# What's the best way to interconvert between molarity and percentage?

Let's say that I want to convert concentration of an acid from % to molarity. One way I found to do this is by using the following formula: percentage = (molarity * molar mass) / 10.

Another way is to use an equivalency chart like this one (see bottom of page): http://www.ochemonline.com/Acid_and_Base_Concentrations

For example, to convert 38% HCl to molarity I calculated: 38% = (molarity * 36.46) / 10 and molarity = 10.42 = 10 M. However, the equivalency chart says that 38% is actually 12.4 M.

The same discrepancy happened when I tried to convert 68% nitric acid molarity. Using the equation, I got molarity to equal 11 M, however, according to another source 68% nitric acid is closer to 15 M.

So my questions are:

1) Is there something I'm doing wrong that I get such a huge discrepancy?

2) What's the best way to figure out molarity from % and vice versa?

• Yes, you are doing it wrong. The equation you are using is not based on known principles. It may be from an observation for 1 acid (or base). As for the best way, there is only one way to do it correctly. Take a known amount of your solution (like 1 kg), determine the mass of the components, convert the mass to the number of moles, and then determine the concentration. You will need to make an adjustment for the volume using the density. – LDC3 Oct 29 '17 at 21:50
• It looks like your values are off by a factor of the density of the conc acid. See LDC3's comment. – airhuff Oct 29 '17 at 21:52
• What do you mean by percentage? en.wikipedia.org/wiki/Concentration en.wikipedia.org/wiki/… – aventurin Oct 29 '17 at 22:38
• Oh my goodness! Thank you, @LDC3! You've answered my question, which I wish I could accept as an answer. I'll post the answer to my own question but if I can give credit to you instead let me know how! – Olga B Oct 29 '17 at 22:39

Thanks to LDC3's comments, I understand that the best way to convert from % to Molarity is as follows:

1) multiply density by volume at hand to get weight

2) solve ratio for x, where x is grams of solute (in volume at hand)

(given% / 100%) = (x / weight from step 1)

3) multiply x by molar mass to get moles (in volume at hand)

4) if needed, multiply resulting moles/volume in hand to get Molarity

Example: Let's say I have 100 mL of 38% HCl.

1) 1.19 g/ml * 100 mL = 119 g

2) (38%/100%) = (x/119) where x = 45.22 g

3) 45.22 g * 1mol/36.46g = 1.240 moles

4) 1.240 mol/100 ml = 12.40 mol / 1000 mL = 12.4 M

Now, the result matches the equivalency chart!

There is no easy way to transform the (mass) percentage of a solution into molarity. This is because percentage relies on the relative mass ratios while molarity is a measure per volume. These two values are connected via the solution’s density; while we can approximate the density of an acidic solution to be approximately that of water for low concentrations, highly concentrated acidic solutions have very different and unpredictable densities.

For example, $\pu{38\%}\ \ce{HCl}$ has a density $\rho=\pu{1.189g/ml}$. This allows us to calculate:

• $\pu{1.000l}$ of $\pu{38\%}\ \ce{HCl}$ contains $\pu{1.189kg}$.

• $\pu{38\%}$ of this is $\ce{HCl}$, so $m(\ce{HCl}) = \pu{0.45kg}$

• Therefore, the amount of $\ce{HCl}$ in $\pu{1.000l}$ is $n(\ce{HCl})=\pu{0.012kmol} = \pu{12mol}$.

• Therefore, $\pu{38\%}\ \ce{HCl}$ is $\pu{12M}$.

Your formula does not contain any reference to a density and is thus doomed to fail.