Chemistry - Mass Percentage

Problem Set

How do we express the concentration of a solute in a solution? We use Mass Percentage.

Mass Percentage is one of the way of representing the concentration of a solute in a solution. Mass percentage is calculated as the mass of a solvent divided by the total mass of the solution, multiplied by $100\%$ .

So it is easy to find out the formula:

$\text{Mass Percentage }=\frac{m_{solute}}{m_{solution}}\times100\%$

A $200g$ beer contains $20g$ alcohol. So, the mass percentage of alcohol in the beer is:

$\frac{20g}{200g}\times100\%=10\%$ .

Some other notes:

The mass percentage does not change with the temperature and air pressure.
There are other ways to express the concentration of a solute in a solution such as volume percentage, molality and amount-of substance concentration.
Some reference books express mass percentage as $g/g$ . However, try not to use that to prevent a confusion with Solubility.

Now, there is an example:

I want to dilute a $50g$ , $98\%$ $H_2SO_4$ to a $20\%$ solution. How much water should I add in grams to do this?

Solution:

$50g$ is the mass of solution.

$98\%$ is the mass percentage before adding water.

$20\%$ is the mass percentage after adding water.

Note that the mass of $H_2SO_4$ does not change before and after the water added.

So, let the fixed mass of $H_2SO_4$ be $c$ and the mass of water should add be $x$

Before adding water,

$\frac{c g}{50g}\times100\%=98\%$

$c=\frac{98}{100}\times50$

After adding water,

$\frac{c g}{(50+x)g}\times100\%=20\%$

$c=\frac{20}{100}\times(50+x)$

Compare the two equations we have,

$\frac{98}{100}\times50=\frac{20}{100}\times(50+x)$

$x=195g$

So, the amount of water we should add is $195g$ .

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Chemistry - Mass Percentage

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