4PY019 Workbook 1-4 Moles and MW PDF

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4PY019

Workbook 1:4 Moles and relative molecular masses

Pharmaceutical Calculations Workbook Year 1, workbook 4 Calculations involving Moles and Relative Molecular Masses/Mr/Molecular Weight Student Study Guide

Facilitators

Required References

Names:

Required:

Michelle Key [email protected]

IMPORTANT For educational use only – Not to be used for the treatment of patients

This workbook

BNF

Chapter

9:

Section

on

Fluids

and

Electrolytes Optional:

Pack produced by: Dr Jan Daly

Introduction to Pharmaceutical Calculations

Dr Mike Daly

Rees, Judith A; Smith, Ian; Watson, Jennie

Jayne Nicholls Fourth edition

2015

Learning Outcomes

By the end of this workbook you should be able to: express the strength of solutions in terms of molar concentration perform calculations involving moles and relative molecular masses required to prepare electrolyte solutions  calculate the amount of a compound (expressed as the base) in a salt or hydrate of that compound, and vice versa

 

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4PY019

Workbook 1:4 Moles and relative molecular masses

PHARMACEUTICAL CALCULATIONS WORKBOOK 1:4

PHARMACEUTICAL CALCULATIONS INVOLVING MOLES, MILLIMOLES AND MICROMOLES

Introduction The determination of the concentration of a substance in a biological fluid is central to many areas of clinical practice (e.g. electrolytes in serum or glucose in urine). It is important, therefore, that concentrations are expressed in clear unambiguous terms. By far the most important unit defining an amount of a biological substance is the mole, with the corresponding concentration being the molarity. The mole is the SI base unit for ‘amount of substance’. One mole of a substance is the relative molecular mass, Mr, (also known as molecular weight or formula mass) of that substance, expressed in grams. As far as possible, the concentrations of specific compounds in serum, urine etc. are now expressed as molarities in clinical laboratories. Take glucose, for example: the Mr of glucose is 180, so: 1 mole of glucose = 180 g 1 mmol glucose = 180 mg 1 micromol glucose = 180 micrograms 100 micromol glucose = 18,000 micrograms = 18 mg Note the abbreviations: 1mmol = 1 millimole. In clinical practice, many medical laboratories use the abbreviation µmol (micromole) but during pharmaceutical calculations micromoles will always be written in full to avoid confusion between mmol and µmol (especially when hand-written). Concentrations in molarities are given by expressing the number of moles of the substance present in a defined volume of solution: a 1 molar (1 M) solution contains 1 mole of substance per litre (1 mol/L) a 1 millimolar (1 mM) solution contains 1 millimole per litre (1 mmol/L) mol and moles mean the same thing (an amount) moles/litre is a concentration and is expressed as mol/L, or M

 Ensure you know the difference between amounts and concentrations  WB1:4_v6_2020

4PY019

Workbook 1:4 Moles and relative molecular masses

So, if you dissolve 0.5 mol of a compound in solvent to produce a final volume of one litre, the concentration of the compound is 0.5 mol/L, or 0.5 M. 100 mL (one tenth) of this solution contains 0.05 mol (one tenth of 0.5mol). Prefix notation: the prefix notation is useful, as it allows rapid mental calculations to be performed (after some practice!). The following concentrations are all the same – just expressed in different units: 0.5 M = 0.5 mol/L; 0.5 mmol/mL; 0.5 micromol/microlitre; 0.5 nanomol/nanolitre By scaling both the units (amount and volume) in the expression of concentration up or down by a factor of 1000, the value of the concentration remains the same. Worked Examples 

How many micromoles (µmol) are contained in 2L of a 20 mM solution? By definition: 20 mM means 20 mmol/L, so 2 L will contain 2 x 20 = 40 mmol. 40 mmol = 40,000 micromoles



The Mr of NaCl is 58.5 How many mg are in 50 micromoles of NaCl? 1 mol of NaCl is 58.5 g, so 1 micromole is 58.5 micrograms, so 50 micromoles is 50 x 58.5 = 2,925 micrograms = 2.925 mg.



What is the molarity of a 1% w/v solution of anhydrous glucose? (molecular weight, Mr = 180) 1% (w/v) contains 1 g in 100 mL and, therefore, 10 g in 1L. A 1 M solution of glucose contains 180 g/L, so 10 g/L represents a molarity of 10/180 = 0.056 M (or 56 mM)

Electrolytes in Clinical Practice The molecules of chemical compounds in solution may remain intact, or they may dissociate into ions, which carry an electric charge. Substances with varying degrees of dissociation in solution are called electrolytes. Electrolyte ions in blood plasma include the cations Na+, K+, Ca2+ and Mg2+, and the anions Cl-, HCO3-, HPO42-, SO42-, organic acids and protein. Although urea and glucose are strictly speaking nonelectrolyes (as they are not dissociated in solution) they are sometimes listed under electrolytes in clinical texts. Electrolytes in body fluids play an important role in maintaining the acid-base balance. They also play a part in controlling body water volumes and help to regulate metabolism. In clinical practice, moles are ‘too big’ for everyday use as biological systems usually operate at the millimole down to the picomole scale. As electrolytes are dissociated in WB1:4_v6_2020

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Workbook 1:4 Moles and relative molecular masses

solution, molar concentrations (as millimoles per litre [mmol/L] and micromoles per litre [micromol/L]) are used in the SI system to express concentrations of electrolyte ions in a solution. Clinical biochemistry laboratories publish specific ‘reference ranges’ for various laboratory investigations, including electrolytes. Individual NHS Trusts may publish their own handbook of clinical biochemistry reference ranges. There is a section in Chapter 9 of the BNF called “Fluids and Electrolytes”, which is recommended reading. It includes a section on the electrolyte concentrations of the common intravenous fluids. For example, the following intravenous fluids contain: Sodium Chloride 0.9%w/v (this is isotonic with blood plasma) Sodium 150 mmol/L Chloride 150 mmol/L Compound Sodium Lactate (Hartmann’s) Sodium 131 mmol/L Potassium 5 mmol/L Bicarbonate 29 mmol/L Chloride 111 mmol/L Calcium 2 mmol/L Typical Adult Reference Ranges for Electrolytes in Serum (serum is blood plasma without the clotting agents) Electrolyte Calcium Chloride Magnesium Potassium Sodium

Reference ‘normal’ range (mmol/L) 2.10 – 2.55 96 – 110 0.70 – 1.00 3.6 – 5.0 137 - 145

FORMULAE USED IN CALCULATIONS INVOLVING MOLES For those of you who prefer to use formulae (rather than a stepwise approach as in the worked examples above) the following formula triangles are useful. Formula triangle 1

Cover the one you want to find and you will see how to calculate it, so mass (grams) = no. of moles x Mr (mass of one mole in grams) and no. of moles = mass (g) Mr WB1:4_v6_2020

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Workbook 1:4 Moles and relative molecular masses

As previously mentioned, in clinical practice, moles are too big for everyday use as biological systems most commonly operate at the millimole and micromole scale. Formula triangle 1 can therefore be ‘scaled down’ as follows: If mass is expressed in mg then you will be calculating millimoles of substance If mass is expressed in micrograms then you will be calculating micromoles of substance

Formula triangle 2 (not used as often as Formula triangle 1)

No. of moles = concentration (mol/L) x Volume (L)

In pharmaceutical practice, calculations concerning moles and molecular weights (Mr) often involve working out the amount of substance required to prepare an electrolyte replacement solution or supplement. As the use of calculators is not permitted for pharmaceutical calculations in Year 1, calculations involving molecular weights for 4PY019 will be limited to those calculations readily achieved without a calculator. However, calculations of

increasing

complexity

involving

molecular

weights

will

feature

in

pharmaceutical calculations across Years 2-4, where calculators may be used.

A typical question might be as follows: Worked Example A prescription is received for 500 mL of an infusion containing 60 mmol potassium chloride. How many grams of potassium chloride are needed to prepare this infusion? (Mr for potassium chloride is 74.5) First convert to the appropriate unit: we want to work out GRAMs of KCl therefore we should be working in moles, so 60 mmol = 0.06moles

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Workbook 1:4 Moles and relative molecular masses

Using the formula, mass (g) = no. of moles x Mr (mass of one mole in g) Amount of potassium chloride required = 0.06 x 74.5 = 4.47 g EXERCISE 1

1) How many millimoles of sodium are there in a 200 mL infusion containing sodium bicarbonate 8.4%w/v? (Mr for sodium bicarbonate is 84)

Answer =

millimoles

2) A child is prescribed an oral electrolyte solution containing 2.5 mmol potassium per 5 mL dose. The doctor requests 200 mL to be prepared. How many grams of potassium chloride are required to prepare this solution? (Mr for potassium chloride is 74.5). Give your answer to 2 decimal places

Answer =

grams

3) Given that the Relative Molecular Mass (Mr) of sodium chloride is 58.5, how many grams of sodium chloride are required to prepare 500 mL of a solution containing 100 millimoles/L? Give your answer to 3 decimal places

Answer =

grams

OTHER CALCULATIONS INVOLVING Mr (MOLECULAR WEIGHTS) a) Hydrates Many drugs and medicinal substances exist as both the anhydrous form and as hydrates. Consider glucose (Mr = 180). The molecular formula for glucose is C6H12O6 so the Mr is calculated by adding up the atomic masses of its components: WB1:4_v6_2020

4PY019

Workbook 1:4 Moles and relative molecular masses

C6 = 12 x 6 = 72 H12 = 1 x 12 = 12 O6 = 6 x 16 = 96 Total = 180 However, glucose is usually available as the monohydrate, i.e. it has one molecule of water attached to the molecule, commonly written as C6H12O6·H2O The Mr for glucose monohydrate is 180 + 18 (Mr for water) = 196 It therefore follows that equal weights of the anhydrous and monohydrate forms will contain different amounts of glucose In the BNF monograph for Glucose it states that it is usually available as Glucose BP (which is the monohydrate) but doses are usually expressed in terms of anhydrous glucose. The monograph gives a dose equivalence/conversion, stating that: 75 g of anhydrous glucose is equivalent to 82.5 mg of Glucose BP This can be checked, using simple proportion, as follows, knowing the Mr for both the anhydrous form and the monohydrate: (1 mole) anhydrous glucose(g) glucose monohydrate (g)

180 198

quantity required

75 ?

? = 198 x 75 = 82.5 g 180

This principle can be used for any medicinal compound to take into account the water of hydration in the molecule. EXERCISE 2 1) Glucose intravenous infusion contains 5% w/v glucose expressed as anhydrous glucose. How many grams of glucose monohydrate (Mr = 198) would be required to prepare 250 mL of glucose intravenous infusion 5% w/v? (Mr anhydrous glucose = 180)

Answer =

grams

The calculations that follow in this workbook are more easily done with the use of a calculator. Please continue to work through this workbook but be assured that you will not be examined on any of the following types of calculations in your Year 1 assessment. However, you WILL be examined on this material in Years 2-4. WB1:4_v6_2020

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Workbook 1:4 Moles and relative molecular masses

2) Ferrous sulfate paediatric mixture contains 60 mg of hydrated ferrous sulfate (FeSO4·7H2O) in each 5 mL of product. You are asked to prepare 100 mL of the mixture but only have the anhydrous form of ferrous sulfate available. How many mg of anhydrous ferrous sulfate are required (Mr of hydrate = 278, Mr of anhydrous ferrous sulfate = 152)? Express your answer to the nearest whole number.

Answer =

milligrams

b) Medicinal compounds and their salts Some medicines are prepared with different salts attached to the basic drug molecule. Iron salts are such an example. Although different iron salts may be prescribed it is normal to specify an equivalent amount of the base drug (elemental iron). Iron preparations are a common cause of accidental overdose in children and as little as 20 mg/kg of elemental iron can lead to symptoms of toxicity. If switching a patient from one form of iron to another it is therefore important to ensure that as close to an equivalent dose as possible is prescribed. The BNF recommends that when prescribing an iron preparation for children it is good practice to express the dose in terms of elemental iron and iron salt and to select the most appropriate preparation; both the iron salt and the formulation should be specified on the prescription. Consider a patient who has been taking one tablet containing 200 mg ferrous sulfate (dried) daily, for prophylaxis of iron deficiency anaemia, but needs to be switched to ferrous gluconate. Worked example How many mg of ferrous gluconate will provide an equivalent dose of elemental iron as 200 mg dried ferrous sulfate ? Note: dried ferrous sulfate is not the anhydrous form but is primarily the monohydrate, with a trace of tetrahydrate, so its ‘average’ Mr = 172, Ar elemental iron = 56, Mr ferrous gluconate [C12H22FeO14] = 446). Express your answer to 2 decimal places. First determine the amount of elemental iron in 200 mg of dried ferrous sulfate. The formula for ferrous sulfate indicates that it contains one atom of iron so: 1 mmol ferrous sulfate (mg) elemental iron (mg)

172 56

quantity required

200 ?

? = 56 x 200 = 65.12 mg (to 2 d.p.) 172 WB1:4_v6_2020

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Workbook 1:4 Moles and relative molecular masses

Now work out the quantity of ferrous gluconate that contains 65.12 mg of elemental iron. Again, the formula for ferrous gluconate indicates that it contains one atom of iron, so: 1 mmol ferrous gluconate (mg) elemental iron (mg)

446 56

quantity required

? 65.12

? = 446 x 65.12 = 518.63 mg (to 2 d.p.) 56

For some medicines the dose is expressed in terms of the base drug rather than the actual salt used. Such an example is erythromycin (a macrolide antibiotic), which is available as a number of different salts, depending on the formulation. In the BNF the dose is expressed in terms of erythromycin base. Worked example How many milligrams of erythromycin ethyl succinate are equivalent to 250 mg of erythromycin base? Express your answer to the nearest milligram. Mr erythromycin = 734, Mr erythromycin ethyl succinate = 862 1 mmol erythromycin base (mg) erythromycin ethyl succinate (mg)

734 862

quantity required

250 ?

? = 862 x 250 = 293.5967 mg 734

= 294 mg expressed to the nearest milligram

The salt forms of some drug bases may contain two or more of the base units attached to one salt unit. Consider quinine base and its salt quinine sulfate (which is available as the dihydrate). The molecular formula for quinine (base) is C20H24N2O2 [Mr 324] whereas quinine sulfate has the molecular formula C40H50N4O8S.2H2O [Mr = 783]. Although not strictly an accurate representation of the chemical structure it is helpful to represent the molecular formula of the sulfate as (C20H24N2O2)2.H2SO4.2H2O, from which it can be seen that one molecule of the sulfate contains two molecules of the base. This must be taken into account in calculations, as illustrated in the following example. Worked example What is the equivalent amount of anhydrous quinine base (in mg) in each 300 mg tablet of quinine sulfate (as the dihydrate)? Mr anhydrous quinine base = 324, Mr quinine sulfate dihydrate = 783. Express your answer to the nearest mg. From the formulae above, one molecule of quinine sulfate contains two molecules of quinine base. Therefore two molecules of quinine base (equivalent Mr = 2 x 324 = 648) contain the same amount of quinine as one molecule of the sulfate (Mr = 783). Therefore by simple proportion: quinine sulfate (mg) quinine base (mg)

783 648

300 ?

? = 648 x 300 = 248.276 mg 734

= 248 mg expressed to the nearest whole mg WB1:4_v6_2020

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Workbook 1:4 Moles and relative molecular masses

EXERCISE 3 1) How many mg of salbutamol sulfate are present in each tablet containing 4 mg of salbutamol base? Give your answer to 1 decimal place. Mr salbutamol (C13H21NO3) = 239; Mr salbutamol sulfate (C13H21NO3)2·H2SO4 = 577.

Answer =

milligrams

2) How many mg of erythromycin stearate are equivalent to 500 mg of erythromycin base? Express your answer to 1 decimal place. Mr erythromycin = 734, Mr erythromycin stearate = 1018.

Answer =

milligrams

Remember: 1 kg = 1000 grams 1 gram = 1000 milligrams 1 milligram = 1000 micrograms 1 microgram = 1000 nanograms 1 mole = 1000 millimoles 1 millimole = 1000 micromoles 1 micromole = 1000 nanomoles Did you struggle with any of the concepts covered in this workbook? Don’t forget there is maths support at the University of Wolverhampton The maths support centre is open to all students from any school studying any subject in any years who want help with any level of mathematics, statistics or numeracy. Further information can be found at: https://www.wlv.ac.uk/current-students/student-support/faculty-student-services/maths-support-centre/

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