13.1 Metric System, Percent Solutions & Dilutions

Key Takeaways

  • Percent weight per volume is grams of solute per 100 mL of finished solution, so grams needed equals percent multiplied by volume in millilitres divided by 100.
  • 10 percent formalin means a 1:10 dilution of commercial 37 to 40 percent formaldehyde stock, which yields only 3.7 to 4.0 percent actual formaldehyde.
  • A 1:10 ratio dilution is one part solute plus nine parts diluent for a total of ten parts, not one part plus ten parts.
  • The dilution equation C1V1 equals C2V2 gives the volume of concentrated stock needed; the diluent volume is the final volume minus that stock volume.
  • Section thickness in micrometres converts to millimetres by dividing by 1000, so a 4 micrometre section is 0.004 mm thick.
Last updated: September 2026

13.1 Metric System, Percent Solutions & Dilutions

ASCP HT Exam Focus: The ASCP BOC content outline lists Laboratory Mathematics with three sub-topics: metric system, percent solutions/dilutions, and molar solutions. Calculation items are scored like any other question, and they are the most reliably earnable points on the examination because the answer is deterministic.


The Metric System

PrefixSymbolMultiplier
kilok$10^{3}$
decid$10^{-1}$
centic$10^{-2}$
millim$10^{-3}$
micro$\mu$$10^{-6}$
nanon$10^{-9}$

Conversions used daily in histology

  • $1\text{ L} = 1000\text{ mL}$; $1\text{ mL} = 1000,\mu\text{L}$
  • $1\text{ g} = 1000\text{ mg}$; $1\text{ mg} = 1000,\mu\text{g}$
  • $1\text{ mm} = 1000,\mu\text{m}$, so a 4 micrometre section is 0.004 mm, and a 5 mm grossed slice is 5000 micrometres
  • $1\text{ ppm} = 1\text{ mg/L} = 1,\mu\text{g/mL}$ - the unit used for formaldehyde and xylene air monitoring
  • Temperature: $^{\circ}\text{C} = (^{\circ}\text{F} - 32) \times 5/9$. A 60 degrees C paraffin oven is 140 degrees F.

Exam trap: cassette and mold dimensions are given in millimetres, microtome settings in micrometres. A specimen grossed at 3 mm is 750 times thicker than the 4 micrometre section cut from it, which is exactly why penetration and processing schedules are thickness-driven.


Percent Solutions: Three Conventions

ConventionDefinitionTypical histology use
% weight/volume (w/v)Grams of solute per 100 mL of finished solutionSolid dyes and salts: silver nitrate, ammonium oxalate, potassium metabisulfite
% volume/volume (v/v)Millilitres of liquid solute per 100 mL of finished solutionAcid alcohol, dilute acetic acid, graded alcohols
% weight/weight (w/w)Grams of solute per 100 g of solutionHow manufacturers label concentrated acids

The w/v formula

Grams=% × final volume in mL100\text{Grams} = \frac{\%\ \times\ \text{final volume in mL}}{100}

Worked example. Prepare 800 mL of 2.5 percent (w/v) potassium metabisulfite for a silver method: $\text{Grams} = (2.5 \times 800)/100 = \mathbf{20.0\text{ g}}$, dissolved in water and brought to 800 mL.

The v/v formula

mL solute=% × final volume in mL100\text{mL solute} = \frac{\%\ \times\ \text{final volume in mL}}{100}

Worked example. Prepare 500 mL of 1 percent (v/v) acid alcohol from concentrated hydrochloric acid and 70 percent ethanol: $(1 \times 500)/100 = \mathbf{5.0\text{ mL}}$ of concentrated HCl added slowly to 495 mL of 70 percent ethanol. Always add the acid to the alcohol, never the reverse.

"Bring to volume" versus "add to": a true percent solution is made by dissolving the solute in part of the solvent and then bringing the total to the final volume in a graduated cylinder or volumetric flask. Adding solute to a full final volume of solvent overshoots the volume and under-concentrates the reagent. For dilute working solutions the difference is small; for concentrated silver and dye solutions it is not.


The Formalin Trap

10 percent formalin is not 10 percent formaldehyde.

  • Commercial "formalin" is a saturated aqueous solution containing 37 to 40 percent formaldehyde gas with methanol as a stabilizer.
  • 10 percent formalin means 100 mL of that commercial stock diluted to 1000 mL, which delivers 3.7 to 4.0 percent actual formaldehyde.
  • 10 percent neutral buffered formalin (NBF) is the same 1:10 dilution made up in phosphate buffer to pH 6.8 to 7.2.

This distinction is tested directly, and it also explains why a fixative labelled "4 percent paraformaldehyde" and one labelled "10 percent formalin" are chemically equivalent in formaldehyde content.


Ratio and Serial Dilutions

A ratio dilution written 1:10 means one part of the substance in a total of ten parts, which is one part solute plus nine parts diluent. The dilution factor is 10, so the concentration falls to one tenth.

Worked example - antibody titre. An immunohistochemistry protocol calls for a 1:400 working dilution in a final volume of 12 mL:

Vstock=12 mL400=0.03 mL=30μLV_{\text{stock}} = \frac{12\text{ mL}}{400} = 0.03\text{ mL} = \mathbf{30\,\mu\text{L}}

Pipette 30 microlitres of antibody concentrate into 11.97 mL of antibody diluent.

Serial dilution. Antibody titration is performed as a doubling or tenfold series: 1:100, 1:200, 1:400, 1:800. Each tube carries forward a fixed volume into an equal or ninefold volume of diluent, so the total dilution is the product of the individual steps (a 1:10 followed by a 1:20 is a 1:200).


The Dilution Equation

C1V1=C2V2V1=C2×V2C1C_1V_1 = C_2V_2 \quad\Rightarrow\quad V_1 = \frac{C_2 \times V_2}{C_1}

where $C_1$ is the stock concentration, $V_1$ the stock volume required, $C_2$ the desired concentration, and $V_2$ the desired final volume. The diluent volume is $V_2 - V_1$.

Worked example - graded alcohols. Prepare 1000 mL of 70 percent ethanol from 95 percent stock:

V1=70×100095=736.8 mL of 95 percent stock+263.2 mL waterV_1 = \frac{70 \times 1000}{95} = 736.8\text{ mL of 95 percent stock} + 263.2\text{ mL water}

Worked example - from absolute alcohol. Prepare 500 mL of 80 percent ethanol from 100 percent stock:

V1=80×500100=400 mL absolute+100 mL waterV_1 = \frac{80 \times 500}{100} = 400\text{ mL absolute} + 100\text{ mL water}

Worked example - silver stock. Prepare 250 mL of 2 percent silver nitrate from a 20 percent stock:

V1=2×25020=25 mL stock+225 mL deionized waterV_1 = \frac{2 \times 250}{20} = 25\text{ mL stock} + 225\text{ mL deionized water}

Unit discipline: $C_1$ and $C_2$ must share one unit, and $V_1$ and $V_2$ must share another. Mixing percent with molar, or millilitres with litres, is the single most common calculation error on the examination.

Test Your Knowledge

A histotechnician must prepare 400 mL of 4 percent (w/v) aqueous ferric chloride for Verhoeff differentiation. How much ferric chloride is weighed, and how is the solution brought up?

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Test Your Knowledge

A laboratory needs 2000 mL of 10 percent neutral buffered formalin and has commercial 37 percent formaldehyde stock and phosphate buffer on hand. Which preparation is correct, and what concentration of actual formaldehyde results?

A
B
C
D
Test Your Knowledge

An immunohistochemistry protocol requires 6 mL of a primary antibody working solution at a 1:250 dilution. How much antibody concentrate is pipetted, and how much diluent?

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B
C
D