Pharmacy Calculations## Pharmacy Calculations | Molecular Weights and Moles

**Knowing the theory behind and how to calculate molecular weights and moles is important. **

**Many pharmaceutical calculations depend upon a rich knowledge of these two concepts. The concepts themselves are not difficult. Rather than using kilograms or pounds, we’re using molecular weights and moles. You don’t need to visualize the concepts but you should have an idea of what we mean when we use these terms. The more you understand about the theory behind molecular weights and moles, the easier the pharmacy calculations become.**

Below, we review twelve examples to help you get to grips with the step-by-step approach needed to effectively tackle these questions. We also covered the **theory and practice of moles, molecular weights and Avogadro’s number** too. Both articles complement each other, be sure to check it out when you’ve finished the below review.

Section 1

**How many milligrams of sodium ions are in a 600mg tablet of sodium chloride?**

**Step 1** – Find the molecular weight of sodium chloride

Na + Cl = 23 + 35.5 = 58.5

Not sure where to find molecular weights? You can find these values, for each element, on the periodic table.

**Step 2** – Find total number of sodium ions

Of the molecular weight, 23 out of 58.5 consists of sodium – now, we simply need to multiply that value by the 600mg tablet:

23/58.5 x 600mg = 235.9mg

Answer – 235.9 milligrams of sodium ions in a 600mg tablet of sodium chloride.

**What percentage of a 600mg tablet of lithium carbonate contains lithium?**

**Step 1** – Find the empirical formula for lithium carbonate and calculate molecular weight.

Formula for lithium carbonate = Li_{2}CO_{3}

Molecular weight: 2(Li) + 1(C) + 3(O) = 2(7) + 1(12) + 3(16) = 74

**Step 2** – Calculate the percentage of lithium inside the tablet.

Lithium occupies 14 parts of the 74 part compound – meaning lithium occupies 18.92% of the compound.

**What weight of ferrous gluconate contains the same quantity of iron as a tablet containing 200mg of ferrous sulphate (dried)?**

First, we need to determine how much iron is in a 200mg tablet of ferrous sulphate.

Empirical formula for ferrous sulphate is FeSO_{4} – meaning it has a molecular weight of **152** [Fe (56) + S (32) + O_{4} (16 x 4)].

Fe occupies 56 parts of 152 parts of ferrous sulphate, a percentage of 36.84%.

We multiply this percentage by 200mg to determine how much iron is in ferrous sulphate. Thus, 36.84% of 200mg works out to be 73.68mg.

We have 73.68mg of iron in a 200mg tablet of ferrous sulphate.

Now, we can answer the rest of the question. We need to find a weight of ferrous gluconate that contains the same amount of iron as the amount found in ferrous sulphate i.e. 73.68mg.

The empirical formula of ferrous gluconate is C_{12}H_{22}FeO_{12}.2H_{2}O.

Ferrous gluconate, then, has a molecular weight of 482. From the formula, we identify that only 1 molecule of iron i.e. 56 parts of the 482, consist of it. We use proportional sets to calculate the equivalent quantity of iron in ferrous gluconate.

Ferrous gluconate | Iron | |
---|---|---|

Amount of iron (mg) | z | 73.68 |

Molecular weight | 482 | 56 |

z/482 = 73.68/56}

z/482 = 1.316

z = 1.316 x 482

z = 634.2

Answer – 634.2mg of ferrous gluconate contains the same quantity of iron in 200mg of ferrous sulphate.

**Calculate the amount of adrenaline acid tartrate needed in each milliliter of adrenaline injection. Adrenaline injection contains 100 micrograms of adrenaline (as the acid tartrate) per mL. The molecular weights of adrenaline and adrenaline acid tartrate are 183 and 333, respectively.**

Let’s tabulate the question using proportional sets.

Amount (micrograms) | Molecular weight | |
---|---|---|

Adrenaline acid tartrate | z | 333 |

Adrenaline | 100 | 183 |

z/100 = 333/183}

z = 182

Answer – each milliliter of injection contains 182 micrograms of adrenaline acid tartrate – itself equivalent to 100 micrograms of adrenaline.

**Using the BNF equivalents for quinine (as anhydrous base), quinine bisulphate and quinine dihydrochloride and given that quinine anhydrous base has a molecular weight of 324, estimate the molecular weights of the other two salts.**

We can also resolve this problem using proportional sets:

Equivalents | Molecular weight | |
---|---|---|

Quinine anhydrous base | 100 | 324 |

Quinine bisulphate | 169 | X |

Quinine dihydrochloride | 122 | Y |

x/342 = 169/100 = 548

y/324 = 122/100 = 395

Answer – we estimate molecular weights for quinine bisulphate and quinine dihydrochloride as 548 and 395 respectively.

**How much codeine phosphate sesquihydrate is equivalent to 30mg of codeine phosphate?**

According to codeine phosphate BP, the compound presents as the hemihydrate. The molecular weights of the codeine phosphate hemihydrate and codeine phosphate sesquihydrate are 406 and 424, respectively. Using proportional sets:

Amount (mg) | Molecular weight | |
---|---|---|

Codeine phosphate hemihydrate | 30 | 406 |

Codeine phosphate sesquihydrate | z | 424 |

z/30 = 424/406

z = 31.33

Answer – a 30mg codeine phosphate tablet contains 30mg of codeine phosphate hemihydrate or 31.33mg of codeine phosphate sesquihydrate.

Section 2

What is “a mole”?

The mole is the SI unit for the amount of a substance. This ‘substance’ can be atoms, molecules or ions. Two examples help to illustrate the point:

- The atomic weight of iron is 56 – meaning 1 mole of iron weighs 56g.
- The molecular weight of NaCl is 58.5 – meaning 1 mole of NaCl is 58.5g. As NaCl is composed of two atoms, Na and Cl – there is 1 mole of sodium ions (which weighs 23g) and 1 mole of chloride ions (which weighs 35.5g).

1 mole is the same as 1,000 millimoles (mmol) and 1 millimole is equivalent to 1,000 micromoles etc.

If 1 mole of a substance is dissolved in a solvent and the resulting solution is made up to 1L, this solution is referred to as a molar solution and expressed as 1 mol/L or 1 mol L-1.

For example: if we dissolve 58.5g of NaCl into water to produce 1L, we have made a 1 mol/L aqueous solution of sodium chloride.

Calculations involving molecular weights and moles don’t need to be difficult. Once you grasp the basic concepts, the questions become that much easier to work out.

Let’s consider a few examples:

**Calculate the number of moles of chloride ion in 500mg of calcium chloride?**

First, let’s consider the molecular formula of calcium chloride – CaCl

This means that 1 mole of calcium chloride consists of 1 mole of calcium ions and 2 moles of chloride ions.

To calculate the number of moles of chloride ion in 500mg of calcium chloride, we need to know the number of moles of calcium chloride.

The molecular weight of calcium chloride is 110, meaning (we convert 500mg to 0.5g of calcium chloride):

Number of moles | z | 1 |

Amount of calcium chloride (g) | 0.5 | 110 |

z = 0.5/110 = 0.0045

0.0045 moles is the same as 4.5 millimoles.

500mg of calcium chloride contains 4.5 millimoles of calcium chloride.

It also contains 4.5 millimoles of calcium and 9 millimoles of chloride ions (given there was twice as many chloride ions).

**Calculate the number of moles of water in 500mg of CaCl _{2}.6H_{2}O?**

This molecule has a molecular weight of 219 – contains 6 moles of water (stated in the formula). Don’t forget to convert 500mg to grams:

Number of moles of water | y | 6 |

Amount of hydrated calcium chloride (g) | 0.5 | 219 |

y/0.5 = 6/219

y = 0.0137

Answer – 0.0137 moles is the same as 13.7 millimoles – meaning there are 13.7 millimoles of water in 500mg of CaCl_{2}.6H_{2}O.

**How many moles of solute are there in 20mL of a 1 mol/L solution?**

Let’s use proportional sets:

Number of moles | y | 1 |

Volume of molar solution (mL) | 20 | 1,000 |

y/20 = 1/1000

y = 0.02

Answer – there is 0.02 mole of solute in 20mL of a 1 mol/L solution.

**How much NaCl is needed to make 25mL of a 0.5 molar solution? How many moles of sodium ion will be in the final solution?**

Molecular weight of Na = 23

Molecular weight of NaCl = 58.5

Number of moles | y | 0.5 |

Volume of solution (mL) | 25 | 1,000 |

y = 0.0125

There is 0.0125 moles of sodium chloride in 25mL.

If we multiply the molecular weight of NaCl (namely, 58.5) by this molar quantity of 0.0125, we determine that 0.73125g of sodium chloride is contained within this 25mL (in a 0.5 molar solution).

Given that 1 mole of sodium chloride contains 1 mole of sodium ions, 25mL of a 0.5 molar solution of NaCl contains 0.0125 moles of sodium ions.

**How many millimoles of sodium and chloride ions are contained in 1L of sodium chloride infusion BP? Sodium chloride infusion BP contains 0.9% sodium shloride.**

Again, let’s tabulate the data within a proportional set:

Number of moles | x | 1 |

Mass of NaCl (g) | 9 | 58.5 |

x = 9/58.5

x = 0.154

There is 0.154 moles of NaCl in 1L of sodium chloride infusion BP – meaning there is 0.154 moles of sodium ions and 0.154 of chloride ions.

0.154 moles is the same as 154 millimoles.

Answer – 154 millimoles of both sodium and chloride ions are contained in 1L of sodium chloride infusion BP.

**How much sodium fluoride is needed to make 250mL of a mouthwash containing 0.012 mol/L of sodium fluoride?**

The molecular weight of sodium fluoride is 42 – meaning 1 mol/L solution contains 42g in 1L.

Using this information, we can now set up a proportional set.

Number of moles | 0.012 | 1 |

Amount of sodium fluoride (g) | x | 42 |

x = 42 0.012

x = 0.504

0.504g of sodium fluoride is needed to make 1L of 0.012 mol/L mouthwash. Now, we need to learn how much is needed to make 250mL. To do so, we simply divide by four.

Answer – 0.126g of sodium fluoride is needed to make 250mL of mouthwash.

**In another article, we covered the theory behind moles, molecular weights and Avogadro’s number – an article that complements much of what you have learned here. For now though, there are a few take home points to consider. First, you should have a solid idea of what we mean by molecular weights and moles. **

You should know how to find a compounds molecular weight, know how many ions are present for each element and know how to calculate the number of moles in that compound. The above twelve examples are an excellent introduction to molecular weight and moles.

**If you’re still not sure – keep practicing! Without practicing any pharmacy calculations, you’re doomed to fail. With the right approach, you can ace these questions in your next pharmaceutical calculations exam!**

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