Pharmaceutical Calculations Flashcards

A 72 kg patient is ordered dopamine at 7.5 mcg/kg/min. The pharmacy dispenses a bag of 400 mg dopamine in 500 mL D5W. What infusion rate (mL/hr) should be programmed into the pump?

Answer: 40.5 mL/hr

Step 1 — Convert concentration: 400 mg ÷ 500 mL = 0.8 mg/mL = 800 mcg/mL. Step 2 — Calculate dose: 7.5 mcg/kg/min × 72 kg = 540 mcg/min. Step 3 — Calculate rate: 540 mcg/min ÷ 800 mcg/mL = 0.675 mL/min × 60 min/hr = 40.5 mL/hr. Choice A (20.25) results from doubling the concentration by misreading the bag as 400 mg/250 mL. Choice C (54.0) results from substituting 60 kg instead of 72 kg. Choice D (81.0) results from misreading the label as 400 mcg (not 400 mg), making the apparent concentration 400 times weaker.

A pharmacist needs to prepare 240 mL of a 40% w/v alcohol solution using 95% w/v alcohol and 15% w/v alcohol. How many mL of 95% alcohol are required?

Answer: 75 mL

Using alligation alternate: parts of 95% alcohol = desired% − lower% = 40 − 15 = 25 parts; parts of 15% alcohol = higher% − desired% = 95 − 40 = 55 parts; total = 80 parts. Volume of 95% alcohol = (25/80) × 240 mL = 75 mL. Verification: (75 × 0.95) + (165 × 0.15) = 71.25 + 24.75 = 96 g alcohol in 240 mL = 40% ✓. Choice D (165 mL) is the volume of the 15% alcohol — the two components are frequently swapped. Choices B and C reflect misapplied alligation ratios.

The standard adult dose of a chemotherapy agent is 25 mg. A 9-year-old child weighs 28 kg and is 130 cm tall. Using the Mosteller BSA formula [BSA (m²) = √(height(cm) × weight(kg) / 3600)] and an adult reference BSA of 1.73 m², what is the appropriate pediatric dose (rounded to the nearest mg)?

Answer: 15 mg

Step 1 — Child's BSA: √(130 × 28 / 3600) = √(3640 / 3600) = √1.0111 = 1.006 m². Step 2 — Proportional dose: (1.006 m² / 1.73 m²) × 25 mg = 0.5815 × 25 = 14.5 mg ≈ 15 mg. Choice C (18 mg) results from omitting the divide-by-3600 step, producing an inflated BSA. Choice D (22 mg) reflects use of Clark's rule (weight-based, 28/70 × 25 ≈ 10 mg) misapplied with adult weight of 30 kg. Choice A (10 mg) corresponds to direct application of Clark's rule with the standard 70 kg reference.

A pharmacist prepares 60 mL of a 3% w/v cocaine hydrochloride solution for topical ophthalmic use. The sodium chloride equivalent (E-value) of cocaine HCl is 0.16. How many mg of NaCl must be added to render the preparation isotonic with lacrimal fluid (equivalent to 0.9% NaCl)?

Answer: 252 mg

Step 1 — NaCl needed for a fully isotonic 60 mL: 0.9 g/100 mL × 60 mL = 0.54 g = 540 mg. Step 2 — NaCl equivalence already contributed by cocaine HCl: 3 g/100 mL × 60 mL = 1.8 g drug; 1.8 g × 0.16 = 0.288 g = 288 mg NaCl equivalent. Step 3 — Deficit to add: 540 − 288 = 252 mg NaCl. Choice A (540 mg) ignores the osmotic contribution of the drug. Choice B (288 mg) is the NaCl equivalent of the drug, not the amount to add. Choice C (396 mg) results from applying the E-value to the wrong quantity (e.g., per 100 mL rather than the full 60 mL volume).

An 80-year-old male patient weighs 58 kg with a serum creatinine of 2.2 mg/dL. His IBW equals his actual body weight. A drug product monograph specifies: 500 mg q8h (CrCl > 60 mL/min); 500 mg q12h (CrCl 30–60 mL/min); 500 mg q24h (CrCl 10–29 mL/min); contraindicated if CrCl < 10 mL/min. Using the Cockcroft-Gault equation, which regimen is appropriate?

Answer: 500 mg q24h

Cockcroft-Gault (male): CrCl = [(140 − age) × weight] / (72 × SCr) = [(140 − 80) × 58] / (72 × 2.2) = (60 × 58) / 158.4 = 3480 / 158.4 ≈ 22 mL/min. This falls in the 10–29 mL/min range → 500 mg q24h. A common error is ignoring the patient's age (80 years), which dramatically reduces CrCl and shifts the result from the 30–60 bracket into the 10–29 bracket. The drug is not contraindicated because CrCl is above 10 mL/min.

A TPN order requires 30 mEq of calcium gluconate. The pharmacy stocks calcium gluconate 10% w/v injection (MW of calcium gluconate = 430 g/mol; calcium is divalent). How many mL of this injection are required?

Answer: 64.5 mL

Step 1 — Equivalent weight (EW): For a divalent ion, EW = MW / valence = 430 g/mol ÷ 2 = 215 g/eq = 215 mg/mEq. Step 2 — Concentration in mEq/mL: 10% w/v = 100 mg/mL; 100 mg/mL ÷ 215 mg/mEq = 0.465 mEq/mL. Step 3 — Volume: 30 mEq ÷ 0.465 mEq/mL = 64.5 mL. Choice A (30.0 mL) incorrectly assumes 1 mEq/mL. Choice B (43.0 mL) results from using a valence of 3, making EW = 143 mg/mEq. Choice D (129.0 mL) results from treating calcium as monovalent (valence = 1), doubling the equivalent weight to 430 mg/mEq and halving the mEq/mL.