# Essential Maths for Medics and Vets Reference Materials Module 2. Amount and Concentration.

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1 2 Amount and concentration: making and diluting solutions 2 Amount and concentration; making and diluting solutions... 2A Rationale... 2B Distinguishing between amount and concentration, g and %w/v... 2C Distinguishing between amount and concentration, moles and molar...2 2D Practice converting g/l to M and vice versa...3 2E Diluting Solutions...5 2F Practice calculating dilutions...6 Summary of learning objectives...7 2A Rationale Biological and biochemical investigations rely completely upon being able to detect the concentration of a variety of substances. For example, in diabetics it is important to know the concentration of glucose in the blood and you may also need to be able to calculate how much insulin would need to be dissolved in a certain of saline so as to give the right amount in a ml injection. It is also vitally important that you perform these calculations quickly, reliably and accurately. This section aims to help you understand how to go about doing these calculations and to give you plenty of practice so that you can do them quickly and reliably. 2B Distinguishing between amount and concentration, g and %w/v The amount of a substance is often measured by how much it weighs i.e. in g or mg or kg etc. The concentration is the amount divided by the it is dissolved in. amount concentration= For example you might dissolve 0.9g of sodium chloride in 00mL of water. Sometimes this is referred to as 0.9% w/v since it is 0.9g weight divided by 00mL and w/v stands for weight per. The concentration of this solution could be described as g/ml or 9 mg/ml. Alternatively you could notice that if 0.9g is dissolved in 00mL, then 9g would be dissolved in 000ml (=L) and the concentration could also be written as 9g/L Page of 7

2 2C Distinguishing between amount and concentration, moles and molar Another way of representing the amount of a substance is to use the concept of mole. A mole is simply x 0 23 molecules (this is Avogadro s number ) Eggs come in dozens (2), molecules come in moles (6 x 0 23 ) The mass of a mole of molecules is the formula weight in grams. This is sometimes called molar mass with symbol M. Sometimes it is referred to as molecular weight. Sometimes it is called relative molecular mass since the mass of a carbon element is designated to be 2 and other molecules or elements are given masses relative to that. The molar mass of a small molecule such as an amino acid (eg glycine is NH 2 CH 2 COOH and has molecular mass = g.mol - ) is much less than the molar mass of a protein (eg haemoglobin has molecular mass ~ g.mol - ) x 0 23 glycine molecules (or mole) would weigh g whereas x 0 23 haemoglobin molecules (or mole) would weigh ~64500g (or 64.5kg) The important thing to remember for calculations is that the molecular weight/molecular mass/molar mass call it what you will has the units of g.mol - To convert between an amount in g and an amount in moles, Amount (in g) = Amount (in mol) x Molar Mass (g.mol - ) Sometimes the term Dalton is used for molecular mass. For example the molecular mass of glycine is 75 Da while the molecular mass of haemoglobin is 64.5kDa (or Da). The concentration of a solution of mole dissolved in L is M ( one molar ) M = mole litre Don t confuse this symbol M for mol/l with M for molar mass. It s unfortunate they have been used for two quite different things. In the case of glucose, a M solution would have mole = 80 g dissolved in L. So another way of saying that is M glucose solution = 80 g/000 ml = 0.8 g/ml Alternatively you could call it 8 % (w/v). You need to be able to convert easily and accurately between different ways of expressing concentration. Page 2 of 7

3 2D Practice converting g/l to M and vice versa Example : Converting g/l to M You have weighed out 20 g KCl (formula weight of KCl is g.mol - ) and dissolved it in 500 ml, what is the concentration in M? Work out how many moles are equivalent to 20 g. Rearranging: Amount (in g) = Amount (in mol) x Molar Mass (g.mol - ) Amount (in moles) = amount (in g) molar mass (in g.mol - ) 20g gmol. = = mol Now work out the concentration by dividing the amount (in mol) by the (in L) Concentration (M) = amount (in mol) (in L) = mol 0.5L = M Example 2. Converting M to g/l You have a M solution of NaCl, what is the concentration in % (w/v)? M = mol / L mol of NaCl = g.mol - Concentration (M) = mol L = 58.44g L = 5.844g 00 ml = 5.8% (w/v) Page 3 of 7

4 Example 3. You are given 580 mg of adrenaline and need to make a mm solution. What of water do you need to dissolve it in? The molar mass of adrenaline is 83.2 g.mol - First you need to convert 580 mg to moles: Amount (in moles) = amount (in g) molar mass (in g.mol - ) 0.58g = mol 83.2 gmol. = Then work out the by rearranging the equation: Concentration (M) = amount (in mol) (in L) becomes: Volume (in L) = amount (in mol) concentration (in M) Volume (in L) = mol 0.00 mol.l - = 3.66 L Example 4. You have a mm solution of adrenaline and wish to dispense a which contains 0 mg. What do you need to dispense? First convert 0mg to moles: Amount (in moles) = amount (in g) molar mass (in g.mol - ) 0.0g 83.2 gmol. = 5.46 x 0-5 mol = Page 4 of 7

5 = = = amount concentration 5.46 x 0-5 mol x 0-3 mol.l x 0-2 L = L = 54.6 ml Example 5. You want to make 0 ml of a 5 mm solution of adrenaline, how much should you weigh out? Amount = x concentration = 0.0 L x 5 x 0-3 mol.l - = 5 x 0-5 mol = 5 x 0-5 mol x 83.2 g.mol - = 96 x 0-5 g = 9.6 mg 2E Diluting Solutions Simple Dilutions Make a :0 dilution (say this as a one in ten dilution ). Take part of your stock solution and add 9 parts of solvent (usually water but sometimes alcohol or other organic solvent). If your original sodium chloride solution is M, and you dilute it :0, then you might take 0ml of M NaCl and 90ml water, ( = 00) or you could take ml of M NaCl and 9ml water, ( + 9 = 0) or 0.2 ml M NaCl and.8ml water ( = 2) In all cases you are diluting by the same factor. The concentration of the resulting solution is M /0 = 0.M where 0 is the dilution factor. [Although it is really quite obvious, you can convince yourself that the resulting concentration is 0.M by the following: Take 0 ml of M NaCl, how much NaCl is there in that? Answer: amount concentration= amount = concentration Amount in 0ml = 0ml x mol.l - x 0.00 L.mL - = 0.0mol So imagine that you are taking 0ml which has 0.0mol salt in it, and you put it into a total of 00ml. Page 5 of 7

6 amount 0.0mol concentration = 0.M = 0.00L = ] For example, a :5 dilution = unit of diluent + 4 unit s of solvent; dilution factor = 5 :2 dilution = unit of diluent + unit of solvent; dilution factor = 2 :7.4 dilution = unit of diluent unit s of : dilution factor = 7.4 Serial Dilutions Serial dilutions are a quick and convenient way to prepare a wide range of concentrations. Typically, they are done in powers of 0 or 00 (but not always). So for example: Take a mm stock solution of adrenaline, dilute it :0 (0.ml + 0.9ml) to give a 0.mM solution dilute this :0 to give a 0.0mM solution dilute this :0 to give a 0.00mM solution dilute this :0 to give a 0.000mM solution and so on This approach allows you to prepare very dilute solutions from a concentrated stock much more accurately than the alternative. 2F Practice calculating dilutions The most common task you will be asked to do is to prepare a particular concentration from a stock solution. So say you are asked to prepare 0.32M sucrose from a M sucrose stock. There are two ways of approaching this and these are illustrated below. Think about them both and then choose whichever makes the most sense for you. Method : Step : work out what the dilution factor is.. /0.32 = 3.25 Step 2: turn this into of diluent and solvent.. part of M sucrose plus 2.25 parts water gives 3.25 parts of 0.32M sucrose Page 6 of 7

7 you could write this as ml x M sucrose ml water 3.25ml x 0.32M sucrose or you could scale it up to whatever you want, say 32ml x M sucrose + 68ml water 00 ml x 0.32M sucrose Method 2: The alternative method is this: Start by noticing that amount = concentration x or A = C.V Now if you take a V of your concentrated solution (concentration = C) Then the amount you have in your pipette is A. Now if you put that amount into a new V2, the amount is still constant, but now because you have a bigger, the concentration C2 = A/V2. But from before, A = C.V, so C.V = C2.V2 What this is saying is that if the amount is the same in both solutions, then their concentrations and s must be related. You can use this relationship to calculate the required V2 to get a particular concentration. Taking our earlier example, C = M, V = ml, C2 (the desired concentration) = 0.32M what is the resulting? M x ml = 0.32M x V2. V2 = 0.32 ml V2 = 3.25ml What about if you were asked to make 00ml of 0.32M sucrose? Now you have C=M, C2 = 0.32M, V2 = 00ml and V is the unknown. M x V = 0.32M x 00ml V = 32ml Summary of learning objectives At the end of this section you should be able to: a. To understand the difference between amount and concentration b. To be able to calculate the concentration of a solution in a variety of formats including M, g/l, %v/v, %w/v c. To be able to quickly and reliably convert from one unit to another d. To be able to quickly and reliably calculate the appropriate way to prepare a dilution Page 7 of 7

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