Calculating Percentage Solutions

For some reason, percentage solution calculations cause students some problems. However, hopefully, once you have read this blog post, you should have a good understanding of percentage solutions and how to do the calculations.

If you want to practice your 'science maths', then have a look at Maths4Biosciences

What are percentage solutions?

A percentage solution is an amount or volume of something per 100 ml or 100 g of a solution. It is as simple as that. It is a percentage.

Why are they used?

Percentage solutions are a convenient and easy way to record solution concentrations. One advantage is that you don’t need to know anything about the compound in terms of molecular weight; all you need is the percentage of the required solution.

Why are there three types of percentage solutions?

This is slightly difficult to explain. However, there are three types of percentage solutions:

• Percentage weight by volume (w/v)
• Percentage volume by volume (v/v)
• Percentage weight by weight (w/w)

The percentage weight by volume (w/v) is the number of grams of compound per 100 ml of solution. This type of percentage solution is used when describing the amount of powder in a solution. For example, 5 g of powder made up to a final volume of 100 ml would be a 5% (w/v) solution. Likewise, 2.5 g of powder made up to 50 ml would also be a 5% (w/v) solution, as you would have 5 g in 100 ml.

The percentage volume by volume (v/v) is the number of ml of some liquid per 100 ml of the solution. This type of percentage solution is usually used to describe a solution made by mixing two liquids. For example, 5 ml of a liquid made up to a final volume of 100 ml would be a 5% (v/v) solution. Likewise, 2.5 ml of a liquid made up to 50 ml would also be a 5% (v/v) solution, as you would have 5 ml in 100 ml.

Finally, the percentage weight by weight (w/w). This one is more difficult to understand, but the principles, as explained above, are still valid. A percentage weight by weight (w/w) solution can be the weight of a powder or a liquid made up in a solution to the final weight of the solution. So, for example, 5 g of a powder (or a liquid) made up in a solution with a final weight of 100 g would be a 5% (w/w) solution. Likewise, 2.5 g of a liquid or powder made up to give a solution that weighed 50 g would also be a 5% (w/w) solution, as you would have 5 g of the powder or liquid in 100 g of a solution.

 

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Additional Resources

•  📗 - Maths and Chemistry Refresher for Life and Biomedical Scientists - Google Play Book Store
•  📗 - Catchup Maths and Stats - (affiliate link)
•  📗 - Catchup Chemistry - (affiliate link)
•  🌎 - You can practice your lab calculations at: https://maths4biosciences.drnickmorris.com/

Calculating Dilutions

I have received a number of emails about dilution calculations.

The type of question is:

You have been asked to dilute a 10 mg/ml solution to give 5 ml of a 1.5 mg/ml solution.

How many ml of water would you use?

How many ml of the stock protein solution would you use?

The way I would work this out is by saying that 5 ml of 1.5 mg/ml contains 5 x 1.5 mg = 7.5 mg.

If the stock is 10 mg/ml then I need to take a volume that would give 7.5 mg. So, the number of ml of the stock I need is 7.5 / 10 = 0.75 ml (that is, 0.75 ml of a 10 mg/ml solution contains 7.5 mg).

Therefore, I have 0.75 ml of the stock, and the final volume is 5 ml, so we need to add 5 - 0.75 = 4.25 ml of water.

The answer:

How many ml of water would you use? 4.25 ml

How many ml of the stock protein solution would you use? 0.75 ml

Although I prefer to work it out as above you can use a formula:

M₁V₁ = M₂V₂

Where:
M1	=	Concentration 1
V1	=	Volume 1
M2	=	Concentration 2
V2	=	Volume 2

Therefore:

M1	=	Concentration 1	=	10 mg/ml
V1	=	Volume 1	=	Unknown X
M2	=	Concentration 2	=	1.5 mg/ml
V2	=	Volume 2	=	5 ml

or:
M₁V₁ = M₂V₂ 10 mg/ml x Unknown X = 1.5 mg/ml x 5 ml

Rearranging to solve for Unknown X:

Unknown X = (1.5 mg/ml x 5 ml) / 10 mg/ml Therefore, Unknown X = 0.75 ml

Using the formula is the same as the first solution, but in the first solution, I don't have to remember the formula....

If you struggle with 'Science Maths' then you may like to look at Maths4Biosciences.

If you would like to support my blogging efforts, then please feel free to buy me a coffee at https://www.buymeacoffee.com/drnickm

Additional Resources

• 📗 - Maths and Chemistry Refresher for Life and Biomedical Scientists
• 📗 - Catchup Chemistry (affiliate link to the book)
• 📗 - Catchup Maths and Stats (affiliate link to the book)