# Concentration Calculations in Analysis – A Primer

Most of us were exposed to chemical calculations in school and college days. As we pursued higher degrees and progressed in our career as analytical scientists we took more interest in learning new techniques and picking up operational details of analytical instruments. It cannot be denied that our pursuit provided us greater self confidence of operating a number of modern sophisticated instruments. However, it is equally important that we have a sound knowledge of concentration units and calculations involving them as even sophisticated instruments provide correct results only if you are capable of avoiding mistakes and arithmetic errors while doing calculations and carrying out dilutions.

The present article is intended to serve as a primer to help you refresh the different units of expressing concentration and gain confidence in calculations. Concentration of a solute in a solution is expressed in a number of units depending on the requirements of analysis and the concentration levels at which you are required to report the results. Now we shall examine the commonly used concentration units for expressing concentration and also calculations involving preparation of solutions and subsequent dilutions.

#### Percent concentration by mass

Percent by mass refers to mass of solute present in a given mass of solution. The mass of solution is the sum total of mass of solute and solvent.

#### Concentration of Mass = \(\frac{Mass of Solute}{Mass of Solution} X 100 \)

**Caution:** total mass of solution is to be considered in the calculation. Some analysts by oversight consider mass of solvent.

**Example:** Find the percent by mass of a solution when 5 g of salt is dissolved in 100 gram of water.

Mass of salt = 5 g

Maas of solution = 100 gram water +5 g salt = 105 g

Percentage concentration = \(\frac{5}{105} X 100 = 4.76%\)

Percentage of concentration by volume

Concentration is generally expressed in such units when both solute and solvent happen to be liquids.

#### Concentration of Mass = \(\frac{Volume of Solute}{Volume of Solution} X 100 \)

The same caution should be exercised when volume of solution is to be taken instead of volume of solvent

**Example:** Find percentage by volume when 30 ml of methanol is added to 70 ml water

Percentage of methanol is equal to \(\frac{30}{( 70+30)} X 100 = 30%\)

#### Normality

Normality is commonly used in volumetric calculations. It is equal to gram equivalent weight of solute per litre of solution, for example, 1N solution of H2 SO 4 will contain 98 /2 or 49 g of pure acid.

#### Molarity (M)

Molarity is the most commonly used unit of expressing concentration. This is the number of gram moles of solute per litre of solution.

**Example:**How much 18M Sulphuric acid is required to make 100ml of 1.0M acid

Use the standard formula \(M_1V_1\)=\(M_2V_2\)

\(V_1\) x18=100x 1

\(V_1\) Add slowly 5.5 ml of 18M \(H_2SO_4\) slowly to about 50ml of water in a volumetric mark and then dilute the solution slowly with mixing to the 100 ml mark.

#### Molality (m)

An uncommon unit which expresses gram moles of solute in one kg of solvent

**parts per million(ppm) and parts per billion(ppb)**

ppm or ppb units are used to express concentration for trace level concentrations.

1 ppm = gms per \(10^-^6\) gms =μg/gm

1ppb = ng per/ltr

In terms of volume 1 ppm = mg/liter

**Example:** how much in NaCl is to be weighed to make 1000 ppm of NaCl solution

Formula weight of NaCl= 23+35.5.= 58.5

1 g of Na in formula weight is present in 58.5 / 23 = 2.5 g

Therefore 2.5 g of NaCl dissolved in 1000 ml will make 1000 ppm of Na in solution

#### Dilutions

Frequently dilutions become necessary when you have to bring down concentration to levels measurable by the required analysis technique. The general relation applicable to dilutions is

\(C_1V_1\) = \(C_2V_2\)

Where \( C_1 \) and \(V_1\) refer to concentration and volume before dilution and \(C_2\) and \(V_2\) represent concentration and volume after dilution

**Example: **How many ml 3M NaOH is required to prepare 500 ml of 1.0 M NaOH solution

3 X \(V_1\) =1.0 X 500

\(V_1\) = 1 X 500 /3 = 166.6 ML

The primer will serve to refresh your concepts which will help you to perform your analysis using highly sophisticated instruments with the required confidence levels.

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