Units of measurement, Chemistry tutorial


The object in this case is generally a salt, a compound, a reagent, water, or glass container. The consequence of weighing is the mass in weight of the object. Since chemistry is a scientific discipline, weighing must be definitive, reproducible and comparable by other weightings done in additional laboratories. Therefore, determined or weighed quantities must be expressed in appropriate and comparable chapter for proper understanding of the object in relation by other quantity. If we forget to provide the chapter when reporting the consequence, the measurement will be meaningless as it will be hard to interprete. It is hence a must that all measurements must be accompanied by the appropriate chapter.

Units of measurement:

Due to the fluctuating tasks we would be carrying out in the laboratory experiments, we will find out that we shall be measuring these quantities as the mass or weight, volume and length of materials or objects as the situation demands. Though for the purpose of this chapter, we shall dwell more on mass and volume.

Unit of mass (m)

The base unit of a measured solid material says a bottle, or a salt is the kilogram, abbreviated as kg. Though, its lower fractional chapter, the gram g, is commonly utilized. Multiples for example metric ton, of it are as well in employ especially in a chemical industrial weighing operation. The connection between the diverse kilogram units is presented therefore.

1866_Unit of mass (m).jpg

The apparatus utilized for weighing is termed a balance. There are many balances varying in both precision and accuracy. During the course of your program, you will be exposed to such varied balances as analytical balance, balm balance and top - loading balance. Don't be disturbed it you might not have to employ any of such balances in this course. Weighings might have been done for us prior to the laboratory exercise. As we go higher in your programme however, you will definitely make use of a balance yourself.

 Units of Volume (V)

The base unit of measured liquid, that is termed volume, is the cubic metre, abbreviated as m3 .All other waits are fractions or multiple of it. The more generally utilized ones are cubic decimeter, dm3 and cubic centimeter, cm3 we are as well probably aware of such terms as litre, 1 and millilitre, ml. The relationships between such units of volume are:

1790_nits of Volume (V).jpg

The value equivalent of such units is 1000cm' = 1dm3   = 1 litre = 1000m.

Amount of substance:


We are familiar with weighings and their units; the next thing we should concern ourselves by is the concentration of a substance

SAQ So what is concentration of a substance?

Answers the word concentration of a substance terms to the amount or quantity of the substance either wholly (for instance 10g of salt) in relative to another substance (for example 10g of salt in water).

Though, much as this is explicit sufficient, in pure chemistry words, another expression termed the mole is regarded as the base unit of the quantity or amount of the substance in chemistry. Let us now focal point our attention on what a mole is.


What hence is a mole? Why must we express our concentration word in moles? A mole, abbreviated as mol, is the amount of substance that encloses as many elementary units as there are atoms in 12.0g of carbon-12. Such elementary units might be atoms, molecules, ions, electrons etc. in effect we could have one mole of sodium ions, or sodium atoms or one mole ethanol molecules.

The number of such elementary units in a mole of a substance is equal to 6.02 x 10-23 particles. This 6.02 x 1023 is termed the Avogadro number. We require to express our determined quantity (concentration of a substance) in mole since we have found out that in chemistry, substance react through each other on the bases of moles equivalent rather than on amount in grams though this can be worked out if we know the moles.

What we are actually pronouncing is that even as it is true that we can determine in grams (solids) or in cm3 (liquids). Such units must be expressed in the mole unit.

Mole as applied to solid:

To do this, let us suggest another definition of a mole. The mole of an element or compound is the atomic, ionic, or molecular/ formular mass of the substance expressed in grams (if solid).

That is

Mole = Mass measure in grams /Molar mass in grams per mole

SAQ1: An undergraduate student utilized a top loading balance to weigh some quantity of sodium hydroxide. If the weight measure is 10g, illustrate this as mole. (Na = 23, 0 = 16, H =1).

The mole as applied to solutions

If the weight of the sodium hydroxide weighed above is now dissolved in a definite amount (volume) of water, it might then be needed to recognize the concentration of the solution in terms of the number of particles (formula units) present in addition to in terms of the mass of solute in the solution. For such reasons, a number of processes of expressing the amount of solute in solution have been derived and widely employed. This suggests the concentration might be given in more than one way (unit) depending on the units in that the quantity of components or quantity of sample is expressed. For example, a solution by a relatively small amount of solute is said to be diloute. Another in which large amount or quantity of solute say 200g of NaOH in 1000cm3 of water is said to be a concentrated solution. Effectively we can simply express concentration quantitatively via calling a solution dilute or concentrated. We can now redefine concentration as given: if a sample mixture contains components A1, A2, etc then the concentration of component A1 in the mixture is

 Quantity of A1 in the sample/ Quantity of sample

A comparable equation can be written for any other components in the model mixture. But more often, the precise concentration words must be utilized. That is, solutions must be expressed quantitatively, several methods are utilized.

i. Molarity:

The molarity of a solution is defined as the number of moles of solute in one cubic decimeter (dm3) or one litre (L) of solution.

Molarity = Moles of solute /Volume of solution in 1 dm3

Thus, a solution is one molar with respect to a given solute if it contains one mole of that solute in one cubic decimeter (dm3) m of solution.

ii. Molality:

The molality of a solution is defined as the number of moles of solute in one kilogram (kg) of solution. Let us remind ourselves here of what a molal or indeed 1 molal is: The weight in grams of a substance   Formula weight in grams of the substance

iii. Normality (n):

Normality is the number of gram - equivalent weight of solute in 1 dm3   of solution. This term is no longer in use but you will probably encounter it in some chemistry textbooks.

Normality = Equivalent weight of solute / Volume of solution in dm


Which is of a higher concentration: a one molar solution or a one molal solution?

Other units of concentration:

Other units of expressing concentration are also encountered in chemistry. This is particularly so when the formula or relative molecular mass of the individual component or a mixture is not known. In these cases it is impossible to use either the molar or molal units. The concentration must therefore be expressed in other units.

I. Percentage of weight (w/w %)

This is described as the ratio of the weight of solute to the total weight of solute plus solvent multiplied via 100.

Wt/wt% = wt of solute x 100 /wt of solute + wt of solvent

ii. Percentage by volume (v/v %)

This is identified as the ratio of the volume of solute to the total volume of solute plus solvent multiplied by 100.

Vol /vol% = volume of solute (cm3) x 100 /Volume of solute + volume of solvent (cm3)

iii. Percentage weight by volume

This is described as the number of grams of solute in one hundred (100) cm3 of solution multiplied by 100. Wt / vol % = weight of solute in gram x 100 Volume of solutions in cm3  

These are the most generally utilized ways of expressing concentration in chemistry and specifically in the chemistry laboratory. But there are others such as mole fraction, pH, and parts per thousand (ppt), parts per million (ppm), gram per unit volume, specific gravity and volume ratios.

Several of such we will come across in our programme in the chemistry laboratory courses, some we will come across in chemistry texts. It must be stressed that, the units of expressing concentrations of solutions discussed here are thoroughly understood by you.

Understanding them is extremely basic to understanding this practical chemistry course is all about. It must as well be stressed that regardless of concentration unit, the concentration of a solution is better illustrate in molarity. This becomes significant whenever we remind that volumes of reacting solutions of similar concentrations bear simple ratios to one another. That is, reactions between reactants to provide products take place in mole ratios to one another.


A solution contains 2.65g of anhydrous NaCO3 in 250cm3 of solution. Discover the concentrations of the solution in:

i. Grams per dm 

ii. Weight / weight percent

iii. Weight/ volume percent

iv. Mol per dm3  

(The formula weight of Na2CO3 = 106, l cm3   of water weighs 1g.)

Standard solution:

We will recall that we are now in a position to recognize the concentration expressions of a solution once we know the relative amounts of the solute and the solvent making up the solution. Now let us illustrate a standard solution. A standard solution is a solution of which the concentration is accurately acknowledged. A standard solution is prepared via weighting a pure solute, dissolve it in a suitable solvent, usually water and make up the solution to a definite volume in a volumetric flask.

Principle of dilution:

It is common practice in a chemistry laboratory to contain a solution of recognized concentration from which other solutions as well of known concentrations can be attained with no reweighing the solute. Whenever this is needed, the 1st solution is usually of higher concentrations than the subsequent solution(s) from which the solutions are derived. That is, the subsequent solutions are of successive lower concentrations and are only attained from the higher concentration solution via dilution by water.

Whenever a indeed volume, V1 of a solution of concentration C1 is diluted through pure water to a new volume V, the amount, n, of the substance in the solution remains similar. Though the concentration transforms via decreasing to C2. This exercise is easily understood using this instance:

A solution in a bottle is 1.20 molar concentrated. What volume of it must be diluted by water to gain 600cm3 of 0.05 molar solutions.

Note that I have specifically not stated the name of the solution. Whether acid, base, salt, or a specific name whether HCI, NaOH and so on. Such are not really needed. What is vital knows the volume and molar concentrations of the 2 solutions.

For this example in this assessment question what do we know.

i. We recognize the original concentration to be 1.20 moledm-3.  Let us call it C1.

ii. We know the concentration of the new solution we would like to prepare. Let's call it

iii. We know the volume of the new solution we would like to prepare. Let's call it V2

What we do not know is the volume of the original solution, which we want to dilute. Let's call this V,

Therefore          C1 V1 = C2 V2

1.2V1 = 0.05 x 600

V1 = 0.05 x 600 /1.2      = 300 /1.2


That is, we will take 25cm3 of C1 and add 575cm3 of water to make 600 cm3. The new solutions will be.a 0.05 molar solution.

This particular means of preparing another solution of a different concentration is particularly useful whenever we are working by acids. This is since acids are naturally provided for us in a solution form in a bottle and it is from this bottle that we take a certain volume which we dilute (or add water) to get our desired concentration. For instance, what will be the molar concentration of a solution we would have obtained if we dilute 10.0cm3   of a solution of conc. H2SO4 that is 18.0 molar concentrated? Using exactly what we did above, we would use

C1 V1 = C2 V 2

Wherever 1 is the original solution and 2 is the new solution we are preparing.

Substituting the values therefore, C1 = 18.0molar, V1 = 10 cm3, C2 = unknown and V2 = 1dM3   (1000cm3)

18 x 10 = C2 X 1000

C2 = 180 /1000 = 0.18 molar

The usual problem encountered via the student is not in calculating the above concentration, but rather in knowing the amount of acid to get that will provide the desired concentration on dilution. This has to do with the fact that the actual concentration of the acid might not actually have been provided. Instead we are just following a bottle of the acid and we are told to get ready a certain concentration in a certain volume of water. On the label of the bottle of acid, certain extremely significant information are generally given.

The label will provide the subsequent information: specific gravity; concentration via weight%. Using this 2 information, we can now calculate the actual concentration of the acid we have been provided through. Let us demonstrate this through an instance. The label on a stock bottle of an acid reads: 56% by mass and 1.25 specific gravity. If the molar mass of the acid is 56, find (i) the concentration in grams per dm3 (ii) the moles per de (iii) the volume of this acid that is needed to prepare 250cm3 of 1.0 molar concentration of the acid.

i. The specific gravity means that 1 cm3 of solution weighs 1.25g or 1000cm3 (1 dm3) weighs 1.25 x 1000g = 1250g. As well 56% by weight of the acid means 100g of the acid solution contain 56g of the acid solute. Since 1 dm3 weighs 1250g 1250g of solution will contain 56 x 1250 of the acid 100

for example l dm3   will contain  = 700g of the acid.

Hence mass concentration is 700g/dm3  

ii. The formula mass of the acid is given as 56g.

Molar concentration    = mass concentration / molar mass

= 700gdm-3  

= 56gmo1-1  

= 12.5moldm-3  

So it has taken us this long to determine the actual concentration of the acid in the bottle following to us. To discover the volume which we shall take out of this bottle to prepare our new solution, we go back to our dilution principle.

iii. C1 V1 = C2 V2

Where 1 is the original solution and 2 is the new solution we are preparing. Substituting the values hence, C 1   = 12.5 molar, V 1   = unknown, C2 = 1.0 molar and V2   = 250cm3

12.5x V1 = 1.0 x 250

V1   = 25/12.5

       = 2.0cm3

that is, we take 2cm3 of the acid in the bottle, put in a 250cm3 volumetric flask and fill it up to mark through distilled water.

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