Displacement Value & Suppository Calculations

Overview

  • Subject matter: Accurate formulation of medicated suppositories when ingredients are weighed (g, mg) but dispensed by volume (mL) into fixed‐volume mould cavities.
  • Central challenge: Each Active Pharmaceutical Ingredient (API) has its own density; when mixed with a suppository base it displaces a characteristic mass of base. Correct dosing therefore requires density-based corrections.

Key Definitions

  • Suppository: Solid dosage form intended for insertion into body cavities (rectal, vaginal …); melts/softens at body temperature to release API.
  • Suppository base: Usually cocoa-butter (Theobroma oil) in examples, can also be synthetic fats or water-soluble PEGs. Supplies bulk and melting characteristics.
  • Mould calibration value (y): Average weight (g) of blank suppositories produced by a particular mould when filled with molten base under routine conditions. Usually given per cavity; typical capacities: 1 g or 2 g.
  • Displacement Value (DV): Volume of drug that displaces 1 g of a specified suppository base. Numerically:
    DV=weight of drug that displaces 1g base1gDV = \frac{\text{weight of drug that displaces }1\,\text{g base}}{1\,\text{g}}
    ‑ If drug density = base density ⇒ DV=1DV = 1 (equal masses exchanged).
    ‑ Drug density > base density ⇒ DV<1DV < 1 (heavier drug displaces less base by weight). ‑ Drug density < base density ⇒ DV>1DV > 1 (lighter drug displaces more base by weight).

Conceptual Points About DV

  • Fixed mould volume means added drug must push out an equivalent volume of base; otherwise cavities under- or over-fill.
  • DV data for many common APIs in cocoa-butter are tabulated in pharmacopeias & compounding references.
  • Two categories of calculations:
    1. Type 1 – Single drug with known DV.
    2. Type 2 – Multiple drugs, each with its own DV.
    3. Type 3 – Drug given as % w/w of final suppository; DV not required (density effect inherently included in percentage).

Cocoa-Butter Displacement Values (selected)

  • (From Table 9-1; all values relative to cocoa-butter)
    • Aminophylline – 1.5
    • Aspirin – 1.1
    • Bismuth subgallate – 3.0
    • Bismuth subnitrate – 6.0
    • Boric acid – 1.5
    • Chloral hydrate – 1.5
    • Dimenhydrinate – 1.3
    • Hydrocortisone – 0.7
    • Menthol – 2.0
    • Morphine HCl – 1.5
    • Potassium bromide – 2.2
    • Quinidine HCl – 3.0
    • Resorcinol – 1.0
    • Salicylic acid – 1.3
    • Tannic acid – 1.6
    • Zinc oxide – 5.0
    • Zinc sulfate – 2.8
      (Full table in transcript; always verify with latest reference when compounding.)

Core Formulae

  • Single-drug base requirement:
    Amount of base (g)=(N×y)N×DDV\text{Amount of base (g)} = (N \times y) - \frac{N \times D}{DV}
    where
    NN = total suppositories prepared ("order + 2" rule: always compound ⁺2 to cover losses)
    yy = mould calibration value (g)
    DD = individual dose of drug (g)
    DVDV = displacement value of that drug.

  • Two-drug scenario:
    Amount of base (g)=N×y(N×D<em>1DV</em>1+N×D<em>2DV</em>2)\text{Amount of base (g)} = N \times y - \left(\frac{N \times D<em>1}{DV</em>1} + \frac{N \times D<em>2}{DV</em>2}\right)

  • Percentage strength (Type 3):
    • Determine total fill weight =N×y= N \times y.
    • API mass =percent×total fill weight= \text{percent} \times \text{total fill weight}.
    • Base mass =total fill weightAPI mass= \text{total fill weight} - \text{API mass}.
    (DV unnecessary because the percentage already expresses the final w/w proportion.)

Worked Examples – Type 1 (Single Drug)

Example 1: 6 × 250 mg Bismuth Subgallate
  • Order: 6 suppositories; add 2 extra ⇒ N=8N = 8.
  • Data: D=0.25gD = 0.25\,g, DV=2.7DV = 2.7, y=0.94gy = 0.94\,g.
  • Base:
    Base=(8×0.94)8×0.252.7=7.520.741=6.7796.78g\text{Base} = (8\times0.94) - \frac{8\times0.25}{2.7} = 7.52 - 0.741 = 6.779 \approx 6.78\,g
  • Final formula:
    • Bismuth subgallate – 2 g (8 doses × 0.25 g)
    • Cocoa-butter base – 6.78 g.
Example 2: 8 × 500 mg Zinc Oxide
  • Add 2 extra ⇒ N=10N = 10.
  • Data: D=0.5gD = 0.5\,g, DV=4.7DV = 4.7, y=1.0gy = 1.0\,g.
  • Base:
    Base=(10×1)10×0.54.7=101.0638=8.9362g\text{Base} = (10\times1) - \frac{10\times0.5}{4.7} = 10 - 1.0638 = 8.9362\,g
  • Final formula:
    • Zinc oxide – 5 g
    • Base – 8.94 g.
Example 3: 4 × 200 mg Metronidazole
  • Add 2 ⇒ N=6N = 6.
  • Data: D=0.2gD = 0.2\,g, DV=1.7DV = 1.7, y=1.0gy = 1.0\,g.
  • Base:
    Base=(6×1)6×0.21.7=60.7058=5.2942g\text{Base} = (6\times1) - \frac{6\times0.2}{1.7} = 6 - 0.7058 = 5.2942\,g
  • Final: Drug 1.2 g; Base 5.29 g.
Example 4: 8 × 500 mg Paracetamol (2 g mould)
  • Add 2 ⇒ N=10N = 10.
  • Data: D=0.5gD = 0.5\,g, DV=1.5DV = 1.5, y=2.04gy = 2.04\,g.
  • Base:
    Base=(10×2.04)10×0.51.5=20.43.33=16.71g\text{Base} = (10\times2.04) - \frac{10\times0.5}{1.5} = 20.4 - 3.33 = 16.71\,g
  • Final: Paracetamol 5 g; Base 16.71 g.

Suppositories with Two Drugs (Type 2)

  • Strategy: Calculate each drug’s displaced base separately, subtract both from total mould fill.
Example 1: 15 × (150 mg Cortisone + 560 mg Zinc Oxide)
  • Add 2 ⇒ N=17N = 17.
  • Doses: D<em>1=0.15gD<em>1 = 0.15\,g, D</em>2=0.56gD</em>2 = 0.56\,g.
  • DV: DV<em>1=1.5DV<em>1 = 1.5, DV</em>2=4.7DV</em>2 = 4.7.
  • Calibration y=2.04y = 2.04.
  • Drug masses:
    • Cortisone = 17×0.15=2.55g17\times0.15 = 2.55\,g
    • ZnO = 17×0.56=9.52g17\times0.56 = 9.52\,g.
  • Base:
    Base=17×2.04(2.551.5+9.524.7)=34.68(1.70+2.03)=30.95g\text{Base} = 17\times2.04 - \left(\frac{2.55}{1.5} + \frac{9.52}{4.7}\right) = 34.68 - (1.70 + 2.03) = 30.95\,g
  • Final: Cortisone 2.55 g; ZnO 9.52 g; Base 30.95 g.
Example 2: 18 × (200 mg Metronidazole + 250 mg Paracetamol)
  • Add 2 ⇒ N=20N = 20.
  • Doses: 0.2 g & 0.25 g.
  • DV: 1.7 & 1.5; mould y=2.04y = 2.04.
  • Drug masses: Metronidazole 4 g; Paracetamol 5 g.
  • Base:
    Base=20×2.04(41.7+51.5)=40.8(2.35+3.33)=35.12g\text{Base} = 20\times2.04 - \left(\frac{4}{1.7} + \frac{5}{1.5}\right) = 40.8 - (2.35 + 3.33) = 35.12\,g
  • Final: Metronidazole 4 g; Paracetamol 5 g; Base 35.12 g.

Percentage-Strength Calculations (Type 3)

  • DV not used; base weight derived from desired % w/w.
Example 1: 8 × 18 % w/w Zinc Oxide, 1 g mould
  • Add 2 ⇒ N=10N = 10. Total fill = 10 g.
  • Drug = 18 % of 10 g = 1.8 g.
  • Base = 10 – 1.8 = 8.2 g.
Example 2: 18 × 5 % Metronidazole
  • Add 2 ⇒ N=20N = 20. Fill = 20 g.
  • Drug = 5 % of 20 g = 1 g.
  • Base = 19 g.
Example 3: 8 × 10 % Aspirin
  • Add 2 ⇒ N=10N = 10. Fill = 10 g.
  • Drug = 1 g; Base = 9 g.
Example 4: 6 × 3 % Paracetamol
  • Add 2 ⇒ N=8N = 8. Fill = 8 g.
  • Drug = 0.24 g; Base = 7.76 g.

Practical & Pharmacological Implications

  • Uniformity: Correct base substitution ensures each suppository occupies mould volume without voids ≡ dose accuracy & regulatory compliance.
  • Stability: Some drugs alter melting point; large percentages may require base modification or alternative (e.g., PEG) to prevent softening below room temperature.
  • Patient comfort: Excessive solids may create gritty texture; guidelines often limit solid content to about 30 % w/w.
  • Manufacturing loss: The “+2” rule hedges against wastage during pouring & trimming; number can be adjusted for larger runs.
  • Ethics/Regulation:
    • Pharmacist must verify DV values from current compendia; outdated data can lead to sub-therapeutic or toxic doses.
    • Documentation: Batch records should detail calibration runs, DV source, calculations, and actual weights measured.
  • Real-world tie-ins: Similar displacement principles apply in capsule packing (tapped density) and parenteral admixtures (e.g., concentrated drug volumes displacing saline in IV bags).

Tips for Exam Preparation

  • Memorise the base formula; practise rearranging for unknowns.
  • Keep typical DV values of common drugs (ZnO, Bismuth salts, Paracetamol, Metronidazole) at fingertips.
  • Double-check units (mg ↔ g) before substitution.
  • Always include extra suppositories in calculations; exam questions nearly always require it.