Concentration is an expression of how much solute is dissolved in a solvent in a chemical solution. There are multiple units of concentration.

Which unit you use depends on how you intend to use the chemical solution. The most common units are molarity, molality, normality, mass percent, volume percent, and mole fraction.

Here are step-by-step directions for calculating concentration, with examples.

Molarity is one of the most common units of concentration. It is used when the temperature of an experiment won’t change. It’s one of the easiest units to calculate.

Calculate Molarity: moles solute per liter of solution (not volume of solvent added since the solute takes up some space)

symbol: M

M = moles / liter

Example: What is the molarity of a solution of 6 grams of NaCl (~1 teaspoon of table salt) dissolved in 500 milliliters of water?

First, convert grams of NaCl to moles of NaCl.

From the periodic table:

Na = 23.0 g/mol

Cl = 35.5 g/mol

NaCl = 23.0 g/mol + 35.5 g/mol = 58.5 g/mol

Total number of moles = (1 mole / 58.5 g) * 6 g = 0.62 moles

Now determine moles per liter of solution:

M = 0.62 moles NaCl / 0.50 liter solution = 1.2 M solution (1.2 molar solution)

Note that I assumed dissolving the 6 grams of salt did not appreciably affect the volume of the solution. When you prepare a molar solution, avoid this problem by adding solvent to your solute to reach a specific volume.

Molality is used to express the concentration of a solution when you are performing experiments that involve temperature changes or are working with colligative properties. Note that with aqueous solutions at room temperature, the density of water is approximately 1 kg/L, so M and m are nearly the same.

Calculate Molality: moles solute per kilogram solvent

symbol: m

m = moles / kilogram

Example: What is the molality of a solution of 3 grams of KCl (potassium chloride) in 250 ml of water?


First, determine how many moles are present in 3 grams of KCl. Start by looking up the number of grams per mole of potassium and chlorine on a periodic table. Then add them together to get the grams per mole for KCl.

K = 39.1 g/mol

Cl = 35.5 g/mol

KCl = 39.1 + 35.5 = 74.6 g/mol

For 3 grams of KCl, the number of moles is:

(1 mole / 74.6 g) * 3 grams = 3 / 74.6 = 0.040 moles

Express this as moles per kilogram solution. Now, you have 250 ml of water, which is about 250 g of water (assuming a density of 1 g/ml), but you also have 3 grams of solute, so the total mass of the solution is closer to 253 grams than 250. Using 2 significant figures, it’s the same thing. If you have more precise measurements, don’t forget to include the mass of solute in your calculation!

250 g = 0.25 kg

m = 0.040 moles / 0.25 kg = 0.16 m KCl (0.16 molal solution)

Normality is similar to molarity, except it expresses the number of active grams of a solute per liter of solution. This is the gram equivalent weight of solute per liter of solution.

Normality is often used in acid-base reactions or when dealing with acids or bases.

Calculate Normality: grams active solute per liter of solution

symbol: N

Example: For acid-base reactions, what would be the normality of 1 M solution of sulfuric acid (H2SO4) in water?

Sulfuric acid is a strong acid that completely dissociates into its ions, H+ and SO42-, in aqueous solution. You know there are 2 moles of H+ ions (the active chemical species in an acid-base reaction) for every 1 mole of sulfuric acid because of the subscript in the chemical formula. So, a 1 M solution of sulfuric acid would be a 2 N (2 normal) solution.

Mass percent composition (also called mass percent or percent composition) is the easiest way to express concentration of a solution because no unit conversions are required. Simply use a scale to measure the mass of the solute and the final solution and express the ratio as a percentage. Remember, the sum of all percentages of components in a solution must add up to 100%

Mass percent is used for all sorts of solutions but is particularly useful when dealing with mixtures of solids or anytime physical properties of the solution are more important than chemical properties.

Calculate Mass Percent: mass solute divided by mass final solution multiplied by 100%

symbol: %

Example: The alloy Nichrome consists of 75% nickel, 12% iron, 11% chromium, 2% manganese, by mass. If you have 250 grams of nichrome, how much iron do you have?

Because the concentration is a percent, you know a 100-gram sample would contain 12 grams of iron. You can set this up as an equation and solve for the unknown “x”:

12 g iron / 100 g sample = x g iron / 250 g sample

Cross-multiply and divide:

x= (12 x 250) / 100 = 30 grams of iron

Volume percent is the volume of solute per volume of solution. This unit is used when mixing together volumes of two solutions to prepare a new solution. When you mix solutions, the volumes aren’t always additive, so volume percent is a good way to express concentration. The solute is the liquid present in a smaller amount, while the solute is the liquid present in a larger amount.​

Calculate Volume Percent: volume of solute per volume of solution (not volume of solvent), multiplied by 100%

symbol: v/v %

v/v % = liters/liters x 100% or milliliters/milliliters x 100% (doesn’t matter what units of volume you use as long as they are the same for solute and solution)

Example: What is the volume percent of ethanol if you dilute 5.0 milliliters of ethanol with water to obtain a 75-milliliter solution?

v/v % = 5.0 ml alcohol / 75 ml solution x 100% = 6.7% ethanol solution, by volume.

Mole fraction or molar fraction is the number of moles of one component of a solution divided by the total number of moles of all chemical species. The sum of all mole fractions adds up to 1. Note that moles cancel out when calculating mole fraction, so it is a unitless value. Note some people express mole fraction as a percent (not common). When this is done, the mole fraction is multiplied by 100%.

symbol: X or the lower-case Greek letter chi, χ, which is often written as a subscript

Calculate Mole Fraction: XA = (moles of A) / (moles of A + moles of B + moles of C…)

Example: Determine the mole fraction of NaCl in a solution in which 0.10 moles of the salt is dissolved in 100 grams of water.

The moles of NaCl is provided, but you still need the number of moles of water, H2O. Start by calculating the number of moles in one gram of water, using periodic table data for hydrogen and oxygen:

H = 1.01 g/mol

O = 16.00 g/mol

H2O = 2 + 16 = 18 g/mol (look at the subscript to note there are 2 hydrogen atoms)

Use this value to convert the total number of grams of water into moles:

(1 mol / 18 g ) * 100 g = 5.56 moles of water

Now you have the information needed to calculate mole fraction.

Xsalt = moles salt / (moles salt + moles water)

Xsalt = 0.10 mol / (0.10 + 5.56 mol)

Xsalt = 0.02

There are other easy ways to express concentration of a chemical solution. Parts per million and parts per billion are used primarily for extremely dilute solutions.

g/L = grams per liter = mass of solute / volume of solution

F = formality = formula weight units per liter of solution

ppm = parts per million = ratio of parts of solute per 1 million parts of the solution

ppb = parts per billion = ratio of parts of solute per 1 billion parts of the solution.



Volumetric analysis is a practical approach towards accurate measurement of concentration, molecular mass, purity percentage, formula of compounds, percentage composition of an element and stoichiometry of a chemical equation.

  • It involves 3 important techniques. The first one is the use of apparatus like burette, pipette and volumetric flasks. These are specially made to offer the highest degree of accuracy.
  • The second aspect is the use of balance in weighing. Most balances come with a minimum and a maximum range – both in terms of weight measured and decimal places correctly reported.
  • The third aspect is the use of appropriate indicators.

Volumetric analysis can further be classified into 3 techniques depending on the nature of reactions:

Acid-Base titrations which involve the reaction of an acid and a base.

Redox titrations which includes redox reaction between analyte and titrant as the key reaction.

Complexometric titration which involves the formation of a coloured complex compound.

Key Points:

Volumetric analysis depends on the use of accurate apparatus like burette, pipette and volumetric flasks.

Accurate weighing of substances is the key to accurate results.

Indicators are often required for establishing the end-point in a volumetric analysis.

Acid-Base titrations, Redox titrations and Complexometric titrations are the major techniques in volumetric analysis.


Essential terms for volumetric analysis

In order to carry out accurate and precise volumetric analyses, one needs to know the following terms associated with it:

MOLARITY: “The number of gram moles of a solute dissolved per liter of solution”.

Molarity (M) = Wsolute/Msolute × 1/Vsovent (L)

Where, Msolute = Gram molecular weight of solute.


MOLALITY: “The number of moles of the substance dissolved in 1kg of the solvent”.

Molality (m) = Wsolute/Msolute × 1/Wsovent (Kg)


NORMALITY: “The number of gram equivalents of the substance dissolved per liter of the solution.

Normality = Wsolute/Esolute × 1/Vsovent (L)

Where, Esolute = Gram equivalent weight of solute.

TITRATION: “The procedure of ascertaining the volume of one solution essential to react entirely with a definite volume of another solution of known concentration”.

TITRANT: “The solution of known concentration (strength)”.

TITRATE: “The solution whose concentration (strength) to be ascertained”.

INDICATOR: “The reagent which specifies the endpoint or equivalent point of the titration”.


Apparatus and essential reagents

Volumetric analysis is always performed with solutions made with distilled or deionized water.

Volumetric analysis requires highly accurate graduated apparatus like burette, pipette, graduated flasks etc. which should be rinsed properly with distilled or deionized water.

These must not contain any contaminants and must not be rinsed with hot water. Hot solutions should not be used for measurement with burette or pipettes.

Burette is used to deliver accurate volumes of liquid within its range. Before using the burette, it must be checked against presence of any air bubble within the liquid.

Burette can be read up to two decimal places and the normal convention is to read the lower meniscus for clear liquids/ solutions and upper meniscus for dark colored solutions/ liquids.

A pipette can be used to draw a definite volume of liquid and transfer it to the conical flask/ beaker etc. The liquid should be allowed to drain out from the pipette on its own and some portion remains inside it which should be drained out by touching the tip of the pipette to the mouth of the conical flask/ beaker.

The most important thing to notice is that one should never blow the liquid that remains.

Graduated flask (Volumetric flask) is used to prepare solutions of known strength.

Indicators are used to determine the end points i.e. the completion of reaction. Most of the titrations require external indicators for detecting the end points which essentially means the change in pH of the system. Some of the common examples of indicators are:

Phenolphthalein for strong acid vs strong base reactions.

Methyl orange for strong acid vs weak base reactions.

Starch for titrations involving iodine and thiosulfate.

Potassium chromate and fluorescein for Silver nitrate titrations.

Many non-acid-base titrations require the maintenance of a constant pH throughout the titration and in such cases, buffer solutions may be used for the purpose.


After the experimental part is done, the results are calculated using the volume of reactant consumed. The amount of analyte is determined using the formula:

Ca = Ct Vt M / Va


Ca is the concentration of analyte, expressed in molarity.

Ct is the concentration of titrant, expressed in molarity.

Vt is the consumed volume of the titrant, expressed in liters.

M is the molar ratio of the analyte versus the reactant, obtained from the balanced chemical equation.

Va is the pipetted out volume of the analyte, expressed in liters.



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