# Question c2038

Mar 31, 2016

Here's what I got.

#### Explanation:

A compound's percent composition essentially tells you how many grams of each constituent element you get per $\text{100 g}$ of compound.

In order to find methyl ethyl ketone's (MEK) percent composition, you need to pick a sample of this compound and try to figure out how many grams of carbon, $\text{C}$, hydrogen, $\text{H}$, and oxygen, $\text{O}$, it contains.

Notice that one mole of MEK contains

• four moles of carbon, $4 \times \text{C}$
• eight moles of hydrogen, $8 \times \text{H}$
• one mole of oxygen, $1 \times \text{O}$

This means that if you use the mass of one mole of MEK, i.e. its molar mass, you can determine how many grams of each constituent element it contains by using the molar masses of the three elements.

MEK has a molar mass of ${\text{72.11 g mol}}^{- 1}$, which means that one mole of MEK has a mass of $\text{72.11 g}$.

If you pick a $\text{72.11-g}$ sample of MEK, you can say that it contains

4 color(red)(cancel(color(black)("moles C"))) * overbrace("12.011 g"/(1color(red)(cancel(color(black)("mole C")))))^(color(purple)("molar mass of C")) = "48.044 g C"

8color(red)(cancel(color(black)("moles H"))) * overbrace("1.00794 g"/(1color(red)(cancel(color(black)("mole H")))))^(color(blue)("molar mass of H")) = "8.064 g H"

1color(red)(cancel(color(black)("mole O"))) * overbrace("16.0 g"/(1color(red)(cancel(color(black)("mole O")))))^(color(green)("molar mass of O")) = "16.0 g O"

To get the percent composition of an element in the compound, use

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% element" = "mass of element"/"mass of compound} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

"% C" = (48.044color(red)(cancel(color(black)("g"))))/(72.11color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"66.6% C"color(white)(a/a)|)))

"% H" = (8.064color(red)(cancel(color(black)("g"))))/(72.11color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"12.2% H"color(white)(a/a)|)))

"% O" = (16.0color(red)(cancel(color(black)("g"))))/(72.11color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"22.2% O"color(white)(a/a)|)))#