THE MOLE CONCEPT AND RELATED

CALCULATIONS

‘THE MOLE AS A UNIT OF MEASUREMENT

The standard unit is caled one mole of the substmncs. One mole of each of thse iirent OO and is found t be 6.0% 107° The value 6.0 x 10°* is called Avogadro's constant or Avogadro's

carbon-12 is 1.993 « 107g. Then, the number of atoms present in 12g of carbon-12 is derived as follows:

12gx1 ”

1993×107

Therefore, the number of atoms in 12g of carbon-12 and hence the number of particles in a mole are 602» 10” atoms Hence, Avogadro's number is the number of atoms in exactly 12g of earbon-12. isotope. One mole of any substance contains as many as many elementary particles as the Avogadro’s number (constant).

So, from the above explanation, the mole can be defined as the amount of a substance that contains as many elementary particles as the mumber of atoms present in 12g of carbon-12 isotope.

Relative formula Mass of one mole This. mass (1. mole)

Substance Formula

mass, M- (molar mass) contains 60 * 10" carbon Carbon c 12 12g atoms Iron Fe 56 56g 6.0 107 iron atoms

60% 107 formula

units Magnesium 6.0 x 10 formula

oxide units Caleium 6.0» 107? formula

carbonate units 6.0 * 10 formula

. . 60 * 107 iron(II) Fe Fe" 56 56g

e e s 6.0 107 electrons The other substances, which also exist as molecules, include ozone molecule (gas). Os; phosphorus molecule (solid), Py; sulphur molecule, Sp, ete In real life, when dealing with large numbers of small objects, i is usual to count them in groups The objects are grouped and counted in unit amounts. For example, we buy a carton of soap, 8 gallon of kerosene, a erate of soda, a dozen of pencils, a ream of papers, ete

Some units of measurement Unit Number of objects per unit

x10" molecules

Molar Quantities of Different Substances

Measure molar quaniites of different substances The mass of one mole of any substance (its molecular mass) is the atomic mass or molecular mass expressed in grams (or kilograms). For convenience, chemists prefer to express mass in grams, although the SI unit of mass is the kilogram. This is because the amount of substances which chemists usually work with in science laboratories, is quite small and if their masses are

expressed in kilograms, the numbers used would be extremely small You can calculate the molar mass (M) of any substance by summing up the relative atomic weights ofits constituents atoms, For example, ethanol, C2HsOH, contains two carbon atoms, six hydrogen atoms and one oxygen atom. So, the molar mass of ethanol can be calculated thus

In a similar way, molar masses of other compounds can be calculated, For example, the molar mass of sodium chloride, NaC, is calculated by adding together the relative atomic masses of the

It is important to note that relative aomic mass or relative molecular mass has no unit while molar masses are always expressed in grams or kilograms The molar mass of a compound is the same as the relative molecular mays and the molar ‘mass of an element is the same as the relative atomic mass (A,) of that element, The only difference lies in the units

Example 1

• M(COs) = 442 (or g mol) = molar mass of carbon dioxide
• M(CO3) = 44 = relative molecular mass oF carbon dioxide
• M(Fe) = 562 (org mot") molar mass of ion
• M,(Fe) = S6= Relative atomie mass of iron

Similarly, the molar masses of each of the following substances can be calculated using values for the relative atomic masses of the elements Molar masses of different substances

Substance Formula Molar mass

Ammonium chloride NELCI 14+ (Ind) +355 ~53.5g

Application of the Mole Concept

Application of the Mole Concept

Known Masses of Elements, Molecules or lons to Moles Convert known masses of elements, molecules or ions to moles In experimental work, chemists work with varying masses. They cannot always use one mole of a substance, The equation that links the mass ofa substance to the number of moles present is Mass

Molar mass

Example 2

Convert 49g of sulphuric acid, HsSQ., into moles.Given:Mass ~ 49g: molar mass ~ 98g Formula Mass

Molar mass

Solution 4p of H:8Oc~ 49/98 0.5 mol

Known Volumes of Gases at S.1-P to Moles Convert known volumes of gases at S.7P 1o moles The volume occupied by one mole of a gas at standard condition of temperature and pressure has been scientifically determined, and it is found to be 224 dm’ ‘This volume is called the molar volume of a gas. The molar volume of a gas, therefore, has the value of 22.4 dm’ at s.tp.

to all gases, ‘Therefore, at s.tp. 32g of oxygen (Q:) or 17g of ammonia (NHL) or 44g of carbon dioxide (CO2) or 40g of argon (Ax) will occupy a volume of 22.4 dm? This makes it easy to convert the volume of any gas at s.tp. into moles, or moles into volume. However, i is important to note that as the conditions of temperature and pressure change the molar volume will also change.

The number of moles of a given sample of gas is obtained by dividing the volume of the gas by molar yolume (22.4 dm).

E Volume

Molar volume

I the volume of the gas is given in em, then it should be divided by the molar volume of @ gas

Altematively, the volume may, first, be converted to dim* and then divides by the molar valume,

Masses of Solids or Volumes of Known Gases to Actual Number of Parties Change masses of solids or volumes of known gases to actual number of parties The number of particles in one mole of any substance is 6.02 « 10°. To find the number of particles ina substance, we use the expression:

• N=, where
• N=the number of particles in that substance;
• n= the amount of substance (moles); and
• L= the Avogadro's constant (6.02 * 10°)

This conversion requires two steps: first convert the mass of solid or volume of gas to moles, and then multiply the number of moles by the Avogadro's constant, For example, to convert 3.6 ‘dm’ of ammonia gos to the actual number of ammonia (NH;) molecules, change 5.6 dm' of

3.0 «10 molecules Altematively, we may find out the number of particles by converting the iven volume to the umber of molecules straight forward without passing through the number of moles first. We

• 1079722.4= 1.5 « 108 molecules

Molar Solutions of Various Soluble Substances

Prepare molar solutions of various saluble substances ‘A molar solution is solution which contains one of the compound in one lite (1 dm’ or 1000 cm) of the solution Let us consider the case of sodium hydroxide, NaOH The molecular weight of this compound is 40g. Therefore, a molar solution of sodium hydroxide will contain 40g in 1000 emn'(1 dm’) of the solution Also, consider anhydrous sodium carbonate, Na,COs. mole of this carbonate weights 1063

Hence, its molar solution will contain 106g of the anhydrous salt in 1000 em’ of solution Af however, 0.1 moles (10.6g) of the solute 5 dissolved in 1.0 dm, the solution is 0.1 molar. But i 0.1 moles is dissolved in 0.1 dm’ of the solution, the solution is stil 1.0 molar (since 1 dm’ of

solution would contain 1.0 mole of the solute),

The molecular weights of some common substances are shown below Compound Molecular weight (1 mole) Potassium hydroxide, NaOH 365 Hydrochloric acid, HCI 3658 Sulphuric acid, F380, 8g Soditum ehloride, NaCl s85g

Sodium bicarbonate, NaHCO; Sg

Caleium hydroxide, Ca(OH)> 74g The molar solution of each of these substances ean be prepared by dissolving one mole of each substance in 1000 em? (I dim*) of distilled water. We see, therefore, that 40g of sodium hydroxide in 1000 cm? of solution will give a 1.0M solution. Hence, 20g of the hydroxide should give a 0.5M solution. In a similar way, we can make derivative solution concentrations ranging as follows: 0.1M, 0.2M, 0.3M, 04M….IM, 2M, ete

However, in each case the amount of solution should always be 1000 em? The concentration ranges like these are known as molarities of solutions, Hence, 05M sodium carbonate can also be read as “a sodium carbonate solution with a molarity of 0 5M"

The Concentration of Solutions

When a chemical substance (the solute) is dissolved in a given volume of solvent, we can measure the “quantity” of solute in two ways, we can measure either its mays (in grams) or its amount (in moles), The final volume ofthe solution is usually measured in dm.

When we measure the mass ofthe solute in grams, we obtainthe mass concentration in gidm” Mass of solute(g’ Concentration (g/dm?) = —_ Mere Volume of solution(dm” )

Example 3

Caleulaie the concentration (gidm’) of sodium chloride solution (NaC!) that contains 20g of sodium chloride in a final solution of 100 em?

Solution

First, convert the given volume to din?

‘Then, work out the concentration of the solution by dividing the mass (weight) of solute (g) by the volume (dm) : Mass of solute(z) Concentration (gdm?) = Volume of solution(dm”)

Alternatively, we could calculate the concentration straightforward without having to convert the given volume into dm‘, e.g-:1f 20g of the solution are contained in 100 em’ of the solution, then the amount of solute in 1000 em? (1 dm?) of the solution would be

Caleulations Based on the Mole Concept Perform calculations based on the mole concept A chemist always wants to know how much of one substance would react with a given amount of another substance. This is achieved by use of balanced chemical equations. Such equations are called stoichiometric equations AA stoichiometric equation is the one in which the reactants and the products are correctly balanced; all the atoms, ions and electrons are conserved, Such an equation gives correct mole

ratios of reactants and products in chemical reactions. This quantitative relationship is called stoichiometry Consider an equation forthe reaction between hydrogen and nitrogen to produce ammonia BEng + Nagy 2NHs This can be read as follows: three moles of hydrogen reacts with ome mote of nitrogen to yield twa moles of ammonia The sumbers 3, 1 and 2 are called stoichiometric coofficients, ‘They tell us the proportions in

Which the substances react and in which the products are farmed

Example 4

What volume of carbon diocide (COs) measured at scp. will be produced when 21.0 of sodium Indrogencarbonate (NaHCOs) is completely decomposed according tothe equation 2NaHCO)-+ NasCOsy + COr) + H20n)

Solution

First, find the weight of carbon dioxide that will be produced by the hydrogencarbonate:

• Mass of NaHCO; =2 » 84 = 168g
• Mass of CO: = 44g,

The weight of carbon dioxide produced can be obtained fiom the following relation:

Then, convert this weight of CO: to volume at s.tp.We know that one mole (44g) of carbon dioxide at stp. occupies 22.4 dm’