# What is the percent potassium, chromium, and oxygen in K_2Cr_2O_7?

Jul 28, 2016

${\text{K"_2"Cr"_2"O}}_{7}$ is $\text{26.58 % K}$, $\text{35.35 % Cr}$, and $\text{38.07 % O}$ 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 $\text{K, Cr}$, and $\text{O}$ in a given mass of the compound.

Let's choose 1 mol of ${\text{K"_2"Cr"_2"O}}_{7}$.

$\text{Mass of K"color(white)(l) =color(white)(m) "2 × 39.10 g" = color(white)(ll)"78.20 g}$
$\text{Mass of Cr" = color(white)(l)" 2 × 52.00 g" = "104.00 g}$
$\text{Mass of O"color(white)(ll) = color(white)(m)"7 × 16.00 g" = "112.00 g}$
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 %