A 50 year old female is seen in the emergency room with signs of toxicity. Her l

Question

A 50 year old female is seen in the emergency room with signs of toxicity. Her lab tests are as follows: Test result expected value Na 143 (135 - 145) mEq/L K 4.3 (3.5 - 5.0) mEq/L Cl 102 (98-108) mEq/L CO_2 21.0 (24 - 30) mEq/L BUN 21 (5 - 26) mEq/L Creat. 1.3 (0.1 - 1.5) mEq/L Gluc 126 (70 - 105) mEq/L Ca 10.0 (9.0 - 11.0) mg/Dl T.Prot. 8.5 (6.0 - 8.5) g/dL Albumin 5.1 (3.5 - 5.5) g/dL T.Bilirubin 0.7 (0.1 - 1.2) mg/dL Alk. Phos 95 (30 - 115) U/L SGOT 38 (1 - 40) U/L SGPT 24 (1 - 40) U/L GFR 58 (>=60) mL/min/1.73m^2 Anion Gap 20.0 (

Explanation / Answer

The calculated or estimated osmolality(CO) is determined by the formula

CO = 2Na + BUN/2.8 + glucose/18 + isopropyl alcohol/6.0

= 2*143+ 21/2.8+ 126/18+51/6

=286+7.5+7.0+8.5

=309

Therefore, calculated osmolality is 309 mosm/kg

Osmolal gap(OG) = Measured osmolality(MO) - calculated osmolality(CO)

here, MO (from lab results is 332 mosm/kg)

Therefore, OG= 332- 309 = 23

I think, isoproponal level accounts for this difference.

Osmolal gap is increased in the patient may be due to isopropanol ingestion.

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