The equation for the combustion reaction is
#"2K"_2"C"_4"H"_4"O"_6 + "5O"_2 → "2K"_2"O" + "8CO"_2 + "4H"_2"O"#
Part 1. Empirical formula.
#("C, H, O, K") + "O"_2 → "CO"_2color(white)(l) +color(white)(l) "H"_2"O"color(white)(l) +color(white)(l) "K"_2"O"#
#color(white)(ml)"0.5000 g"color(white)(mmmmm) "0.389 g"color(white)(ll) "0.0796 g"color(white)(ll) "0.208 g"#
We can calculate the masses of #"C, H"#, and #"K"# from the masses of their oxides.
#"Mass of C" = 0.389 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.1062 g C"#
#"Mass of H" = 0.0796 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.008 91 g H"#
#"Mass of K" = 0.208 color(red)(cancel(color(black)("g K"_2"O"))) × "78.20 g K"/(94.20 color(red)(cancel(color(black)("g K"_2"O")))) = "0.1727 g K"#
#"Mass of O" = "mass of (C, H, O, K) – mass of (C + H + K)" = "0.5000 g – (0.1062 + 0.00891 + 0.1727) g" = "(0.5000 – 0.2878) g" = "0.2122 g O"#
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
#"Element"color(white)(X) "Mass/g"color(white)(Xm) "mmol"color(white)(mll) "Ratio" color(white)(mll)"Integers"#
#stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)0.1062 color(white)(Xml)"8.843"
color(white)(Xll)2.002color(white)(Xmm)2#
#color(white)(ll)"H" color(white)(XXXXl)0.00891 color(white)(mll)"8.839" color(white)(Xll)2.001 color(white)(XXX)2#
#color(white)(ll)"K" color(white)(XXXml)0.1727color(white)(mml)"4.417" color(white)(Xll)1 color(white)(XXXmlll)1#
#color(white)(ll)"O" color(white)(XXXXl)0.2122 color(white)(mll)"13.26" color(white)(Xml)3.002color(white)(XXX)3#
The empirical formula is #"C"_2"H"_2"O"_3"K"#.
Part 2. Molecular formula.
The relative empirical formula mass of #"C"_2"H"_2"O"_3"K"# is 113.13.
The relative molecular mass must be an integral multiple of the empirical formula mass.
#"MM" = n × "EFM"#, or
#n = "MM"/"EFM" = 226.27/113.13 = 1.9992 ≈ 2#
The molecular formula must be twice the empirical formula.
#"MF" = "EF"_2 = ("C"_2"H"_2"O"_3"K")_2 = "C"_4"H"_4"O"_6"K"_2#
Part 3. The chemical equation
The first experiment is a combustion reaction.
The equation for the reaction is
#"2K"_2"C"_4"H"_4"O"_6 + "5O"_2 → "2K"_2"O" + "8CO"_2 + "4H"_2"O"#
