Chemistry - Calculating Percentage Solutions

Introduction

Percentage weight by volume is just another way of stating the concentration of a solution.

The advantage of using percentages to express concentration is that it tells you how much of the compound should be dissolved to make up the solution. There is no need to work out moles or molarity, and you do not need to know the molecular weight of the compound.

What is a percentage weight by volume (w/v)?

Percentage weight by volume (w/v) is just the number of grams of compound per 100 ml final volume of solution. However, it should be noted that the final volume, as we shall see, does not have to be 100 ml. You could make only 30 ml of the solution, or you could make 200 ml.

How do I calculate percentage weight by volume (w/v)?

Percentage weight by volume (w/v) is just the number of grams of compound used, divided by the final volume in ml, multiplied by 100.

Expressed as a formula we get:

% = (m / v) x 100

Where:

% = percent
m = mass of compound (g)
v = final volume (ml)

Rearranging the above equation we can calculate the number of grams you would need to make up a volume of solution for a given percentage. Rearranging the above equation we get:

m = (% / 100) x v

Where:

m = mass of compound (g)
% = percent
v = final volume (ml)

You can explore your understanding of percentage weight by volume (w/v) by looking at the worked examples in the next lecture.

Below you will find some of the examples from the above video worked through again, as well as some new example calculations.

Example 1: What would be the percentage (w/v) of a solution in which 4 g of NaCl had been made up to a final volume of 100 ml?

For this you need the following equation from the previous lecture:

% = (m / v) x 100

Where:

% = percent
m = mass of compound (g) = 4 g
v = final volume (ml) = 100 ml

Hence we get:

% = (4 / 100) x 100

% = 0.04 x 100

% = 4 % (w/v)

Therefore, 4 g of NaCl dissolved in water, and made up to a final volume of 100 ml, would give a 4% (w/v) solution.

Example 2: What would be the percentage (w/v) of a solution in which 4 g of NaCl had been made up to a final volume of 50 ml?

For this you will need the equation:

% = (m / v) x 100

Where:

% = percent
m = mass of compound (g) = 4 g
v = final volume (ml) = 50 ml

Hence we get:

% = (4 / 50) x 100

% = 0.08 x 100

% = 8% (w/v)

Therefore, 4 g of NaCl dissolved in water, and made up to a final volume of 50 ml, would give a 8% (w/v) solution.

Example 3: What would be the percentage (w/v) of a solution in which 2 g of NaCl had been made up to a final volume of 50 ml?

For this you will need the equation:

% = (m / v) x 100

Where:

% = percent
m = mass of compound (g) = 2 g
v = final volume (ml) = 50 ml

Hence we get:

% = (2 / 50) x 100

% = 0.04 x 100

% = 4% (w/v)

Therefore, 2 g of NaCl dissolved in water, and made up to a final volume of 50 ml, would give a 4% (w/v) solution.

Example 4: What would be the percentage (w/v) of a solution in which 2.4 g of NaCl had been made up to a final volume of 30 ml?

For this you will need the equation:

% = (m / v) x 100

Where:

% = percent
m = mass of compound (g) = 2.4 g
v = volume (ml) = 30 ml

Hence we get:

% = (2.4 / 30) x 100

% = 0.08 x 100

% = 8% (w/v)

Therefore, 2.4 g of NaCl dissolved in water, and made up to a final volume of 30 ml, would give a 8% (w/v) solution.

Example 5: What would be the percentage (w/v) of a solution in which 6 g of NaCl had been made up to a final volume of 200 ml?

For this you will need the equation:

% = (m / v) x 100

Where:

% = percent
m = mass of compound (g) = 6 g
v = final volume (ml) = 200 ml

Hence we get:

% = (6 / 200) x 100

% = 0.03 x 100

% = 3% (w/v)

Therefore, 6 g of NaCl dissolved in water, and made up to a final volume of 200 ml, would give a 3% (w/v) solution.

Example 6: What would be the percentage (w/v) of a solution in which 9 g of NaCl had been made up to a final volume of 225 ml?

For this you will need the equation:

% = (m / v) x 100

Where:

% = percent
m = mass of compound (g) = 9 g
v = final volume (ml) = 225 ml

Hence we get:

% = (9 / 225) x 100

% = 0.04 x 100

% = 4% (w/v)

Therefore, 9 g of NaCl dissolved in water, and made up to a final volume of 225 ml, would give a 4% (w/v) solution.

Example 7: What would be the percentage (w/v) of a solution in which 25 g of NaCl had been made up to a final volume of 0.5 l?

First, you need to convert 0.5 l in to ml. So, as there are 1,000 ml in 1 l, then 0.5 l is 500 ml.

For this you will need the equation:

% = (m / v) x 100

Where:

% = percent
m = mass of compound (g) = 25 g
v = final volume (ml) = 500 ml

Hence we get:

% = (25 / 500) x 100

% = 0.05 x 100

% = 5% (w/v)

Therefore, 25 g of NaCl dissolved in water, and made up to a final volume of 0.5 l, would give a 5% (w/v) solution.

Below you will find some of the examples from the above video worked through again and some new example calculations.

Example 1: How many grams of NaCl would you need to make 100 ml of a 5% (w/v) solution?

For this you need the equation we discussed earlier:

m = (% / 100) x v

Where:

m = mass of compound (g)
% = percent = 5%
v = final volume (ml) = 100 ml

Hence we get:

m = (5 / 100) x 100

m = 0.05 x 100

m = 5 g

Therefore, the answer is we need 5 g of NaCl dissolved in water and made up to a final volume of 100 ml.

Example 2: How many grams of NaCl would you need to make 50 ml of a 5% (w/v) solution?

For this you need the equation:

m = (% / 100) x v

Where:

m = mass of compound (g)
% = percent = 5%
v = final volume (ml) = 50 ml

Hence we get:

m = (5 / 100) x 50

m = 0.05 x 50

m = 2.5 g

Therefore, the answer is we need 2.5 g of NaCl dissolved in water and made up to a final volume of 50 ml.

Example 3: How many grams of NaCl would you need to make 50 ml of a 2.5% (w/v) solution?

For this you need the equation:

m = (% / 100) x v

Where:

m = mass of compound (g)
% = percent = 2.5%
v = final volume (ml) = 50 ml

Hence we get:

m = (2.5 / 100) x 50

m = 0.025 x 50

m = 1.25 g

Therefore, the answer is we need 1.25 g of NaCl dissolved in water and made up to a final volume of 50 ml.

Example 4: How many grams of NaCl would you need to make 30 ml of an 8% (w/v) solution?

For this you need the equation:

m = (% / 100) x v

Where:

m = mass of compound (g)
% = percent = 8%
v = volume (ml) = 30 ml

Hence we get:

m = (8 / 100) x 30

m = 0.08 x 30

m = 2.4 g

Therefore, the answer is we need 2.4 g of NaCl dissolved in water and made up to a final volume of 30 ml.

Example 5: How many grams of NaCl would you need to make 200 ml of a 3% (w/v) solution?

For this you need the equation:

m = (% / 100) x v

Where:

m = mass of compound (g)
% = percent = 3%
v = final volume (ml) = 200 ml

Hence we get:

m = (3 / 100) x 200

m = 0.03 x 200

m = 6.0 g

Therefore, the answer is we need 6.0 g of NaCl dissolved in water and made up to a final volume of 200 ml.

Example 6: How many grams of NaCl would you need to make 225 ml of a 4% (w/v) solution?

For this you need the equation:

m = (% / 100) x v

Where:

m = mass of compound (g)
% = percent = 4%
v = final volume (ml) = 225 ml

Hence we get:

m = (4 / 100) x 225

m = 0.04 x 225

m = 9.0 g

Therefore, the answer is we need 9.0 g of NaCl dissolved in water and made up to a final volume of 225 ml.

Example 7: How many grams of NaCl would you need to make 0.5 l of a 5% (w/v) solution?

First, you need to convert 0.5 l in to ml. So, as there are 1,000 ml in 1 l, then 0.5 l is 500 ml.

For this you need the equation:

m = (% / 100) x v

Where:

m = mass of compound (g)
% = percent = 5%
v = final volume (ml) = 500 ml

Hence we get:

m = (5 / 100) x 500

m = 0.05 x 500

m = 25.0 g

Therefore, the answer is we need 25.0 g of NaCl dissolved in water and made up to a final volume of 0.5 l.

Making a w/v solution is fairly easy.

You take the number of grams of the compound you need and then add it to a volume of liquid that is LESS than the final volume required. Once the compound has been thoroughly mixed in you make the solution up to the final required volume. For example, you have to make 100 ml of 5% (w/v) sodium chloride (NaCl) solution:

1. Add approximately 80 ml of water to a 100 ml beaker.

2. Stir (use a magnetic stirrer is possible).

3. To the moving liquid add 5 g of NaCl, with stirring.

4. Stir until all the NaCl is dissolved.

5. Add water to make up to a final volume of 100 ml.

The reason you don't add the compound to the final required volume is because the compound, even if a powder, once mixed in will take up some volume.

Introduction

Percentage volume by volume (v/v) is just another way of stating the concentration of a solution. The advantage of using percentages to express concentration is that it tells you how much of the compound should be dissolved to make up the solution. There is no need to work out moles or molarity, and you do not need to know the compound's molecular weight.

What is a percentage volume by volume (v/v)?

Percentage volume by volume (v/v) is simply the number ml of one liquid made up to 100 ml with another liquid. The final volume doesn't have to be 100 ml as you can make v/v solutions of any volume.

How do I calculate percentage volume by volume (v/v)?

Percentage volume by volume (v/v) is just the number of ml of one liquid, divided by the final volume in ml, multiplied by 100.

Expressed as a formula we get:

% = (v1 / vt) x 100

Where:

% = percent
v1 = volume of liquid (ml)
vt = final total volume (ml)

The above equation can be rearranged to give you the number of ml you would need to make up a volume of a given percentage solution. Rearranging the equation, we get:

v1 = (% / 100) x vt

Where:

v1 = volume of liquid (ml)
% = percent
vt = final total volume (ml)

What should be kept in mind is that the final volume, vt, in the above equations is the sum of v1 plus the volume of the second liquid used to dilute v1. For further information on this, see "How do I make a v/v solution?" later in this section.

Essentially, what is being stated is:

vt = v1 + v2

Where:

vt = final total volume (ml)
v1 = volume of liquid 1 (ml)
v2 = volume of liquid 2 (ml)

You can explore your understanding of percentage volume by volume (v/v) by looking at the worked examples in the next lecture.

Example 1: What would be the percentage (v/v) of a solution in which 4 ml of glycerol has been made up to a final volume of 100 ml with water?

For this you need the equation:

% = (v1 / vt) x 100

Where:

% = percent
v1 = volume of liquid (ml) = 4 ml
vt = final total volume (ml) = 100 ml

Hence we get:

% = (4 / 100) x 100

% = 0.04 x 100

% = 4 % (v/v)

Therefore, 4 ml of glycerol dissolved in water, and made up to a final volume of 100 ml, would give a 4% (v/v) solution.

Example 2: What would be the percentage (v/v) of a solution in which 4 ml of glycerol has been made up to a final volume of 50 ml with water?

For this you need the equation:

% = (v1 / vt) x 100

Where:

% = percent
v1 = volume of liquid (ml) = 4 ml
vt = final total volume (ml) = 50 ml

Hence we get:

% = (4 / 50) x 100

% = 0.08 x 100

% = 8 % (v/v)

Therefore, 4 ml of glycerol dissolved in water, and made up to a final volume of 50 ml, would give an 8% (v/v) solution.

Example 3: What would be the percentage (v/v) of a solution in which 4 ml of glycerol has been made up to a final volume of 200 ml with water?

For this you need the equation:

% = (v1 / vt) x 100 [9]

Where:

% = percent
v1 = volume of liquid (ml) = 4 ml
vt = final total volume (ml) = 200 ml

Hence we get:

% = (4 / 200) x 100

% = 0.02 x 100

% = 2 % (v/v)

Therefore, 4 ml of glycerol dissolved in water, and made up to a final volume of 200 ml, would give a 2% (v/v) solution.

Example 1: How many millilitres (ml) of glycerol would you need to make 100 ml of an 8% (v/v) solution?

For this you need the equation:

v1 = (% / 100) x vt

Where:

v1 = volume of liquid (ml)
% = percent = 8% (v/v)
vt = final total volume (ml) = 100 ml

Hence we get:

v1 = (8 / 100) x 100

v1 = 0.08 x 100

v1 = 8 ml

Therefore, 8 ml of glycerol would be needed to make 100 ml of an 8% (v/v) solution.

Example 2: How many millilitres (ml) of glycerol would you need to make 50 ml of an 8% (v/v) solution?

For this you need the equation:

v1 = (% / 100) x vt

Where:

v1 = volume of liquid (ml)
% = percent = 8% (v/v)
vt = final total volume (ml) = 50 ml

Hence we get:

v1 = (8 / 100) x 50

v1 = 0.08 x 50

v1 = 4 ml

Therefore, 4 ml of glycerol would be needed to make 50 ml of an 8% (v/v) solution.

Example 3: How many millilitres (ml) of glycerol would you need to make 250 ml of a 4% (v/v) solution?

For this you need the equation:

v1 = (% / 100) x vt

Where:

v1 = volume of liquid (ml)
% = percent = 4% (v/v)
vt = final total volume (ml) = 250 ml

Hence we get:

v1 = (4 / 100) x 250

v1 = 0.04 x 250

v1 = 10 ml

Therefore, 10 ml of glycerol would be needed to make 250 ml of a 4% (v/v) solution.

Example 4: How many millilitres (ml) of water would be required to make 100 ml of a 4% (v/v) solution of glycerol?

For this you need the equation:

v1 = (% / 100) x vt

Where:

v1 = volume of liquid (ml)
% = percent = 4% (v/v)
vt = final total volume (ml) = 100 ml

Hence we get:

v1 = (4 / 100) x 100

v1 = 0.04 x 100

v1 = 4 ml

Now, remember that the final volume, vt:

vt = v1 + v2

Where:

vt = final total volume (ml) = 100 ml
v1 = volume of liquid 1 (ml) = 4 ml
v2 = volume of liquid 2 (ml)

Hence, v2 = 96 ml

Therefore, 96 ml of water would be mixed with 4 ml of glycerol to make 100 ml of a 4% (v/v) solution.

Example 5: How many millilitres (ml) of water would be required to make 80 ml of a 4% (v/v) solution of glycerol?

For this you need the equation:

v1 = (% / 100) x vt

Where:

v1 = volume of liquid (ml)
% = percent = 4% (v/v)
vt = final total volume (ml) = 80 ml

Hence we get:

v1 = (4 / 100) x 80

v1 = 0.04 x 80

v1 = 3.2 ml

Again, you need to take in to account the volume of the liquid being dilutes, v1, and the final volume, vt:

vt = v1 + v2 [11]

Where:

vt = final total volume (ml) = 80 ml
v1 = volume of liquid 1 (ml) = 3.2 ml
v2 = volume of liquid 2 (ml)

Hence, v2 = 76.8 ml

Therefore, 76.8 ml of water would be mixed with 3.2 ml of glycerol to make 80 ml of a 4% (v/v) solution.

Making a v/v solution is fairly easy.

You take the number ml of compound you need and then you add them to a volume of liquid so that the final volume is the volume required. For example, you have to make 100 ml of 5% (v/v) glycerol solution:

1. Add around 80 ml of water to a 100 ml beaker.

2. Stir.

3. To the moving liquid add 2 ml of glycerol, with stirring (you can add the water to the glycerol as shown in the video).

4. Continue to stir for a few minutes until the glycerol is dissolved.

5. Check that the final volume is 100 ml.

Percentage volume by volume solutions it is slightly easier to make up as you know the volume of each component to be used, and there should be very little need to adjust the final volume.

Introduction

Percentage weight by weight (w/w) is just another way of stating the concentration of a solution. The advantage of using percentages to express concentration is that it tells you how much of the compound should be dissolved to make up the solution. There is no need to work out moles or molarity, and you do not need to know the compound's molecular weight.

However, w/w is (in my opinion) a little more tricky to understand that w/v or v/v (see earlier lectures in this course).

What is a percentage weight by weight (w/w)?

Percentage weight by weight (w/w) is simply the number grams (g) of one compound made up to 100 g total final weight with a liquid (solvent). The final weight doesn't have to be 100 g as you can make w/w solutions to any final weight; for example, you may make up a 50 g of a solution, as we will see below.

How do I calculate percentage weight by weight (w/w)?

Percentage weight by weight (w/w) is just the number of g of one compound, divided by the solution's final weight, multiplied by 100.

Expressed as a formula we get:

% = (m1 / mt) x 100

Where:

% = percent
m1 = mass of the compound (g)
mt = final mass of the solution (g)

The above equation can be rearranged to give you the number of grams of the compound you would need to make up a final mass of a given percentage solution. Rearranging the equation, we get:

m1 = (% / 100) x mt

Where:

m1 = mass of the compound (g)
% = percent
mt = final mass of the solution (g)

What should be kept in mind is that the final mass, mt in the above equations, is the sum of m1 plus the mass of the liquid used to dissolve m1. Essentially, what is being stated is:

mt = m1 + m2

Where:

mt = final mass of the solution (g)
m1 = mass of the compound (g)
m2 = mass of the solvent (g)

If you were making a 2% (w/w) solution of NaCl, you would take 2 g of NaCl and add 98 g of water. The final w/w solution in the beaker would weight 100 g (98 g of water + 2 g of NaCl), but that the solution's volume may not be 100 ml, it is the weight that is important here and not the volume.

You can explore your understanding of percentage weight by weight (w/w) by looking at the worked examples in the next lecture.

Example 1: What would be the percentage (w/w) of a solution in which 4 g of a compound has been dissolved to give a solution with a final weight of 100 g?

For this you will need the equation:

% = (m1 / mt) x 100

Where:

% = percent
m1 = mass of the compound (g) = 4 g
mt = final mass of the solution (g) = 100 g

Hence we get:

% = (4 / 100) x 100

% = 0.04 x 100

% = 4 % (w/w)

Therefore, 4 g of the compound made up to a solution with a final mass of 100 g is a 4% (w/w) solution.

Example 2: What would be the percentage (w/w) of a solution in which 4 g of a compound has been dissolved to give a solution with a final weight of 50 g?

For this you will need the equation:

% = (m1 / mt) x 100

Where:

% = percent
m1 = mass of the compound (g) = 4 g
mt = final mass of the solution (g) = 50 g

Hence we get:

% = (4 / 50) x 100

% = 0.08 x 100

% = 8 % (w/w)

Therefore, 4 g of the compound made up to a solution with a final mass of 50 g is a 4% (w/w) solution.

Example 3: What would be the percentage (w/w) of a solution in which 6 g of a compound has been dissolved to give a solution with a final weight of 250 g?

For this you will need the equation:

% = (m1 / mt) x 100 [12]

Where:

% = percent
m1 = mass of the compound (g) = 6 g
mt = final mass of the solution (g) = 250 g

Hence we get:

% = (6 / 250) x 100

% = 0.024 x 100

% = 2.4 % (w/w)

Therefore, 6 g of the compound made up to a solution with a final mass of 250 g is a 2.4% (w/w) solution.

Example 4: What would be the percentage (w/w) of a solution in which 6 g of a compound is dissolved in 94 g of water?

First you need to work out the final weight of the solution, and for this we can use the equation:

mt = m1 + m2

Where:

mt = final mass of the solution (g)
m1 = mass of the compound (g) = 6 g
m2 = mass of the solvent (g) = 94 g

Hence:

mt = 6 + 94 = 100 g

Next you will need to use the equation:

% = (m1 / mt) x 100 [12]

Where:

% = percent
m1 = mass of the compound (g) = 6 g
mt = final mass of the solution (g) = 100 g

Hence we get:

% = (6 / 100) x 100

% = 0.06 x 100

% = 6 % (w/w)

Therefore, 6 g of the compound mixed with 94 g of water gives a 6% (w/w) solution.

Example 5: What would be the percentage (w/w) of a solution in which 5 g of a compound is dissolved in 50 g of water?

First you need to work out the final weight of the solution, and for this we can use the equation:

mt = m1 + m2

Where:

mt = final mass of the solution (g)
m1 = mass of the compound (g) = 5 g
m2 = mass of the solvent (g) = 50 g

Hence:

mt = 5 + 50 = 55 g

Next you will need the equation:

% = (m1 / mt) x 100 [12]

Where:

% = percent
m1 = mass of the compound (g) = 5 g
mt = final mass of the solution (g) = 50 g

Hence we get:

% = (5 / 55) x 100

% = 0.09 x 100

% = 9 % (w/w)

Therefore, 5 g of the compound mixed with 50 g of water gives a 9% (w/w) solution.

Example 1: How many grams of the compound would you need to make a 5% (w/w) solution if the solution's final total weight was 100 g?

For this you need the following equation:

m1 = (% / 100) x mt

Where:

m1 = mass of the compound (g)
% = percent = 5% (w/w)
mt = final mass of the solution (g) = 100 g

Hence we get:

m1 = (5 / 100) x 100

m1 = 0.05 x 100

m1 = 5 g

Therefore, you would need 5 g of the compound to make a 5% (w/w) solution with a final mass of 100 g.

Example 2: How many grams of the compound would you need to make a 5% (w/w) solution if the solution's final total weight was 50 g?

For this you will need the equation:

m1 = (% / 100) x mt

Where:

m1 = mass of the compound (g)
% = percent = 5% (w/w)
mt = final mass of the solution (g) = 50 g

Hence we get:

m1 = (5 / 100) x 50

m1 = 0.05 x 50

m1 = 2.5 g

Therefore, you would need 2.5 g of the compound to make a 5% (w/w) solution with a final mass of 50 g.

Example 3: How many grams of the compound would you need to make a 2.5% (w/w) solution if the solution's final total weight was 250 g?

For this you will need the following equation:

m1 = (% / 100) x mt

Where:

m1 = mass of the compound (g)
% = percent = 2.5% (w/w)
mt = final mass of the solution (g) = 250 g

Hence we get:

m1 = (2.5 / 100) x 250

m1 = 0.025 x 250

m1 = 6.25 g

Therefore, you would need 6.25 g of the compound to make a 2.5% (w/w) solution with a final mass of 250 g.

Making a w/w solution is fairly easy.

You weigh out the number of grams of the compound you need, and then you add them to the number of grams of the liquid needed (Note: If the liquid is water, then the number of grams required is the same as the volume as water has a density of 1 g/ml).

For example, you have to make 100g of 5% (w/w) glycerol solution:

1. Weigh out 5 g of glycerol.

2. Add 95 g of water.

3. Mix.

Percentage weight by weight solutions can be a little tricky to understand, but they can be very useful in the lab. In the example below, you are making an agar-based growth media that would require heating to dissolve and need to be autoclaved to sterilise (the autoclaving would heat the solution and cause the agar to dissolve). As the solution tends to stick to the glassware, it can be difficult to work with. Using w/w means that everything can be easily added to the one bottle, i.e. add 48 g of water and 2 g of agar, and we know the solution is made up to the correct final concentration.

For example, you have to make 50 g of 4% (w/w) agar solution:

1. Add 48 g of water to a 100 ml bottle (this would be 48 ml as 1 ml of water weighs 1 g).

2. Add 2 g of agar.

3. Autoclave to sterilise.

As I said at the start of the course, for some reason percentage solutions cause students problems. I have no idea why this should be the case, and hopefully, now that you have completed this course you should have a better understanding of percentage solutions in terms of their differences, and how to calculate the amounts of compound and solvents required.

Remember

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

There are also three types of percentage solution:

1. Percentage weight by volume (w/v) - the number of grams per 100 ml

2. Percentage volume by volume (v/v) - the number of ml per final volume of 100 ml

3. Percentage weight by weight (w/w) - the number of grams per 100 g of the solution

An understanding of percentage solutions and how they are calculated is critical in the biosciences and chemistry as the concentration of solutions is often quoted in percentage terms.

Missing something? A mistake?

Have I missed something in this lecture? Is there something you would like to be included? Have I made a mistake?

If the answer is yes to any of the above questions then please leave me a comment and tell me what you would like included and why, or what the mistake is and where it can be found. Thanks.