#"K"_2"Cr"_2"O"_7# is #"26.58 % K"#, #"35.35 % Cr"#, and #"38.07 % O"# by mass.
The formula for percent composition is
#color(blue)(|bar(ul(color(white)(a/a) "% by mass" = "mass of component"/"mass of sample" × 100 %color(white)(a/a)|)))" "#
We must therefore find the masses of #"K, Cr"#, and #"O"# in a given mass of the compound.
Let's choose 1 mol of #"K"_2"Cr"_2"O"_7#.
#"Mass of K"color(white)(l) =color(white)(m) "2 × 39.10 g" = color(white)(ll)"78.20 g"#
#"Mass of Cr" = color(white)(l)" 2 × 52.00 g" = "104.00 g"#
#"Mass of O"color(white)(ll) = color(white)(m)"7 × 16.00 g" = "112.00 g"#
#stackrel(———————————————————)("Mass of K"_2"Cr"_2"O"_7color(white)(mmmmm) =color(white)(l) "294.20 g")#
∴ #"% K" = "mass of K"/("mass of K"_2"Cr"_2"O"_7) × 100 % = (78.20 color(red)(cancel(color(black)("g"))))/(294.20 color(red)(cancel(color(black)("g")))) × 100 % = 26.58 %#
#"% Cr" = "mass of Cr"/("mass of K"_2"Cr"_2"O"_7) × 100 % = (104.00 color(red)(cancel(color(black)("g"))))/(294.20 color(red)(cancel(color(black)("g")))) × 100 % = 35.35 %#
#"% O" = "mass of O"/("mass of K"_2"Cr"_2"O"_7) × 100 % = (112.00 color(red)(cancel(color(black)("g"))))/(294.20 color(red)(cancel(color(black)("g")))) × 100 % = 38.07 %#
