For this module, we will tackle calculations considering moles, molecular weights and associated concentrations as they relate to some common dosage forms in your daily pharmacy practice. These calculations can, at first, seem quite complicated and difficult but with practice and a systematic approach, they can be tackled easily. A good understanding of chemical stoichiometry and equations is important. So, in preparing for these calculations, make sure you have revised this section from your lecture notes and other resources.

Converting between moles and mass appears straightforward for simple atoms, but it can get confusing when compounds are involved. Often, there is confusion about whether the atomic weight of the individual element or molecular weight of compound should be used. As a standard, it is important to read the question carefully and understand what is being asked.

Below, we show you five worked examples demonstrating when the use of atomic weight versus molecular weight is most appropriate and situations where both can be used in slightly different ways to yield the correct answer. We have also included the wrong way round of calculating these questions, where appropriate, to warn you of potential traps.

See example calculations for extended matching questions and dilution and compounding calculations.

Please feel free to use a method that works best for you, and as always we are happy to hear your feedback.

**Question 1: Calculating elemental calcium (Ca2+) in a solid dosage compound**

A 40-year-old man is taking two tablets of Cacit effervescent tablets per day for the treatment of osteomalacia. The summary of product characteristics (smpc) states that “each tablet contains 1.25g calcium carbonate that when dissolved in water provides 500mg of calcium as calcium citrate”.

**How many millimoles of elemental calcium (Ca2+) does he consume in 30 days? Give your answer to the nearest whole number.**

**Atomic masses (g/mol)**

Calcium, Ca = 40

Carbon, C = 12

Hydrogen, H = 1

Oxygen, O = 16

**Formula**

Calcium citrate, Ca3 (C6H5O7)2

Calcium carbonate, CaCO3

Citric acid, C6H8O7

### Click here for the comments, working out and answer

First things first, **read the question carefully **in order to ascertain what is being asked. Only after you have done so will you be able to identify relevant information you need to do the calculation. Note that not all the information provided is relevant depending on how you decide tackle the question. You are required to calculate the number of millimoles of elemental Ca2+ for 30 days. You already know the mass of CaCO3 per tablet (1.25g) and that this yields 500mg of calcium as calcium citrate when dissolved in water. This is where most candidates get confused.

**Correct Method 1:**

Using the molecular weight of calcium carbonate, CaCO3

- Mass of CaCO3 consumed in 30 days = 2x1.25g x30 = 75g or 75000mg
- Molecular mass of CaCO3 = 40+12+ (16x3) = 100g/mol or 100mg/mmol
- 1mole of CaCO3 = 100mg; 75000mg of CaCO3 = X; use proportions
- X/75000mg = 1/100
- X = 75000/100 = 750mmoles
- From knowledge of stoichiometry: moles of Ca2+ = moles of CaCO3
- Moles of Ca2+ = 750mmoles or 0.75moles

**Tips and hints:**

- mg is equivalent to millimoles, as is grams to moles
- Working in grams rather than milligrams can reduce the risk of error

**Correct Method 2:**

Using the atomic weight of calcium, Ca2+: the tricky but easy method

- Amount of calcium, Ca2+ consumed in 30 days = 500mg x2x30 = 30000mg of elemental Ca2+ as citrate
- Please note the patient consumed 30000mg of elemental Ca2+ in calcium citrate NOT the calcium carbonate!
- Atomic weight of calcium, Ca2+ = 40
- 1mmole Ca2+ = 40mg/mmol; 30000mg = X
- X/30 000 = 1/40
- X = 30000/40 = 750mmoles of elemental Ca2+ or 0.75moles

**Wrong Method 1:**

- CaCO3 consumed in 30 days = 75g = 75000mg
- If 1 mole Ca2+ = 40mg
- 75000mg = X
- X/75000 = 1/40
- X = 75000/40
- X = 1875mmoles.
**This is WRONG**because you have calculated the amount of CaCO3 consumed in 30 days and tried to calculate number of moles of Ca using the proportions method with molecular mass of Ca atom. You should have used molecular mass of CaCO3 as above.

**Wrong Method 2:**

- Calcium citrate, Ca3 (C6H5O7)2 consumed in 30 days = 30g or 30000mg
- Molecular weight of calcium citrate = 498g/mol or 498mg/mmol
- If 1mmole of calcium citrate = 498mg; X = 30000mg,
- X/30 000 = 1/498
- X = 30000/498
- X = 60.24mmoles = 60 to nearest whole number.
**This is WRONG**because each tablet contains 500mg of calcium as calcium citrate. The fact that it is calcium citrate is irrelevant.

*Note that the citric acid formula is irrelevant!*

Finally, the correct answer is:

**750 mmoles or 0.75 moles of elemental Ca2+.**

**Reference:**

Accord Healthcare Limited (2019) Cacit 500mg effervescent tablets smpc [last accessed 07/02/2021].

**Question 2:**

**Converting elemental lithium (Li+) contained in a solid dosage form to equivalent elemental lithium (Li+) in a liquid dosage form from two different salts.**

A 40-year-old woman is taking lithium carbonate 1200mg per day. Due to swallowing difficulties, you would like to change her to the liquid version of lithium, Priadel (lithium citrate) 520mg/5ml.

**How many (mL) should she take each day to deliver an equivalent dose of elemental lithium (Li+)? Give your answer to the nearest whole number.**

**Formula**

Lithium carbonate = Li2CO3

Lithium citrate =Li3C6H5O7

**Atomic weights (g/mol)**

Lithium, Li = 7

Carbon, C = 12

Oxygen, O = 16

### Click here for the comments, working out and answer

As before, you need to be clear about what is being asked. You need a dose of lithium citrate that will deliver exactly the same amount of elemental lithium (Li+) as lithium carbonate. This is a very tricky question and you could easily go into the wrong path without even realising it.

**Correct Method 1:**

- Amount of millimoles of Li2CO3 consumed per day = 1200mg/ (molecular weight of Li2CO3)
- Lithium carbonate, Li2CO3 molecular weight = (2x7) + (1x12) + (3x16) = 74mg/mmol or 74g/mol
- mmoles of Li2CO3 = 1200mg/(74mg/mmol) = 16.21mmol
- 1mole of Lithium carbonate yields 2 moles of Li+ as follows: Li2CO3 = 2Li + CO3, therefore mmoles of elemental Li+ = 2 x 16.2 = 32.4mmol
- So you need the liquid to deliver 32.4mmol of elemental Li+
- Each 5mL contains 520mg lithium citrate
- mmoles of lithium citrate = 520mg/molecular weight of lithium citrate
- Molecular weight of lithium citrate = (3x7) + (6x12) + (1x5) + (7x16) = 210mg/mmol
- Lithium citrate = 520mg / (210mg/mmol) = 2.47619mmoles
- Looking at the formula for lithium citrate, Li3C6H5O7: 1mole of citrate = 3 moles of Li+
- mmoles of elemental Li+ = 3 x 2.47619 = 7.4285mmoles/5mL
- Volume of liquid required = 32.4mmol/(7.4285mmol/5mL) = 21.8mL
- 22mL to the nearest whole number.

**Wrong method 1:**

- Daily dose = 1200mg Li2CO3
- Molecular weight Li2CO3 = 74mg/mol
- Molecular weight Lithium citrate = 210mg/mol
- Simple proportions: molecular weight of lithium carbonate/ molecular weight of lithium citrate = mg lithium carbonate/mg lithium citrate
- 74/210 = 1200/X
- X = 1200x210/74
- X = 3405.405mg
- Volume = 3405.405mg/520mg x 5mL
- Volume = 32.74mL.
**This is WRONG**. Lithium carbonate has 2 elemental Li+ per molecule and Lithium citrate has 3, check the formulae. Be careful to check what is being asked.

Final answer is:

**22 mL to nearest whole number.**

**References:**

Sanofi-Aventis (2011) Priadel 520mg/5mL liquid smpc. [last accessed 07/02/2021]

Sanofi-Aventis (2010) Priadel 400mg tablets smpc. [last accessed 07/02/2021]

**Question 3:**

**Calculating millimoles of elemental calcium (Ca2+) in a solid preparation with multiple calcium salts.**

A 60-year-old man is taking Calvive effervescent tablets, 1 tablet BD. Each Calvive tablet contains: calcium lactate gluconate 2263mg and calcium carbonate 1750mg.

**How many millimoles (mmoles) of elemental calcium (Ca2+) does he consume in 28 days? Give your answer to the nearest whole number.**

**Atomic masses (g/mol)**

Calcium, Ca = 40

Carbon, C = 12

Hydrogen, H = 1

Oxygen, O = 16

**Formula**

Calcium carbonate = CaCO3

Calcium lactate gluconate = C9H16CaO10

### Click here for the comments, working out and answer

We cannot emphasize enough reading the question carefully so that you set off in the right path right from the beginning. For this question, you want to calculate the total mmoles of elemental Ca2+ from the two sources, calcium carbonate and calcium lactate gluconate.

**Correct Method 1:**

Molecular weights calculations:

- Calcium carbonate, CaCO3 = 40+12+16x3mg/mol = 100mg/mmol or 100g/mol
- Calcium lactate gluconate, C9H16CaO10 = (12x9) + (16x1) + (40x1) + (10x16) = 324mg/mmol
- Per tablet elemental Ca2+ (mmol): calcium lactate gluconate = 2263mg/(324mg/mmol) = 6.98mmol
- Per tablet elemental Ca2+ in CaCO3 = 1750mg/(100 mg/mmol) = 17.5mmol
- Total mmoles per tablet = 6.98mmol + 17.5mmol = 24.48mmoles
- Dose is 1BD for 28 days = 56 tablets
- mmoles = 56 x 24.48mmol = 1371mmoles to the nearest whole number

Final answer:

**1371mmoles or 1.371moles**

**Reference:**

GlaxoSmithKline Consumer Healthcare (2020) Calvive 1000 effervescent tablets (formerly Sandocal Tablets) smpc. [last accessed 07/02/2021]

**Question 4:**

**When should you use atomic weights versus molecular weights or both?**

A 45-year-old woman is taking two tablets of Adcal per day for the treatment of osteomalacia. The smpc states that “each tablet contains 1500mg calcium carbonate equivalent to 600mg elemental calcium (Ca2+)”.

**How many millimoles of elemental calcium (Ca2+) does she consume in 30 days? Give your answer to the nearest whole number.**

**Formula**

Calcium carbonate = CaCO3

**Atomic masses (g/mol)**

Calcium, Ca = 40

Carbon, C = 12

Hydrogen, H = 1

Oxygen, O = 16

### Click here for the comments, working out and answer

Be clear on what has been asked. It is mmoles of elemental calcium, Ca2+ for a compound for which both the amount of calcium carbonate and elemental calcium, Ca2+ per tablet has been stated. You should know that with this type of calculation, both atomic weight of Ca2+ or molecular weight of CaCO3 can be used to arrive at the same correct answer, but you need to know how to play around with your figures correctly without getting confused.

**Correct method 1:**

- Elemental Ca2+ consumed for 30 days = 600mg x 2 x 30 = 36000mg
- Atomic weight of Ca2+ = 40
- mmoles of Ca2+ = 36000/40
**900mmoles or 0.9 moles**

*The easiest method with only 3 steps if you really know what’s going on - There’s no need to worry about CaCO3 at all, when you have been given amount of calcium per tablet.*

**Correct method 2: using atomic weight of calcium, Ca2+**

- Mass of CaCO3 consumed in 30 days = 2x1.5g x30 = 90g = 90000mg;
- 1500mg CaCO3 = 600mg elemental Ca2+;
- 90 000mg = X;
- X/90 000mg = 600mg/1500mg;
- X = 90 000/1500 x 600mg = 36000mg
- mmoles of Ca2+ = 36000mg/(40mg/mmol) =
**900mmoles or 0.9 moles**

**Note, we have used the atomic weight of calcium, Ca2+ (40mg/mmol) to calculate mmoles!**

See how we can use the molecular weight of calcium carbonate, CaCO3 to arrive at the same solution in the next correct method 2.

**It is amazing if you are fluid and flexible enough to use both and be able recognise when you can use what and where. It comes with practice.**

**Correct Method 3:**

- CaCO3 = Ca2+ + CO3 2-, ratio of CaCO3 to Ca2+ moles is 1: 1 so number of moles of CaCO3 is exactly the same as number of moles of Ca2+
- Molecular weight for calcium carbonate, CaCO3 = 40 + 12 + (16x3) = 100g/mol or 100mg/mmol
- Dose as before for 30 days = 90g or 90000mg
- mmoles of CaCO3 = 90000mg/(molecular weight of CaCO3) = 90000mg/(100mg/mmol) =
**900mmoles or 0.9moles** - This is the same number of mmoles of calcium, Ca2+

**Wrong method 1:**

- Mass of CaCO3 consumed in 30 days = 2x1.5g x30 = 90g = 90000mg;
- 1mmole of Ca2+ = 40mg
- X = 90000mg
- X/90000 = 1/40
- X = 90000/40
- X = 2250mmole.
**This is WRONG**. You have used molecular mass of calcium with mass of CaCO3.

Final correct answer:

**900mmoles**

**Reference:**

Kyowa Kirin Ltd (2016). Adcal 1500mg chewable tablets smpc. [last accessed 07/02/2021]

**Question 5:**

**Conversion of millimoles to millilitres from a given % (w/v) strength.**

A 72-year-old woman is due to receive 4.5mmoles of Calcium Chloride (10% w/v) for hypocalcaemia by slow intravenous infusion.

**Calculate how many millilitres (mL) should be administered? Give your answer to 1 decimal place.**

**Formula**

Calcium chloride = CaCl2. 2 (H2O)

**Atomic masses (g/mol)**

Calcium, Ca = 40

Chlorine, Cl = 35.5

Hydrogen, H =1

Oxygen, O = 16

### Click here for the comments, working out and answer

As before, make sure you understand the question fully. This question is testing your competencies in converting units as well as dealing with mmoles and molecular weights. You have been asked to calculate a volume of drug that would deliver 4.5mmoles of calcium chloride.

**Correct method 1:**

- Convert 10% (w/v) to mg/mL as that would make it easier to calculate
- 10% (w/v) = 10g/100mL
- 10000mg/100mL = 100mg/mL
- Molecular weight for calcium chloride, CaCl2. 2 (H2O) = 40+ (2x35.5) +(1x4)+ (2x16) mg/mmol = 147mg/mmol or 147g/mol
- If 147mg = 1mmole
- Xmg = 4.5mmoles
- X/4.5 = 147/1
- X = 147/1x4.5mg
- X = 661.5mg
- 100mg = 1mL
- 661.5mg = Y
- Y/661.5 = 1/100
- Y = 6.61mL = 6.6mL to 1 decimal place

Correct answer:

**6.6mL**

**Reference:**

Martindale Pharma (2018) Calcium chloride intravenous infusion, 10% w/v smpc. [Last accessed 07/02/2021]

###### Conclusion

As you have seen from the examples above, you can manipulate atomic and molecular weights in different ways to yield your final correct answer. With practice, you should be fluid enough to know what figures to use without worrying about taking a wrong path. We do hope these scenarios will increase your problem solving skills and open up new ways for solving calculations around moles and molecular weights.

Good luck!

**Authors:**

**1. Luso Kumwenda**

MSc Community Pharmacy (Cardiff), B Pharm Hons (Zimbabwe), Independent Prescriber, MRPharms, Mentor at UKBPA & RPS

**2. Prof David R. Katerere**

PhD Pharmaceutical Science (Strathclyde),

Tshwane University of Technology, Platform Research Chair – Pharmaceutical and Biotech Advancement and Development in Africa (PbADA)

Acknowledgements: The questions were kindly provided by:

*Disclaimer: The questions and explanations presented here are for educational purposes only and do not replace your training, knowledge and application of professional judgement as a pharmacist or pre-reg or prov-reg pharmacist. The views in this article are our own and do not represent the views of any organisations we are associated with.*