClO"_3)color(white)(a/a)|)))#