A 54-year-old man is brought to the emergency department by emergency medical services aft… — USMLE Step 2 CK

A 54-year-old man is brought to the emergency department by emergency medical services after his wife found him unresponsive in their garage. She reports that he has a history of major depressive disorder and alcohol use disorder. An open, half-empty bottle of engine antifreeze was found nearby. Vital signs show a temperature of $36.2^\circ\text{C}$ ( $97.2^\circ\text{F}$ ), pulse of $105$ /min, blood pressure of $145/88$ mm Hg, and respirations of $26$ /min. Physical examination reveals a comatose patient with no focal neurological deficits.

Laboratory studies reveal the following:

Parameter	Value
Sodium	$141$ mEq/L
Potassium	$4.2$ mEq/L
Chloride	$104$ mEq/L
Bicarbonate	$8$ mEq/L
Blood urea nitrogen	$42$ mg/dL
Creatinine	$1.8$ mg/dL
Glucose	$126$ mg/dL
Ethanol	Undetectable
Measured serum osmolality	$340$ mOsm/kg $\text{H}_2\text{O}$

Arterial blood gas on room air:
pH: $7.18$
$\text{pCO}_2$ : $20$ mm Hg
$\text{pO}_2$ : $98$ mm Hg

Urinalysis reveals the presence of envelope-shaped crystals.

Based on the provided laboratory values, what is the patient's serum osmolar gap in mOsm/kg $\text{H}_2\text{O}$ ?

Answer: 36 mOsm/kg

Answer

The patient's osmolar gap is

36

mOsm/kg

\text{H}_2\text{O}

. This is obtained by first finding the calculated serum osmolality using the formula:

2[\text{Na}^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8}

. Substituting the provided values yields

2(141) + \frac{126}{18} + \frac{42}{2.8} = 282 + 7 + 15 = 304

mOsm/kg

\text{H}_2\text{O}

. Subtracting this calculated value from the measured osmolality (

340

mOsm/kg

\text{H}_2\text{O}

) gives a gap of

36

mOsm/kg

\text{H}_2\text{O}

Step-by-Step Solution

Determine the formula for calculated serum osmolality.

\text{Calculated Osmolality} = 2[\text{Na}^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8}

This standard calculation is required when assessing a patient for suspected toxic alcohol ingestion.

Input the patient's laboratory values into the formula.

\text{Calculated Osmolality} = 2(141) + \frac{126}{18} + \frac{42}{2.8} = 282 + 7 + 15 = 304

mOsm/kg

\text{H}_2\text{O}

Dividing glucose by 18 and BUN by 2.8 correctly converts these values from mg/dL to mOsm/L so they can be summed with sodium.

Calculate the serum osmolar gap.

\text{Osmolar Gap} = \text{Measured Osmolality} - \text{Calculated Osmolality} = 340 - 304 = 36

mOsm/kg

\text{H}_2\text{O}

The difference between measured and calculated osmolality reveals the concentration of unmeasured osmotically active substances (e.g., ethylene glycol).

Key Concept

The serum osmolar gap is calculated by subtracting the calculated serum osmolality (

2[\text{Na}^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8}

) from the measured serum osmolality. A gap

>10

mOsm/kg

\text{H}_2\text{O}

suggests the presence of unmeasured osmotically active agents such as toxic alcohols (e.g., ethylene glycol, methanol).

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