Density and displacement values play a central role in formulation science and, by extension, pharmaceutical calculations. This section delves into what these terms mean, and why they are relevant to medicine today. This is the fifth part of our seven-part series on pharmaceutical calculations:

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

#### Introduction

The concept of density is a simple one. It’s an **essential characteristic of a substance**, in much the same way volume happens to be. Density is merely the name given to a relationship, a relationship between the mass of a substance on the one hand, and the space it occupies on the other. A substance which has a high mass, but occupies a small volume will, therefore, have a higher density that a substance which has a low mass but occupies a larger volume. Drugs are substances too, meaning these principles of density also apply.

In this section we provide an overview of density and displacement values – first explaining the nomenclature, before going on to analyse some examples of how and why they are used. The examples below are not intended to offer pharmacy students an exhaustive account of this subject, as it only intends to touch on some of the essential points that all pharmacy students should have some knowledge of.

##### What is Density?

At a certain temperature, say 20 degrees Celsius, 100g of water occupies a volume of 100mL – which is a 1:1 relationship. This directly proportional relationship is **not the same for all liquids, solids and semi-solids**. Given that different substances do not have a 1:1 ratio of weight:volume, it may become necessary for the formulator to convert either weight or volume. A given liquid may, for example, be easier to measure by volume as opposed to by weight.

To understand this essential concept, let’s take a look at **Example 1** and **Example 2** below.

##### Example 1

**A pharmacist is presented with two liquids: X and Y. There is 37.5mL of Liquid X and 18.2g of Liquid Y. Liquid X has a weight per mL of 0.65g, whereas Liquid Y has a weight per mL of 1.1g. With these figures in mind, convert Liquid X to its equivalent weight and convert Liquid Y to its equivalent volume in mL.**

First things first, what is the question asking us to do? Recognise that there are two parts to this question – where you must convert the volume of Liquid X to its equivalent weight, and to convert the weight of Liquid Y to its equivalent volume.

In order to answer these two questions, we need to know their respective densities. Mercifully, the questioner has provided these values – where Liquid X has a weight/mL of 0.65g, and Liquid Y has a weight/mL of 1.1g.

Let’s not forget the formula for density either, which is **mass/volume**. The question may be answered using the density formula or, alternatively, by using proportional sets.

Liquid X |
Liquid Y |
---|---|

37.5mL x 0.65g = 24.4g | 18.2g/ 1.1g = 16.5g |

From the table, we can see that 37.5mL of Liquid X has a weight of 24.4g, whereas 18.2g of Liquid Y has a volume of 16.5g.

Knowing the relationship between density, mass, and volume is vital for the formulator. For instance, it allows the formulator to measure a low viscosity liquid by volume as opposed to weighing the liquid itself.

##### Example 2

**The excipients for an ointment are tabled below:**

Liquid Paraffin |
20mL |

Zinc Oxide |
15mg |

Yellow Soft Paraffin to… |
100g |

**Calculate the final formula of 25g of the ointment.**

The first thing you should notice is that the **units are discrepant**.

Solids are measured in **grams**, but the liquid is measured in **mL**.

1mL of liquid paraffin does not occupy a volume of 1g meaning we need to find the density of liquid paraffin in order to make this conversion possible. The density of liquid paraffin is **0.88g/mL**.

With this information in mind, we need only multiply this density value by 20mL of liquid paraffin (as per the question details) to obtain the quantity of liquid paraffin in grams:

20mL x 0.88g/mL = **17.6g**

Thus, 20mL of Liquid Paraffin occupies a volume of 17.6 grams.

Liquid Paraffin |
17.6g | x |

Zinc Oxide |
15mg | y |

Yellow Soft Paraffin to… |
100g | 25g |

We can spot from this table that the difference between the values is **1/4**. (100g/25g)

Hence;

x = **4.4g**

y = **3.75g**

Both x and y give us a combined weight of 8.15g.

Given that we require a weight of yellow soft paraffin to 25g, we need to subtract the weight of both liquid paraffin and zinc oxide to give us the weight of yellow soft paraffin required.

25g – 8.15g = **16.85g** of yellow soft paraffin required.

This example reinforces three essential points. First, that students should analyse units used in each question. Second, that students should convert necessary weights and volumes and, thirdly, that students should understand that not all liquids, solids, and semi-solids share a 1:1 proportional relationship for conversion. These are core principles in understanding density and displacement values.

##### Displacement Values

Before turning to displacement values, we must first understand the concept of **displacement volume**. This is the volume occupied by a solute after it has been added to a given solution. All substances occupy volume when they are added to a solution. Some substances occupy greater, or lesser, volumes than other substances. Knowing how to calculate this volume allows one to formulate exact solutions – meaning there is no remaining excess. It would be akin to baking, in the sense that cakes rise to occupy space: and knowing (in advance) how much the cake will rise allows you to know how much baking soda to place into the cake in the first instance.

Perhaps baking is not, however, the most appropriate analogy one might draw. Let’s consider another example. If 15mL of water is added to 85mL of water, the resulting solution is 100mL – simple, right? But, what if we were to add 10g of sodium chloride to 90mL of water? The total volume will not be 100mL as with the earlier example. Instead, the volume will be greater than 100mL. The dissolved 10g of sodium chloride occupies a space greater than 10mL of water. The sodium chloride acts to **displace water, increasing the solutions volume**. Drugs, like sodium chloride, come with their own unique displacement volumes.

Let’s take a hypothetical drug called Drug D. This drug has a well-defined, measured displacement volume of 0.8mL/g. This means that **1g of Drug D displaces 0.8mL when it is dissolved in water**. Such displacement values are vital in, for instance, reconstituted powders. Precise concentrations and doses depend on the production of an exact volume of solution. If a solution were 10mL more than the solution required, it would dilute the concentration and, as a result, dilute the dose. It is for this reason that density and displacement values play such an important role in pharmacy and pharmaceutical calculations at large.

Let’s consider some sample problems, demonstrating how to use displacement values correctly.

##### Example 3

**Drug X has a displacement volume of 0.07mL/5mg. Calculate the volume of WFI needed to produce 1mL of an injection containing 5mg of Drug X.**

In the question above, we can see that Drug X has a displacement volume of **0.07mL/5mg**. This means that 5mg of Drug X displaces 0.07mL in solution.

We need to calculate the volume of WFI needed to produce 1mL of injection. Given that Drug X displaces 0.07mL, it must mean that 0.93mL is required to produce 1mL of injection. In other words, the displacement of the drugs 0.07mL plus the WFI of 0.93mL is enough to produce the injection required.

Let’s consider what might happen if we didn’t use 0.93mL. If we added 1mL to 5mg of Drug X without taking into consideration its displacement, the resulting solution would be 1.07mL. This would result, in other words, in a diluted dose and reduction in drug concentration – not exactly what we would like.

##### Example 4

**You are asked to calculate the volume of diluent needed to be added to 80mg of Drug X to produce a required concentration of 4mg in 1mL. The displacement volume of Drug X is 0.5mL/40mg.**

**Step 1** – *Find the Final Volume*

If the required concentration is 4mg in 1mL, then 20mL is needed for 80mg of Drug X.

**Step 2** – *Determine the Displacement Volume*

If 40mg displaces 0.5mL of solution, it means that 80mg would displace 1mL.

**Step 3** – *Subtract the Displacement Volume from the Final Volume*

20mL – 1mL = **19mL of diluent required**.

##### Example 5

**68mL of water is required to produce 100mL of a mixture containing 250mg of Drug X in each 5mL dose. Bearing this in mind, calculate the displacement volume for a quantity of powder equivalent to the 250mg dose of Drug X.**

If there is 250mg in each 5mL dose, then 100mL would have 20x this – i.e. 5,000mg.

If 68mL is required, it must mean 5,000mg displaces the remaining 32mL of water.

Thus, 5,000mg displaces 32mL in 100mL; therefore, 250mg displaces 1.6mL in a 5mL dose.

**1.6mL is displaced in a reconstituted 250mg/5mL dose**.

##### Example 6

**You are asked to prepare 2 sets of twelve 1g suppositories, Set A and Set B. Set A needs 12 suppositories each containing 50mg of Drug X whereas Set B needs both 50mg of Drug X as well as 300mg of Drug Y in all 12 suppositories. Theobroma oil is used as the base for both suppository sets.**

Displacement Value of Drug X |
0.7 |

Displacement Value of Drug Y |
1.5 |

**What quantity of theobroma oil base is required in each set?**

**How much does an individual suppository weigh in each set?**

Consider the following…

Drug X and Drug Y are in **milligrams** whereas displacement values are expressed in **grams**. We need to convert milligrams to grams first.

For practical reasons we need to consider **excess** when making suppositories as waste naturally occurs during extemporaneous compounding. For the purposes of this calculation, we will assume production of 15 suppositories as opposed to 12.

For Set A, we need to figure out **how much theobroma oil is displaced**.

(15 suppositories).(0.5g of Drug X) = **0.75g** of Drug X needed.

The next step is to divide 0.75g by its displacement value (0.7) to calculate how much theobroma oil is displaced.

0.75g/0.7 = 1.1g

This means that **0.75g of Drug X displaces 1.1g of theobroma oil**.

Given that 15 suppositories are required at 1g each, we now subtract the displaced theobroma oil from the 15g needed.

15g – 1.1g = **13.9g** of theobroma oil needed for **Set A**.

We can now run the same process for Set B.

15 x 0.3g = 4.5g of Drug Y

4.5g/1.5 = 3g of theobroma oil displaced.

3g of Drug Y + 1.1g from Drug X = 4.1g of displaced theobroma oil from Drug X and Y (for Set B).

15g – 4.1g = **10.9g** of theobroma oil needed for **Set B**.

This means that 13.9g of base is needed for Set A and 10.9g is needed for Set B. This answers the first part of the question.

We now need to determine how much an individual suppository weighs from each set.

Set A |
Set B |
---|---|

0.75g of Drug X | 0.75g of Drug X + 4.5g of Drug Y |

13.9g of Theobroma Oil | 10.9g of Theobroma Oil |

Total Weight: 14.65g | Total Weight: 16.15g |

14.65/15 suppositories => 0.98g |
16.15/15 suppositories => 1.08g |

Discrepancies in weight is explained by the fact that the displacement value for Drug X is less than 1 for Set A, whereas its greater than 1 for Set B.

##### Example 7

**You are asked to prepare twenty 2g glycogelatin pessaries each containing 150mg of Drug X. The displacement value for Drug X is 1.6. How much does each pessary weigh?**

As with the previous question:

1) Ensure units are consistent: in this case, convert mg to grams.

2) Take excess into consideration.

With this in mind, the question now becomes: create twenty-five 2g glycogelatin pessaries – each containing 0.15g of Drug X.

**First step** – calculate the total amount of Drug X required: 0.15g x 25 pessaries = **3.75g**

**Second step** – divide the total amount by the drugs displacement value: 3.75g/1.6 = 2.34g

This means that 3.75g of Drug X displaces 2.34g of base. However, there is a slight problem. If we were using theobroma oil base this would be perfectly valid. However, we’re using glycogelatin base. 1.2g of glycogelatin base is, though, equal to 1g of theobroma oil. Based on this conversion we can determine the quantity of glycogelatin base required:

1.2 [(25 x 2) – 2.34]g = 57.19g of glycogelatin base required.

If we add this 57.19g to the 3.75g of Drug X (calculated above), the total weight of the mixture is 60.94g. Given that we are creating 25 pessaries, it means each pessary weighs 2.44g (60.94/25).