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Unit conversions, the median and absorbance

13 min

  • Convert between compatible mass, volume, and amount concentration units
  • Calculate the median of an ordered set of results
  • Convert percent transmittance to absorbance

Read the full reference

Try first

Try first

A serum creatinine is 0.90 mg/dL. What is the same result in mg/L?

The next section explains it.

The next section explains it.

Right. The next section explains why.

The next section explains it.

The next section explains it.

Get the idea

Three small calculations sit under a great deal of laboratory work: moving a result between units, finding the middle of a set of results, and reading absorbance from a photometer.

Units

A concentration is an amount of substance in a volume, and both parts of the unit can change.

  1. Change the volume first. A liter is 10 deciliters, so a result in mg/dL becomes mg/L when it is multiplied by 10.1
  2. Then change the mass with the molar mass, the mass of one millimole of the substance:

mmol/L = mg/L ÷ molar mass (mg/mmol)

Glucose has a molar mass of 180.16 mg/mmol, so 1,080 mg/L of glucose is 1,080 ÷ 180.16 = 5.99 mmol/L.

The number changes every time the unit does. A result relabeled without changing its number is off:2

  • by a factor of 10 for a volume change
  • by the molar mass for a mass change

Each laboratory reports in the units it has validated, with its own stated factors.

The median

Put the results in rank order and take the middle one.

  • When the count is odd there is one middle value.
  • When the count is even there are two, and the median is their average. For four chloride results of 98, 101, 104 and 109 mmol/L, it is (101 + 104) ÷ 2 = 102.5 mmol/L.

The median depends on rank, so one extreme result moves it far less than it moves the mean.3

Absorbance

A photometer compares the light leaving a sample with the light entering it.

  • That fraction is the transmittance, T.
  • The percent transmittance, %T, is 100 × T.
  • Absorbance is the negative base-10 logarithm of the fraction.

A = −log₁₀(T) = 2 − log₁₀(%T)

The two forms give the same answer as long as each receives its own input:2,4

  • At 10%T, T is 0.10 and A = 1.00.
  • At 100%T, no light was absorbed and A = 0.
  • Taking the logarithm of the percent in the first form gives a negative number, the sign that the percent went into the fraction form.

Absorbance has no unit.2,4

References
  1. Thompson A, Taylor BN. Guide for the Use of the International System of Units (SI). NIST Special Publication 811. 2008 ed. National Institute of Standards and Technology; 2008.
  2. Bishop ML, Fody EP, Van Siclen C, Mistler JM, Moy M. Clinical Chemistry: Principles, Techniques, and Correlations. 9th ed. Jones & Bartlett Learning; 2023.
  3. National Institute of Standards and Technology. Measures of location. NIST/SEMATECH e-Handbook of Statistical Methods. Accessed September 23, 2026.
  4. International Union of Pure and Applied Chemistry. Beer–Lambert law. In: Compendium of Chemical Terminology. 5th ed. Online version 5.0.0. International Union of Pure and Applied Chemistry; 2025. doi:10.1351/goldbook.B00626

Watch one

A glucose result is 90 mg/dL. Your laboratory's comparison table lists glucose in mmol/L with a molar mass of 180.16 mg/mmol. What is the result in mmol/L, to three significant figures?

  1. Change the volume: 90 mg/dL × 10 = 900 mg/L.

    The answer is wanted per liter, so the volume changes from deciliters to liters first.

  2. Divide by the molar mass: 900 ÷ 180.16 = 4.9956 mmol/L.

    Each millimole of glucose weighs 180.16 mg, so the number of millimoles is the mass divided by that.

  3. Round the final value to three significant figures: 5.00 mmol/L.

    Rounding once, at the end, keeps the intermediate error out of the reported value.

  4. Check the size: 5.00 is smaller than 900, as it must be.

    A molar mass above 1 mg/mmol always makes the amount smaller than the mass, so an answer larger than the mass means the mass was multiplied by the molar mass.

The glucose is 5.00 mmol/L.

Your turn

Problem 1 of 3

A serum magnesium is 1.8 mg/dL. Using a molar mass of 24.305 mg/mmol, what is the result in mmol/L, to three significant figures?

Hint
  1. Get the mass per liter first.
  2. A result of 1.8 mg/dL is 18 mg/L.
  3. Divide the milligrams per liter by the mass of one millimole.
Show the answer

0.741 mmol/L

The volume step gives 1.8 mg/dL × 10 = 18 mg/L, and dividing by the molar mass gives 18 ÷ 24.305 = 0.7406 mmol/L, reported as 0.741 mmol/L.

Review Mass and amount unit conversions

Problem 2 of 3

Four ordered control results are 101, 103, 107, and 109 mg/dL. What is their median?

Incorrect. With four results, the median is the average of the two middle values, 103 and 107.

Correct. For an even number of observations, the median is (103 + 107) ÷ 2 = 105 mg/dL.

Incorrect. 107 alone is not the median because the median averages the two central values, 103 and 107.

Hint
  1. Count the results. Is the count odd or even?
  2. With an even count, two values share the middle.

Review Central tendency, control limits, and confidence intervals

Problem 3 of 3

A blank-corrected sample reads 25.0%T. What is its absorbance, to three decimal places?

Show the answer

0.602

A = 2 − log₁₀(25.0) = 2 − 1.398 = 0.602. The same answer comes from the fraction: −log₁₀(0.250) = 0.602.

Review Absorbance and percent transmittance

Keep

Sources checked