Percentage Solutions - Introduction...Percentage Solutions - Introduction Introduction For some...

Percentage Solutions - IntroductionIntroduction

For some reason, percentage solutions cause students problems. I have no idea why this should be the case, and once you have completed this course percentage solutions should no longer cause you any problems.

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.

Why are they used?

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.

Why are their three types of percentage solutions?

This is slightly difficult to explain, particularly before we have tackled the calculation of percentage solutions in the subsequent lessons.

However, there are three types of percentage solution:

1. Percentage weight by volume (w/v)

2. Percentage volume by volume (v/v)

3. Percentage weight by weight (w/w)

So, why?

Well, there are three types of percentage solution to take in to account the problems associated with measuring out material. All three can be used for describing the concentration of a solution, and using the wrong type at the wrong time can result in solutions of an incorrect final concentration.

In the different types, w/v, v/v, and w/w, the first letter represents the solute (the thing being dissolved or diluted, and the second letter represents the solvent (e.g. water) into which the solute is being mixed or dissolved.

When measuring out a powder, e.g. sodium chloride, you would use weight as the material is a solid and measuring out a consistent volume of sodium chloride would be very difficult as the volume would be dependent on crystal size and how those crystals settle. So, for a solid, we tend to use weight, and so would use w/v (weight by volume) or w/w (weight by weight), where the solid is represented by the first w (weight).

When measuring out a liquid, you could use weight (w) or volume (v) to describe the amount of liquid being used. However, it should be noted that weight (w) or volume (v) cannot be used interchangeably (i.e. if it says v/v you can't switch to w/v) in such situations as you will end up with the wrong concentrations.

For example, let's say you have to make a 5% solution of glycerol (we will get into the calculations more in the later lessons, but stick with me as we work through this). If you make this as w/v, v/v or w/w you can end up with different final concentrations.

The molecular weight of glycerol is: 92.09 g/mol

maths4biosciences.com Introduction to Percentage SolutionsPage � of �1 2

http://maths4biosciences.com



And the density (how much 1 ml weighs) of glycerol is: 1.261 g/ml

If you were to make 100 ml of a 5% (w\v) solution (see a later lesson for an explanation of the calculation), you would take 5 g of glycerol and make up to 100 ml with water. This would be equivalent to 50 g in 1 litre, so it would be 50/92.09, or 0.543 M solution (molarity calculations are also explained in another lesson).

If you were to make 100 ml of a 5% (v/v) solution (see a later lesson for an explanation of the calculation), you would take 5 ml of glycerol and make up to 100 ml with water.

As the density of glycerol is 1.261 g/ml, then 5 ml would be 6.305 g per 100 ml or 63.05 g per litre. This would be a 63.05/92.09, or 0.685 M solution, which is clearly different from the concentration of the 5% (w/v) solution (0.543 M) calculated above.

Finally, a 5% (w/w) solution. In this case, it is mass of the compound divided by the mass of the compound plus the mass of liquid (this is explained in more detail in a later lesson). So in the case of glycerol, a 5% (w/w) solution in water would be 5 g of glycerol mixed with 95 ml of water (as 1 ml of water weighs 1 g). As we know the density of glycerol is 1.261 g/ml, the 5 g would be 5/1.261, or 3.965 ml. This means the final volume would be 95 + 3.965, which is 98.965 ml. This 98.965 ml contains 5 g of glycerol, so this would be 50.523 g per litre. As the molecular weight of glycerol is 92.09 g/mol, then the concentration would be 0.549 M.

As can be seen, for a 5% glycerol solution we have 3 possible final concentrations:

• 5% w/v = 0.543 M

• 5% v/v = 0.685 M

• 5% w/w = 0.549 M

Hence, as you can see when working with percentage solutions, it is very important to state whether it is weight by volume (w/v), volume by volume (v/v), or weight by weight (w/w).

Summary

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 lesson? 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 email me at [email protected] and tell me what you would like included and why, or what the mistake is and where it can be found. Thanks.


mailto:[email protected]






Percentage Solutions - IntroductionIntroduction

For some reason, percentage solutions cause students problems. I have no idea why this should be the case, and once you have completed this course percentage solutions should no longer cause you any problems.

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.

Why are they used?


Why are their three types of percentage solutions?

This is slightly difficult to explain, particularly before we have tackled the calculation of percentage solutions in the subsequent lessons.

However, there are three types of percentage solution:

1. Percentage weight by volume (w/v)

2. Percentage volume by volume (v/v)

3. Percentage weight by weight (w/w)

So, why?

Well, there are three types of percentage solution to take in to account the problems associated with measuring out material. All three can be used for describing the concentration of a solution, and using the wrong type at the wrong time can result in solutions of an incorrect final concentration.

In the different types, w/v, v/v, and w/w, the first letter represents the solute (the thing being dissolved or diluted, and the second letter represents the solvent (e.g. water) into which the solute is being mixed or dissolved.

When measuring out a powder, e.g. sodium chloride, you would use weight as the material is a solid and measuring out a consistent volume of sodium chloride would be very difficult as the volume would be dependent on crystal size and how those crystals settle. So, for a solid, we tend to use weight, and so would use w/v (weight by volume) or w/w (weight by weight), where the solid is represented by the first w (weight).

When measuring out a liquid, you could use weight (w) or volume (v) to describe the amount of liquid being used. However, it should be noted that weight (w) or volume (v) cannot be used interchangeably (i.e. if it says v/v you can't switch to w/v) in such situations as you will end up with the wrong concentrations.

For example, let's say you have to make a 5% solution of glycerol (we will get into the calculations more in the later lessons, but stick with me as we work through this). If you make this as w/v, v/v or w/w you can end up with different final concentrations.

The molecular weight of glycerol is: 92.09 g/mol




And the density (how much 1 ml weighs) of glycerol is: 1.261 g/ml

If you were to make 100 ml of a 5% (w\v) solution (see a later lesson for an explanation of the calculation), you would take 5 g of glycerol and make up to 100 ml with water. This would be equivalent to 50 g in 1 litre, so it would be 50/92.09, or 0.543 M solution (molarity calculations are also explained in another lesson).

If you were to make 100 ml of a 5% (v/v) solution (see a later lesson for an explanation of the calculation), you would take 5 ml of glycerol and make up to 100 ml with water.

As the density of glycerol is 1.261 g/ml, then 5 ml would be 6.305 g per 100 ml or 63.05 g per litre. This would be a 63.05/92.09, or 0.685 M solution, which is clearly different from the concentration of the 5% (w/v) solution (0.543 M) calculated above.

Finally, a 5% (w/w) solution. In this case, it is mass of the compound divided by the mass of the compound plus the mass of liquid (this is explained in more detail in a later lesson). So in the case of glycerol, a 5% (w/w) solution in water would be 5 g of glycerol mixed with 95 ml of water (as 1 ml of water weighs 1 g). As we know the density of glycerol is 1.261 g/ml, the 5 g would be 5/1.261, or 3.965 ml. This means the final volume would be 95 + 3.965, which is 98.965 ml. This 98.965 ml contains 5 g of glycerol, so this would be 50.523 g per litre. As the molecular weight of glycerol is 92.09 g/mol, then the concentration would be 0.549 M.

As can be seen, for a 5% glycerol solution we have 3 possible final concentrations:

• 5% w/v = 0.543 M

• 5% v/v = 0.685 M

• 5% w/w = 0.549 M

Hence, as you can see when working with percentage solutions, it is very important to state whether it is weight by volume (w/v), volume by volume (v/v), or weight by weight (w/w).

Summary


Missing something? A mistake?

Have I missed something in this lesson? 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 email me at [email protected] and tell me what you would like included and why, or what the mistake is and where it can be found. Thanks.







Introduction to weight by volume (w/v)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 lesson.

maths4biosciences.com Introduction to weight by volume (w/v)Page � of �1 1



Worked examples (w/v) - what is the percentage?In the following examples we will look at how to calculate the percentage of a solution for a given number of grams of compound in a given final volume.

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.


For this you will need the equation:

% = (m / v) x 100

Where:


Hence we get:

% = (4 / 50) x 100

% = 0.08 x 100

% = 8% (w/v)


maths4biosciences.com Worked examples (w/v) - what is the percentage?Page � of �1 3





% = (m / v) x 100

Where:


Hence we get:

% = (2 / 50) x 100

% = 0.04 x 100

% = 4% (w/v)


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?


% = (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.



% = (m / v) x 100

Where:


Hence we get:maths4biosciences.com Worked examples (w/v) - what is the percentage?

Page � of �2 3



% = (6 / 200) x 100

% = 0.03 x 100

% = 3% (w/v)




% = (m / v) x 100

Where:


Hence we get:

% = (9 / 225) x 100

% = 0.04 x 100

% = 4% (w/v)


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.


% = (m / v) x 100

Where:


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.

maths4biosciences.com Worked examples (w/v) - what is the percentage?Page � of �3 3


Worked examples (w/v) - how many grams?In the following examples we will look at how to calculate how many grams of compound are required to make up a solution..

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.


For this you need the equation:

m = (% / 100) x v

Where:


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.

maths4biosciences.com Worked examples (w/v) - how many grams? Page � of �1 3



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


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




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




m = (% / 100) x v

Where:





Hence we get:

m = (3 / 100) x 200

m = 0.03 x 200

m = 6.0 g




m = (% / 100) x v

Where:


Hence we get:

m = (4 / 100) x 225

m = 0.04 x 225

m = 9.0 g


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.


m = (% / 100) x v

Where:


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.



How to make a w/v solutionMaking a w/v solution is fairly easy.

You take the number of grams of 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 throughly 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.

maths4biosciences.com How to make a w/v solution Page � of �1 1



Percentage weight by weight (v/v) solutions - 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 molecular weight of the compound.

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.


% = (v1 / vt) x 100

Where:

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

The above equation can be rearranged so as to give you the number of ml you would need to make up a volume of solution of a given percentage. 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 in fact the sum of v1 plus the volume of the second liquid that is used to dilute v1. For further information on this see "How do I make a v/v solution?" later in this course.

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 lesson.

maths4biosciences.com Percentage volume by volume (v/v) solutions Page � of �1 1



Worked examples (v/v) - what is the percentage?In the following examples we will look at how to calculate the percentage volume by volume of a solution for a given number of ml of one liquid made up to a final given final volume.

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?


% = (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.



% = (v1 / vt) x 100

Where:


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.

maths4biosciences.com Worked examples (v/v) - what is the percentage? Page � of �1 2





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

Where:


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.

maths4biosciences.com Worked examples (v/v) - what is the percentage? Page � of �2 2



Worked examples (v/v) - how many ml?In the following examples we will look at how to calculate the number of ml we need of one liquid to make up a volume of a percentage liquid.

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


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 milli-litres (ml) of glycerol would you need to make 50 ml of an 8% (v/v) solution?


v1 = (% / 100) x vt

Where:


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.

maths4biosciences.com Worked examples (v/v) - how many ml? Page � of �1 3



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


v1 = (% / 100) x vt

Where:


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 milli-litres (ml) of water would be required to make 100 ml of a 4% (v/v) solution of glycerol?


v1 = (% / 100) x vt

Where:


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 milli-litres (ml) of water would be required to make 80 ml of a 4% (v/v) solution of glycerol?


v1 = (% / 100) x vt

Where:


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

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.



How to make a 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 95 ml of water to a 100 ml beaker.

2. Stir.

3. To the moving liquid add 5 ml of glycerol, with stirring.

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.

maths4biosciences.com How to make a v/v solution? Page � of �1 1



Percentage weight by weight (w/w) solutionsPercentage 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 molecular weight of the compound.

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 final weight of the solution, multiplied by 100.


% = (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 so as to give you the number of grams of the compound you would need to make up a final mass of solution of a given percentage. 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 in fact the sum of m1 plus the mass of the liquid that is 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)

It should be noted that if you were making a 2% (w/w) solution of NaCl that 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 volume of the solution may not be 100 ml, it is weight that is important here, and not volume.

maths4biosciences.com Percentage weight by weight (w/w) solutions Page � of �1 1



Worked examples (w/w) - what is the percentage?In the following examples we will look at how to calculate the percentage weight by weight of a solution for a given number of grams of one compound made up to a final given weight.

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?


% = (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 in to a solution that has a final mass of 100 g is a 4% (w/w) solution.



% = (m1 / mt) x 100

Where:


Hence we get:

% = (4 / 50) x 100

% = 0.08 x 100

% = 8 % (w/w)

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

maths4biosciences.com Worked examples (w/w) - what is the percentage? Page � of �1 3





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

Where:


Hence we get:

% = (6 / 250) x 100

% = 0.024 x 100

% = 2.4 % (w/w)

Therefore, 6 g of the compound made up in to a solution that has 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:


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 solution of 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:


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 solution of 9% (w/w) solution.



Worked examples (w/w) - how many grams?In the following examples we will look at how to calculate the number of grams we need of one compound to make up a solution to a final given weight.

Example 1: How many gram of compound would you need to make a 5% (w/w) solution if the final total weight of the solution 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 that has a final mass of 100 g.

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


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 that has a final mass of 50 g.

maths4biosciences.com Worked examples (w/w) - how many grams? Page � of �1 2



Example 3: How many gram of compound would you need to make a 2.5% (w/w) solution if the final total weight of the solution 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 that has a final mass of 250 g.

maths4biosciences.com Worked examples (w/w) - how many grams? Page � of �2 2



How to make a w/w solution?Making a w/w solution is fairly easy.

You weigh out the number of grams of 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 which would require heating to dissolve, and would also need to be autoclaved to sterilise (the autoclaving would heat the solution and cause the agar to dissolve). As the solution has a tendency to stick to glassware it can be difficult to work with. By using w/w it 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.

maths4biosciences.com How to make a w/w solution? Page � of �1 1



Percentage Solutions SummaryAs 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


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


maths4biosciences.com Percentage Solutions SummaryPage � of �1 1



Percentage Solutions - Introduction...Percentage Solutions - Introduction Introduction For some...

Documents

Transcript of Percentage Solutions - Introduction...Percentage Solutions - Introduction Introduction For some...