# What is the difference between mass composition and percent composition?

Sep 9, 2015

Mass composition is an absolute way to quantify the amount of a particular component in a compound, whether it's an atom in a molecule, an aqueous compound within a larger-scale solution, or something else.

Percent composition is just a relative way to quantify what [insert quantity type here] composition could have quantified, where [insert quantity type here] could be mass, volume, etc.

PERCENT COMPOSITION

One example is "Mg"("NO"_3)_2. Maybe you want to find the percent of magnesium. Then all you do is use the formula weight of magnesium and of the whole compound:

"% Composition Mg" = ("24.305 g Mg")/("24.305 g Mg +2*14.007 g N +3*2*15.999 g O") xx 100%

$\approx$ $\textcolor{b l u e}{\text{16.39% Mg}}$

Or, for a more real-life example, if you take 37% reaction-grade $\text{HCl}$, whose bottle is labeled as such, you know that there is always a percent composition of 37% $\text{HCl w/w}$ in that stock solution to begin with (by the way, it is the equivalent of ~"12.08 M").

That is said to be "thirty-seven percent HCl by weight". This is what you could find in lab.

MASS COMPOSITION

The same solution can be worked with to determine the mass composition. What you have to know is the density of the solution, which is $\text{1.19 g/mL}$.

Now all you have to do is determine what the amounts are for:

$0.37 = \left(\text{g HCl")/("g H"_2"O}\right)$

And then the mass composition of $\text{HCl}$ is the amount you find. Suppose you have $\text{100 mL}$ of solution (this is an important step many students miss).

What you can do is use the density of the stock solution (which you can google) to find the mass of the solution overall, multiply by $0.37$, and find the mass of $\text{HCl}$ in there.

$\left(\text{100" cancel"mL") xx (("1.19 g soln")/cancel"mL}\right)$

$=$ $\text{119.0 g}$

Finally, we just have:

$0.37 \times 119.0 = \textcolor{b l u e}{\text{44.03 g HCl}}$ in $\text{100 mL}$ of $\text{37% HCl w/w}$ stock solution.

And if we know that, we could also verify the molarity I said. :)

((44.03 cancel"g HCl")/"0.100 L soln") xx ("1 mol HCl"/("36.4609" cancel"g")) ~~ color(blue)("12.08 M")