A 36-year-old man is brought to the emergency department by emergency medical services aft… — USMLE Step 3

A $36$ -year-old man is brought to the emergency department by emergency medical services after being found obtunded in his garage. Beside him were several unlabeled containers of antifreeze and an empty bottle of vodka. His temperature is $36.2^\circ\text{C}$ ( $97.2^\circ\text{F}$ ), blood pressure is $110/70\text{ mmHg}$ , pulse is $105/\text{min}$ , and respiratory rate is $22/\text{min}$ . The patient is responsive only to deep pain.

Laboratory evaluation yields the following results:

Test	Result
Serum sodium	$140\text{ mEq/L}$
Serum potassium	$4.0\text{ mEq/L}$
Serum chloride	$102\text{ mEq/L}$
Serum bicarbonate	$14\text{ mEq/L}$
Blood urea nitrogen	$28\text{ mg/dL}$
Serum creatinine	$1.4\text{ mg/dL}$
Serum glucose	$108\text{ mg/dL}$
Blood ethanol	$138\text{ mg/dL}$
Measured serum osmolality	$358\text{ mOsm/kg H}_2\text{O}$
Arterial blood gas pH	$7.24$
Arterial $P\text{CO}_2$	$32\text{ mmHg}$

Based on these laboratory results, what is the patient's calculated osmolar gap?

(Enter your answer as a whole number).

Answer: 32 mOsm/kg

Answer

The osmolar gap represents the difference between directly measured serum osmolality and the calculated serum osmolality. A normal gap is

\le 10\text{ mOsm/kg H}_2\text{O}

. An elevated gap in the setting of a high anion gap metabolic acidosis strongly suggests intoxication with an unmeasured osmotically active substance, such as methanol or ethylene glycol. When calculating the expected osmolality, one must include measured ethanol if it is present, using the formula:

(2 \times [Na^+]) + ([Glucose]/18) + ([BUN]/2.8) + ([Ethanol]/4.6)

. Plugging in the patient's values yields:

280 + 6 + 10 + 30 = 326\text{ mOsm/kg H}_2\text{O}

. Subtracting this from the measured osmolality of

358\text{ mOsm/kg H}_2\text{O}

gives an osmolar gap of

32\text{ mOsm/kg H}_2\text{O}

, indicating the presence of another toxic alcohol (likely ethylene glycol from antifreeze).

Step-by-Step Solution

Set up the equation for calculated serum osmolality.

Calculated\ Osmolality = (2 \times [Na^+]) + \frac{[Glucose]}{18} + \frac{[BUN]}{2.8} + \frac{[Ethanol]}{4.6}

Serum osmolality is primarily determined by sodium and its accompanying anions, along with glucose and blood urea nitrogen. In cases of alcohol ingestion, ethanol significantly contributes and must be accounted for to find the true baseline gap.

Substitute the patient's laboratory values into the osmolality equation.

Calculated\ Osmolality = (2 \times 140) + \frac{108}{18} + \frac{28}{2.8} + \frac{138}{4.6}

Applying the specific measured variables allows for the conversion of mg/dL into mmol/L (mOsm/kg).

Calculate the specific contributions of each component and sum them.

280 + 6 + 10 + 30 = 326\text{ mOsm/kg H}_2\text{O}

This yields the theoretical serum osmolality based solely on standard measured solutes plus the co-ingested ethanol.

Calculate the osmolar gap by subtracting the calculated osmolality from the measured osmolality.

358 - 326 = 32\text{ mOsm/kg H}_2\text{O}

The difference reveals the presence of unmeasured osmotically active substances (in this vignette, likely ethylene glycol from the antifreeze), confirming a toxic alcohol ingestion.

Key Concept

The osmolar gap is calculated by subtracting the calculated serum osmolality (which must incorporate ethanol if present) from the directly measured serum osmolality to identify unmeasured toxic alcohols.

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