CHE3: MOLE CONCEPT

Evidence for existence of particles

Diffusion;

It is spreading movement of particles (molecules) from a region higher concentration to a region of lower concentration or zero concentration. The particles of one substance meet with those of another substance until a uniform mixture is formed.

A bottle of a strong smelling perfume is opened at th front of a dormitory. What happened.

Eventually all the perfume particles diffuse in the dormitory and the students can smell the perfume.

Experiment 2

A tall gas jar is filled with water. A coloured crystal is tied onto a string non lowered in water. The set is left to stand for about 15 minutes.

 Observation

The purple colour spreads slowly through the water. Eventually the water is uniformly purple.

Discussion;

The particles of potassium permangate dissolved in the water and spread out and mixed with water molecule – diffusion took place.

 Experiment 3 

A gas jar of air is inverted over a gas jar of bromine gas. The set up is left for 15 minutes.

Observation

Eventually a light gas is observed in both jars.

 Discussion

Molecules of bromine gas and air diffused and mixed uniformly

 Experiment 4

A long glass is clumped horizontally, one end it of it is plugged with cotton wool soaked in ammonia and the other end plugged with cotton wool soaked in hydrochloric acid. (concentrated).

As shown in the apparatus below. Set up is left for about 10 minutes.

Observation

  • White ring forms at position A.

 Discussion

  • Hydrogen chloride and ammonia gas have diffused in the tube and where they meet a white ring has formed (ammonium hydrocloride)

Note;

The white ring forms nearer to the cotton wool in hydrochloric acid .

This suggests that ammonia diffuses faster than hydrogen chloride.

Question:

Why does ammonia diffuse aster than hydrogen chloride?

Ammonia is lighter than hydrogen chloride gas or is less dense than hydrogen chloride gas.

Relative atomic mass (R.A.M)

R.A.M =

R.A.M  is mass of 1 atom of the element divided by of the mass of one atom carbon – 12

R.A.M is a ratio therefore it has no units.

Relative atomic of same elements

RAM

Elements whose relative atomic mass are not whole numbers consist isotopes

 Relative Formula Mass (R.F.M)

R.F.M =

Note:

R.F.M is a ratio, therefore it has no units.

The term R.F.M applies to ionic and covalent substances. The R.F.M element of compounds can be determined by adding the mass of the constituent elements.

mass, (R.M.M)

R.M.M  

R.M.M is a ratio, therefore it has no units

The term R.M.M only applies to covalent substances consisting of molecules.

Like R.F.M, R.M.M can be determined by adding the masses of atoms of the constituent elements.

THE MOLE

It is the amount of substance that contains as may elementary entities as one atom of 12.0g of carbon 12 isotope.

The mole is an avogabro’s no of partickes r basic unit and is approximately 6.02X1023.

Any amount of matter which contains this number of particles e.g molecules, atoms, ions, proton (electrons) is referred to as the mole of a substance.

Molar mass;

It is the mass of one mole of a substance in grammes.

The molar mass of an element is its prelative atomic mass.

The molar mass of a compound is its relative formula mass.

One mole of chlorine atoms is one chlorine atom, 2 moles of chlorine atoms are 2 chlorine atoms.

One mole of chlorine molecules is one chlorine molecules., 2 moles f chlorine molecules are 2 chlorine molecules.

Use formula

Mass of the element = number of moles X R.A.M

  1. 2moles of oxygen

= number of moles X R.A.M

= 2×64

= 32g

  1. 2 moles of copper metal (Cu = 64)

= number of moles x R.A.M

= 2×64

= 128g

Empirical formula or simplest formula

It is the formula which shows the ratios of the different atoms of an element

Present in a compound.

It can be determined from percentage composition or combining masses.

Example one

A certain componund contain 17.7% hydrogen, and 82.3% nitrogen by mass

Determine its empirical formula

Element                                               N         H

Divide             % composition by R.AM

Divide by the smallest value

Simplest atom ratio                            1      :      3

N1H3 = NH3

A compound contains 40% by mass of carbon, 6.7% by mass of hydrogen and 53.3% by mass of oxygen. Calculate

  1. The empirical formula of the compound

C6

Molecular formula –

It is the actual formula of the compound which shows the number of atms.

A hydrocarbon which contains 85.7% carbon and the orest hydrogen

Determine its empirical formula

Percentage hydrogen = 100% – 85.7%

= 14.3%

C              H

1         :     2

Determine the molecular formula of the hydrocarbon if its molecular mass is 42

(CH2) y

C         H

= 42

= 3

(CH2)3 = C3H6

Write the equation for the complete combustion of the hydro carbon.

2(C3H6  2O2  3CO2  3H2O)

2C3H6 (g)  9O2  6CO2(g)  6H2O(g)

A compound  contains 1.44g, of magnesium metal and 0.96g of oxygen. Calculate the simplest formula. (O = 16, Mg = 24)

Elements                                 Mg                   O

Combining mass

Divide mass by R.A.M

Simplest combining ratio

1            :         1

= MgO

A compound y contains 15.8% alminium, 28% sulphur and 56.2% oxygen. (S = 32, Al = 27, O = 16).

  1. Calculate the empirical formula of y

Al                                S                     O

1                :           3            :         12

= Al2S3O12

= Al2(SO4)3

MOLAR GAS VOLUME

Is the volume occupied by one mole of any gas. At s.t.p (standard temperature and pressure) one mole of any gas occupies 22.4 litres or 22.4dm3 or 22,400cm3.

The standard temperature is 0  (273k), and the standard pressure is 760mm of mercury  1 atmosphere.

At room temperature one mole of any gas occupies 24 litres, 24.0dm3, 24,00cm3. Convert the following volumes of gas to number of molecules.

Mole concept Formulae

In this post, you will find the important formulae for mole concept. Learn how to apply these formulae here.

Mole Concept – Learning from examples

Definitions

The relative atomic mass (Ar) of an element is the mass of one atom of the element, relative to the mass of 1/12 the mass of one atom of carbon- 12.

The relative molecular mass (Mr) is the mass of one molecule, relative to the mass of 1/12 the mass of one atom of carbon- 12.

Mole

The mole is used to represent the amount of particles. One mole refers to 6.02 x 1023 particles.

The Avogadro constant (L) has a value of 6.02 x 1023.
Hence, number of particles = number of moles x Avogadro constant
Another formula for finding number of moles = mass in grams / molar mass (or mass you get from periodic table)
Example
: Find the mass of 5 mol of chlorine atom.

Step 1: Since questions ask for chlorine atom, we make use of Ar. From the periodic table, Ar of chlorine is 35.5.

Step 2: moles = mass/ molar mass ==> mass = molar mass x moles ===> mass = 5 x 35.5 = 177.5g

Example: Find the mass fo 5 mol of chlorine gas.

Step 1: Chlorine gas exists as Cl2. Since the Ar of one chlorine is 35.5, the Mr of Cl2 is 35.5 x 2 = 71.

Step 2: moles = mass/ molar mass ==> mass = molar mass x moles ===> mass = 5 x 71 = 355g

Calculating relative molecular mass

You will need to use the mass number of the elements, to determine relative molecular mass.

Example: Find the relative molecular mass of carbon dioxide.

Carbon dioxide has a formula of CO2. This means that it contains 1 carbon atom, and 2 oxygen atoms per molecule. From the periodic table, the mass number of carbon is 12, and the mass number of oxygen is 16. Hence, the Mr of carbon dioxide = 12 + 16 x 2 = 44.0

Calculating percentage mass (or percentage composition)

Example: Calculate the percentage composition of oxygen in carbon dioxide.

Step 1- Find the Mr of carbon dioxide.

Mr of CO2 = 12 + 16 x 2 = 44.0

Step 2- Determine relative mass of oxygen per molecule.

Since there are 2 oxygen per molecule of carbon dioxide, and the mass number of each oxygen is 16, hence the relative mass of oxygen per molecule is 16 x 2 = 32.

Step 3- Calculate percentage composition, by taking answer from step 2 divided by answer from step 1, and multiply by 100%.

Percentage composition of oxygen = 32/44 x 100% = 72.7% (3 significant figures)

Example: Calculate the mass of aluminium, in 25g of aluminium oxide.

Step 1- Write out the formula of aluminium oxide, since it is not given.

Al2O3.

Step 2- Calculate the Mr of Al2O3.

Mr of Al2O= 2 x 27 + 3 x 16 = 102

Step 3 – Determine the relative mass of aluminium in one compound of aluminium oxide.

There are 2 aluminium atoms per Al2O3. Hence, relative mass of Al in Al2O= 2 x 27 = 54.

Step 4- Step 2 and 3 tell you that for every 102 g of Al2O3, there is 54g of Al. Hence, in 25g of Al2O3, mass of Al = 54/102 x 25 = 13.2 g (3 significant figures).

Calculating empirical and molecular formulae

Empirical formula gives the simplest whole number ratio of each element in a compound.

Molecular formula shows the exact number of each element in a compound.

Calculating empirical and molecular formulae, given mass of each element.

Example. 50 g of a hydrocarbon is make up of 42.9g carbon. Find the empirical formula of this compound. Given that the molecular mass of this hydrocarbon is 42.0, determine its molecular formula.

Step 1. Since this is a hydrocarbon, it means that the compound consists of hydrogen and carbon only. Since it contains 42.9g of carbon, mass of hydrogen = 50 – 42.9 = 7.10 g

Step 2. Set up the table.

C H
mass/g (1) 42.9 7.10
Ar (2) 12 1
Number of moles per 100 g (3)

[Take (1) divided by (2)]

3.58 7.1
Ratio (round off to nearest whole number).

 

From (3), notice that 3.58 is the smallest number. Hence, divide all numbers in (3) by 3.58.

1 2
Empirical Formula CH2

Step 3: Determine molecular formula.

Let the molecular formula be (CH2)n, where n is an integer.

Molecular mass  = (12 + 2 x 1) x n = 42

14 n = 42

n = 3

Hence, molecular formula = C3H6.

A VIDEO ABOUT CALCULATING EMPIRICAL AND MOLECULAR FORMULAE

Calculating empirical and molecular formulae, given percentage composition of each element.

Example. A hydrocarbon consists of 86% by mass of carbon. Find the empirical formula of this compound. Given that the molecular mass of this hydrocarbon is 42.0, determine its molecular formula.

Step 1. Since this is a hydrocarbon, it means that the compound consists of hydrogen and carbon only. Since it contains 86% carbon, percentage composition of hydrogen = 100% – 86% = 14%

Step 2: Set up the table

C H
% composition (1) 86 14
Ar (2) 12 1
Number of moles per 100 g (3)

[Take (1) divided by (2)]

7.17 14
Ratio (round off to nearest whole number).

 

From (3), notice that 7.17 is the smallest number. Hence, divide all number in (3) by 7.17.

1 2
Empirical Formula CH2

Step 3: Determine molecular formula.

Let the molecular formula be (CH2)n,where n is an integer.

Molecular mass  = (12 + 2 x 1) x n = 42

14 n = 42

n = 3

Hence, molecular formula = C3H6.

  1. Gases

number of moles of gases = (volume of gas in dm3)/ (molar volume)

molar volume at room temperature and pressure = 24 dm3

molar volume at standard temperature and pressure = 22.4 dm3.

Note: 1dm3 = 1000 cm3

Example.Find the number of moles of carbon dioxide in 250 cm3of carbon dioxide gas.

Step 1. 250 cm3= 250/1000 dm3 = 0.250 cm3

Step 2. moles = volume/ molar volume = 0.25/24 =0.0104 mol

Solutions

Concentration of solutions can be represented by mol/dm3 or g/dm3.

Concentration in mol/dm3= (number of moles) / (volume in dm3)

Concentration in g/dm3= (mass in grams) / (volume in dm3)

Concentration in g/dm3= Concentration in mol/dmx Mr

Chemical equations and mole ratios

2NaOH (aq)+ H2SO4 (aq)—> Na2SO4 (aq) + 2H2O (l)

From the above equation, it means that 2 moles of NaOH react with 1 mole of H2SO4 to form 1 mole of Na2SO4 and 2 moles of H2O.

Example. 20.0 cmof 0.250 mol/dm3 sodium hydroxide reacts completely with 15 cm3 of sulfuric acid.  Find the concentration of the sulfuric acid in (i) mol/dm3 (ii) g/dm3.

Step 1. Write the equation for the reaction, if it is not given. In this case, the equation is:

2NaOH (aq)+ H2SO4 (aq)—> Na2SO4 (aq) + 2H2O (l)

Step 2. Find the number of moles of NaOH.

  1. of mol of NaOH = 20/1000 x 0.250 = 0.005 mol

Step 3. Use equation to find number of moles of H2SO4.

From equation, 2 mol of NaOH reacts with 1 mol of H2SO4.

Hence, number of moles of H2SO4= 0.005/2 = 0.00250 mol

Step 4. Find concentration of H2SO4.

15 cm3= 15/1000  dm3=  0.015 dm3.

Concentration (mol/dm3) of  H2SO =  0.00250/ 0.015 = 0.167 mol/dm3

Mr of  = 2 x 1 + 32 + 16 x 4= 98.0

Concentration (g/dm3) of  H2SO= 0.167 x 98 = 16.4 g/dm3.

Limiting and excess reactant

The limiting reactant is one that will be used up in the reaction. The excess reactant is one that will not be used up in the reaction.

Example. 20.0 cmof 0.250 mol/dm3 sodium hydroxide is added to 20.0 cm3 of 0.2 mol/dm3  sulfuric acid. Find the mass of sodium sulfate formed.

Step 1: Write the equation for the reaction, if it is not given. In this case, the equation is:

2NaOH (aq)+ H2SO4 (aq)—> Na2SO4 (aq) + 2H2O (l)

Step 2: The question gives sufficient information to find number of moles of both reactants — H2SO4 and NaOH. This means that you need to find limiting and excess reactant.

Step 3: Find number of moles of both reactant.

No. of mol of NaOH =  20/1000 x 0.250 = 0.005 mol

No. of mole of H2SO4 = 20/1000 x 0.2 = 0.004 mol

Step 4. Use values from step 3, and equation to determine limiting and excess reactant.

From equation, 2mol of NaOH will react with 1 mol of H2SO4.

Hence, 0.005 mol of NaOH will react with 0.005/2 = 0.0025 mol of H2SO4. However, 0.004 mol of H2SOis added. This means that H2SOis in excess, and NaOH is the limiting reagent.

Alternatively

From equation,  1 mol of H2SOwill react with 2 mol of NaOH

Hence, 0.004 mol of H2SO4 will react with 0.004 x 2 = 0.008 mol of NaOH. However, only 0.005 mol of NaOH is added. This means that H2SOis in excess, and NaOH is the limiting reagent.

Step 5. Once you found the limiting reagent, use its number of moles for calculating amount or mass of other reactant/ products, because it is the one that reacts completely.

Since NaOH is the limiting reactant, we use it to calculate number of moles of Na2SO4.

No. of mol of NaOH =  20/1000 x 0.250 = 0.005 mol

From the equation, 2 moles of NaOH form 1 mole of Na2SO4.

No. of mol of  Na2SO= 0.005 / 2 = 0.00250

Mr of Na2SO= 23 x 2 + 32+ 16 x 4 =  142

Mass of Na2SO4= 142 x 0.00250 = 0.355 g

A VIDEO ABOUT LIMITING AND EXCESS REACTANT

Percentage yield and purity

% yield = (actual mass)/ (theoretical mass) x 100%

% purity = (calculated mass)/ (mass of impure substance) x 100%

Example. When 100 g of calcium carbonate is heated, 32 g of calcium oxide is obtained. Calculate the percentage yield.

Step 1. Write the equation for the reaction, if it is not given. In this case, the equation is:

CaCO—> CO2 + CaO

Step 2. Find the Mr of CaCO (what is given) and CaO (what you need to caculate).

Mr of CaCO= 40 + 12 + 16 x 3 = 100

Mr of CaO = 40 + 16 = 56.0

Step 3. Find mass of CaO produced from 100g of CaCO.

No. of mol of CaCO= 100/ 100 = 1

From equation, 1 mol of CaCO3  forms 1 mol of CaO.

No. of mol of CaO = 1

Theoretical mass of CaO = 1 x 56 = 56.0 g

Step 4. Calculate percentage yield.

percentage yield = actual mass/ theoretical mass x 100% = 32/56 x 100% = 57.1%

Example. Limestone is consist of calcium carbonate and other impurities. When 100 g of calcium carbonate is heated, 32 g of calcium oxide is obtained. Calculate the percentage purity.

Step 1. Write the equation for the reaction, if it is not given. In this case, the equation is:

CaCO—> CO2 + CaO

Step 2. Find the Mr of CaCO and CaO.

Mr of CaCO= 40 + 12 + 16 x 3 = 100

Mr of CaO = 40 + 16 = 56.0

Step 3. Find mass of CaCOrequired to produce 32 g of CaO.

No. of mol of CaO = 32/ 56 = 0.571 mol

From equation, 1 mol of CaCO3  forms 1 mol of CaO.

No. of mol of CaCOreacted = 0.571 mol

Step 4. Calculate mass of CaCOpresent in limestone.

Mass of CaCO3 = 0.571 x 100 = 57.1 g

Step 4. Calculate percentage yield.

percentage yield = theoretical mass/ mass of impure substance x 100% = 57.1/100 x 100% = 57.1%

A VIDEO ABOUT PERCENTAGE YIELD AND PURITY

ASSIGNMENT : CHE3: MOLE CONCEPT Assignment MARKS : 10  DURATION : 1 week, 3 days

 

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