This is the sixth part of our seven-part series on pharmaceutical calculations, this time dealing with molecular weights and moles.

**Concentrations****Dilutions****Formulations****Calculation of Doses****Density & Displacement Values***Molecular Weights & Moles***Parenteral Solutions & Isotonicity**

#### Section One

##### Example 1

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

**Step 1** – Find the Molecular Weight of Sodium Chloride

\[Na + Cl => 23 + 35.5\Rightarrow58.5\]

**Step 2** – Find the Total Quantity of Sodium Ions

From Step 1, we found that 23 out of 58.5 consists of Na. Thus, for the remainder of the calculation, all we have to do is multiply this by 600mg.

\[\frac{23}{58.5}\times600mg = 235.9mg\]

Therefore, there are 235.9 milligrams of sodium ions in a 600mg tablet of Sodium Chloride.

##### Example 2

*What percentage of a 600mg tablet of Lithium Carbonate contains Lithium?*

**Step 1** – Find the empirical formula for Lithium Carbonate and calculate its molecular weight.

\[\L_2CO_3 – Li(x^2) + C + O(x^3)\to14 + 12 + 48\to74\]

**Step 2** – Calculate the percentage of Lithium inside this Lithium Carbonate compound.

Lithium occupies 14 parts of the 74 part compound. Thus, Lithium occupies 18.92% of the compound.

##### Example 3

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

So to answer this question, we first need to determine how much iron is in a 200mg tablet of ferrous sulphate. The empirical formula for ferrous sulphate is FeSO_{4} which works out as a molecular weight of **152** [Fe (56) + S (32) + O_{4} (16 x 4)].

We can see from the empirical formula that Fe occupies 56 parts of the total 152 parts of ferrous sulphate which works out as a percentage of 36.84%.

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

Thus, we have 73.68mg of iron in a 200mg tablet of ferrous sulphate.

Now we can proceed to answer the remainder of the problem. We need to find a weight of ferrous gluconate which contains the same amount of iron as that in ferrous sulphate i.e. 73.68mg.

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

After much laborious table checking, you’ll determine ferrous gluconate to have a molecular weight of 482. We can also see from the formula that only 1 molecule of iron i.e. 56 parts of the 482, consist of it. We can use proportional sets to work out the equivalent quantity of iron of ferrous gluconate as illustrated below.

Ferrous Gluconate | Iron | |
---|---|---|

Amount of Iron (mg) |
Z | 73.68 |

Molecular Weight |
482 | 56 |

\[\frac{Z}{482} = \frac{73.68}{56}\]

\[\frac{Z}{482} = 1.316\]

\[Z = 1.316\times482\]

\[Z = 634.2\]

In conclusion, 634.2mg of ferrous gluconate contains the same quantity of iron as 200mg of ferrous sulphate.

##### Example 4

*Calculate the amount of adrenaline acid tartrate required in each millilitre 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.*

This is a relatively straightforward example as all we need to do is set up proportional sets.

Amount (micrograms) | Molecular Weight | |
---|---|---|

Adrenaline Acid Tartrate |
Z | 333 |

Adrenaline |
100 | 183 |

\[\frac{z}{100} = \frac{333}{183}\]

\[\frac{z}{100} = 182\]

In conclusion, each millilitre of injection contains 182 micrograms of adrenaline acid tartrate, which is itself equivalent to 100 micrograms of adrenaline.

##### Example 5

*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.*

This, too, can be resolved using proportional sets.

Equivalents | Molecular Weight | |
---|---|---|

Quinine Anhydrous Base |
100 | 324 |

Quinine Bisulphate |
169 | X |

Quinine Dihydrochloride |
122 | Y |

\[\frac{x}{324} = \frac{169}{100}\Rightarrow548\]

\[\frac{y}{324} = \frac{122}{100}\Rightarrow395\]

Hence, we can estimate the molecular weights for quinine bisulphate and quinine dihydrochloride as 548 and 395 respectively.

##### Example 6

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

According to Codeine Phosphate BP, it’s presented as the Hemihydrate. The molecular weights of the Codeine Phosphate Hemihydrate and Codeine Phosphate Sesquihydrate are 406 and 424 respectively.

We can now use this information to set up proportional sets.

Amount (mg) | Molecular Weight | |
---|---|---|

Codeine Phosphate Hemihydrate |
30 | 406 |

Codeine Phosphate Sesquihydrate |
Z |
424 |

\[\frac{z}{30} = \frac{424}{406}\]

\[z = 31.33\]

Thus, a 30mg codeine phosphate tablet would contain 30mg of codeine phosphate hemihydrates or 31.33mg of codeine phosphate sesquihydrate.

#### Section Two

Before we take a look at the final four examples we need to take a look at the concept of a mole. The mole is the SI unit for the amount of a substance. This ‘substance’ can be atoms, molecules or ions. Let’s take a look at two examples to illustrate this point.

- The atomic weight of iron is 56 therefore 1 mole of iron is 56g.
- The molecular weight of NaCl is 58.5 therefore 1 mole of NaCl is 58.5g. But NaCl is composed of two atoms, Na, and Cl. Therefore, 1 mole of sodium ions weighs 23g while 1 mole of chloride ions weigh 35.5g.

In addition, moles can be broken down into smaller equivalents. For example, 1 mole is the same as 1,000 millimoles (mmol), while 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, then this is referred to as a molar solution and expressed as 1 mol/L or 1 mol L-1. If we dissolve 58.5g of NaCl into sufficient water to produce 1L then we have made a 1 mol/L aqueous solution of sodium chloride.

All of these points will rear their head in the next few examples.

##### Example 7

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

As we learned above, 1 mole of calcium chloride consists of 1 mole of calcium and 2 moles of chloride ion.

In order that we can calculate the number of moles of chloride ion in 500mg of calcium chloride, we need to calculate the number of moles of calcium chloride. The molecular weight of calcium chloride is 110 (and given 100 is in grams, we need to convert 500mg to grams too). We can now set up the following proportional set.

Number of Moles |
z |
1 |

Amount of Calcium Chloride (g) |
0.5 | 110 |

\[Z = \frac{0.5}{110}\to0.0045\]

0.0045 moles is the same as 4.5 millimoles. Thus, 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 of chloride ions as calcium ions).

##### Example 8

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

This molecule has a molecular weight of 219 and it contains 6 moles of water (as can be seen from the formula). Again, don’t forget to convert the 500mg to grams!

Number of Moles of Water |
y | 6 |

Amount of Hydrated Calcium Chloride (g) |
0.5 | 219 |

\[\frac{y}{0.5} = \frac{6}{219}\]

\[y = 0.0137\]

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

##### Example 9

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

As usual, proportional sets save the day.

Number of Moles |
y | 1 |

Volume of Molar Solution (mL) |
20 | 1,000 |

\[\frac{y}{20} = \frac{1}{1000}\]

\[y = 0.02\]

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

If you’re not sure about doing a calculation such as this then setting up proportional sets is ideal. We can clearly see from the table that on the right hand side we have 1 mole in a Litre of solution, so we’re looking for the equivalent amount of moles (y) in only 20mL of that solution. Presenting it in this way should make the possibility of error evaporate.

##### Example 10

*How much NaCl is required to make 25mL of a 0.5 molar solution? How many moles of sodium ion will there 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 |

\[y = 0.0125\]

Therefore, there is 0.0125 moles of sodium chloride in 25mL.

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

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

##### Example 11

*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 Chloride.*

This is a straightforward follow-on question from Example 10. The only difference is the relationship with 0.9% Sodium Chloride. This simply refers to the fact that there is 9g of Sodium Chloride in 1L of solution.

Number of Moles |
x |
1 |

Mass of NaCl (g) |
9 | 58.5 |

\[x = \frac{9}{58.5}\]

\[x = 0.154\]

Therefore, there is 0.154 moles of NaCl in 1L of Sodium Chloride Infusion BP. This *must* mean, as in Example 10, that there is 0.154 moles of Sodium ions and 0.154 of Chloride ions.

0.154 moles is the same as 154 millimoles. Therefore, 154 millimoles of both sodium and chloride ions are contained in 1L of Sodium Chloride Infusion BP.

##### Example 12

*How much sodium fluoride will be required to make 250mL of a mouthwash containing 0.012 mol/L of sodium fluoride?*

The molecular weight of sodium fluoride is 42 therefore 1 mol/L solution will contain 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 |

\[x = 42\times0.012\]

\[x = 0.504\]

Therefore, 0.504g of sodium fluoride is required to make 1L of 0.012 mol/L mouthwash. Now we need to find out how much is required to make 250mL. This can be done simply by dividing by 4 (1,000mL/250mL).

Therefore, 0.126g of sodium fluoride is required to make 250mL of mouthwash.

This completes this section on molecular weights and moles. The penultimate section of this seven-part series deals with **Parenteral Solutions & Isotonicity**.