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Learn to quickly solve an osmolarity problem like a medical pro
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Nurses, doctors, and other professionals in health fields use osmolarity to figure out the levels of glucose, sodium chloride, and other particles in blood, urine, and other solutions. To calculate the osmolarity of a solution, the first step is to convert the number to moles per liter. Convert that number to osmoles per liter and add the osmolarity of each particle together to get the solution's osmolarity. Keep reading to learn how to solve any osmolarity problem in just a few simple steps.

Formula for Osmolarity

  • Solutions made of 1 solute (like sodium): Osmolarity = (value given) x (molarity) x (# of particles dissociated from the solute molecule)
  • Solutions made of different kinds of solutes: Osmolarity = (# of particles dissociated from the solute molecule) / (# of liters in a solution).
  • Total osmolarity of a solution = Sum of all osmoles of each solute
Section 1 of 3:

What is the formula for Osmolarity?

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  1. Osmolarity is the number of osmoles (or milliosmoles), or the concentration of all the ions, in 1 liter of a solution.[1] To calculate the osmolarity of a solution with 1 solute, convert the data given to Osmol/L using this basic formula:[2]
    • Osmolarity = (value given) x (molarity of solute) x (# of particles dissociated from solute)
      • This formula can also be written as: mOsmol/L = (weight of substance (g/L) / molecular weight of substance (AMU or g)) x (# of particles dissociated) x 1000[3]
    • To calculate the osmolarity of blood, use the equation: Osmolarity = (Sodium x 2) + (Glucose / 18) + (BUN / 2.8).[4]
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Section 2 of 3:

Calculating the Osmolarity of a Solute

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  1. When you have NaCl in a solution, it dissociates in water to form a Na+ ion and an anion (or negative ion) like Cl- or HCO3-. This means a solution with 1 mol/L NaCl has an osmolarity of 2 Osmol/L.[5]
    • Example 1: Given: 1 mol/L NaCl.
      1 mol/L NaCl = 1 Osmol/L [Na+] + 1 Osmol/L [Cl-] = 2 Osmol/L NaCL.
    • Example 2: Given: [Na+] = 0.140 mol/L (or 140 mEq/L).
      Sodium is one particle, so it doesn’t split up in water. This means mol/L equals Osmol/L.
      [Na+] = 0.140 Osmol/L.
      Each ion of Na+ pairs with Cl- to get 2 Osmol of particles.
      [Cl-] = 0.140 mol/L = 0.140 Osmol/L.
      0.140 Osmol/L [Na+] + 0.140 Osmol/L [Cl-] = 0.280 Osmol/L NaCl.
    • Example 3: Given: 0.15 mol/L NaCl.
      (0.15 mol / 1L) NaCl x (2 Osmol / 1 mol) = 0.30 Osmol/L NaCl.
  2. When you add together the atomic weight of each particle in glucose (C6, H12, and O6), you get 180 atomic mass units (AMU). When you’re given the value of glucose in grams, you can divide it by 180 to get the value in moles, which can then be converted to osmoles.
    • Example 1: Given: Glucose = 900 mg/L.
      Glucose is C6H12O6. To get the atomic weight (AMU) of the entire molecule, multiply each particle’s coefficient by its atomic weight on the periodic table.
      (6 x 12) + (12 x 1) + (6 x 16) = 180 AMU.
      1 mol glucose = 180 grams.
      1 mmol glucose = 180 mg.
      (900 mg / 1 L) x (1 mmol / 180 mg) = 5 mmol / L.
      5 mmol / 1 L x (1000 mmol / 1 mol) x (1 mol / 1 Osm) = 0.005 Osm/L Glucose.
    • Example 2: Given: Glucose = 238 mg / 100 mL.
      238 mg x (1 mg / 1000 g) = 0.238 g.
      (0.238 g / 100 mL) x (1000 mL / 1 L) x (1 mol / 180 g) x (1 Osm / 1 mol) = 0.0132 Osm/L.
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Section 3 of 3:

Calculating the Total Osmolarity of Plasma & Other Solutions

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  1. When calculating the total osmolarity of plasma (centrifuged from blood) or other solutions, convert each value given to osmoles per liter (Osm/L). Once each value is in Osm/L, add them all together to get the osmolarity of the entire solution.[6]
    • Example: [Na+] = 0.320 mol/L; Glucose = 0.463 mg/mL; BUN (Blood Urea Nitrogen) = 0.15 mg/mL
      • [Na+] osmolarity = (0.320 mol / 1 L) x (2 Osmol / 1 mol) = 0.640 Osmol/L [Na+].
      • Glucose osmolarity = (0.0463 mg / 1 mL) x (1000 mL / 1 L) x (1 g / 1000 mg) x (1 mol / 180 g) x (1 Osmol / 1 mol) = 2.42 Osmol/L glucose.
      • BUN osmolarity = (0.15 mg / 1 mL) x (1000 mL / 1 L) x (1 g / 1000 mg) x (1 mol / 28 g) x 1 Osmol / 1 mol) = 0.00536 Osmol/L BUN.
        • BUN only focuses on the 2 nitrogen in the CH4N20 molecule. Since the AMU of nitrogen is 14 g, the AMU of BUN is 28 g.
      • Blood osmolarity = (0.640 Osmol/L [Na+]) + (2.42 Osmol/L glucose) + (0.00536 Osmol/L BUN) = 3.065 Osmol/L or 3065.36 mOsmol/L.

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About This Article

Anne Schmidt
Co-authored by:
Chemistry Instructor
This article was co-authored by Anne Schmidt and by wikiHow staff writer, Cheyenne Main. Anne Schmidt is a Chemistry Instructor in Wisconsin. Anne has been teaching high school chemistry for over 20 years and is passionate about providing accessible and educational chemistry content. She has over 9,000 subscribers to her educational chemistry YouTube channel. She has presented at the American Association of Chemistry Teachers (AATC) and was an Adjunct General Chemistry Instructor at Northeast Wisconsin Technical College. Anne was published in the Journal of Chemical Education as a Co-Author, has an article in ChemEdX, and has presented twice and was published with the AACT. Anne has a BS in Chemistry from the University of Wisconsin, Oshkosh, and an MA in Secondary Education and Teaching from Viterbo University. This article has been viewed 16,081 times.
4 votes - 50%
Co-authors: 2
Updated: November 7, 2024
Views: 16,081
Categories: Science

Medical Disclaimer

The content of this article is not intended to be a substitute for professional medical advice, examination, diagnosis, or treatment. You should always contact your doctor or other qualified healthcare professional before starting, changing, or stopping any kind of health treatment.

Thanks to all authors for creating a page that has been read 16,081 times.

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