Second Term Lesson Note for Week Two
Class : SSS One
Subject : Chemistry
Topic : Mole Concept in terms of Avogadro’s numbers, percentage by mass and empirical formula.
Duration : 80 Minutes
Period : Double Periods
Reference Book :
- Lagos State Unified schemes of work for Senior Secondary School, (SSS 1 – 3).
- Chemistry for Senior Secondary School.
- Online Resources
Instructional Material : Chart showing the definition of terms and formula for calculating empirical and molecular formula.
Learning Objectives : By the end of the lesson learners will be able to :
i. Determine the percentage by mass of element in a compound.
ii. Explains moles in terms of number or particles.
iii.
Content :
MOLE IN TERMS OF THE RELATIVE ATOMIC MASS OR RELATIVE MOLECULAR MASS OF A SUBSTANCE.
The mole can be expressed in terms of the R.A.M of an element or the R.M.M. of a substance/molecule/compound this:
1 mole of any substance = the R.A.M. of the substance or the R.M.M of the substance.
NOTE:
1 mole of Na (g) = 23g mole
1 mole of O2 (aq) = 16g mole
1 mole of O2 (g) = 16 x 2 = 32g/mol
1 mole of CO2 (g) = 12 + (16 x 2) = 44 g/mol
1 mole of H2SO4 (aq) = (2 x 1) + 32 + 64 = 98glmol
Example :
Calculate the number of moles present in 11g of carbon dioxide or carbon (iv) oxide (CO2) gas
Solution:
1 mole CO2 (g) = Rmm of CO2 (g)
1 mole of CO2 = 12 + (16 x 2) = 44g g/mol
:. 44g of CO2 = 1 mole of CO2 (g)
1 g of CO2 = 1/44 mole of CO2 (g)
:. 11g of CO2 = 1/44 x 11 mole of CO2 (g)
= 0.25 mole of CO2 (g)
Determine the number of grammes of substance present in 0.05 of sodium carbonate (Sodium trioxocarboante (iv) ) (Na2CO3).
Solution:
1 mole Na2CO3 = Rmm of Na2CO3
1 mole Na2CO3 = (23 x 2) + 12 + (16 x 3)
= 106glmol
:. 0.05 mole Na2CO3
Of Na2CO3
= 0.055moles
MOLES IN TERMS OF NUMBER
The term number takes into consideration the number of particles such as atoms, ions, molecules, electrons, protons, and neutrons etc, contained by a certain amount of a substance.
NOTE: The number of particles taking part and formed in a chemical reaction can be determined.
Avogadro determined the actual number of atoms of carbon in 12.00g of 126C isotope in various ways. He found out that 12.00g of 126C contains 6.02 x 1023 atoms of carbon. He worked with a large number by elements, compounds and ions and came to the conclusion that :
a. The gram atomic man of all elements always contains the same number of atoms.
b. The gram molar mass of all compounds always contain the same number of molecular.
c. The gram formula mass of all ions also contain the same number ions.
Avogadro established that the number of particles (ions, atoms, molecules, electrons, protons etc.) present in one gramme formula (atomic, ionic, molecular etc.) mass of a substance is 6.02 x 1023.
This value is called the Avogadro’s number or constant NA
NOTE: This number of particle is contain and in one mole of any substance
1 mole = 6.023 x 10 23 particles
Example:
1. Calculate the number of particles
i. 44g of iron (II) sulphide (Fes)
ii. 5.5g of manganese (Mn)
iii. 8g of oxygen molecule (02)
iv. 8g of oxygen atom (0)
(Mn – 55, 0-16, Fe =56, S=32)
Mole = mass of given substance (g)
(n) Gram atomic/molar mass
ii. Mole of Mn = 5.5g
55 g/mole
Number of particles
= NA x n
: .1 mole of Mn = 6.023 x 10 23 atoms
0.1mole Mn
= 0.1 x 6.023 x 1023 atoms
= 6.023 x 10 22 atoms
ii. Mole of Fe = mass in grams of FeS ÷ Molar mass
Given mass = 44g
Molar mass of FeS = (56 +32) = 88g
= 44g / 88g/mole
= 0.5 moles
1mole = 0.5 mole
1 mole FeS (s) = 6.023 x 1023molecules
:. 0.5 mole= 6.023 x 1023 x .05
= 3.012 x 1023 molecules of Fes
iii. Mole of O2 = mass in grams
G.M.M
Gmm of O2 = 16 x 2 = 32 g/mole
:. Mole = 8 g ÷ 32 g/mole
= 0.25 mole
1 mole of O2 = 6.023 x 1023 mole of O2
:. 0.25 moles of O2
= 6.023 x 1023 x 0.25 molecules
= 1.506 x 1023 molecules of O2
iv. G.m.m of oxygen = 16g/mole
Mole of 0 = massing of 0
G.a.m
= 8g
16g/mole
= 0.5mole
1 mole of 0 = 6.023 x 1023 atoms of 0
:. 0.5 mole of 0
= 0.5x 6.023 x 1023 atoms
= 3. 012 x1023 atoms of 0
2. A sample of nitric (trioxonitrate (v) acid contains 1.2 x 1023 molecules of the acid.
Calculate
a. The number of moles
b. The mass of the acid (HN03) in the sample
(NA = 6.023 x 10 23 particles mol-1, H=1, N=14, 0=16)
a. 1 mole of HNO3 acid =
6.02 x 10 23 molecules
1 mole = 1 mole
6.023 x 10 23
:. 1.2 x 1023molecules
= 1 x 1.2 x 1023
6.02 x 1023
= 1.2 mole
6.02
= 0.2 mole
b. mole = mass of substance in g ÷
Molar mass of substance
Molar mass of HNO3
= 1x 14 + (16 x 3)
= 63g/mole
Mass of HNO3 = mole x Molar mass
= 0.2 mole x 63 g/mole
= 12.6g
THE MOLE CONCEPT IN TERMS OF VOLUME
There three states of matter solid, liquid and gas. A good example of a substance that can form
the three states is water.
NOTE: Matter can change its state when there is a considerable change in KE.
The effect of temperature and pressure in which more pronounced in gaseous sate because f the
Volume it occupies compared to the and solid and liquid states which have a definite
Volume.
Experimentally, it has been proved that the gramme molar mass amount of any gaseous substance
will always occupy a volume of 22.4dm3 at standard temperature and pressure (s.t.p) and 24dm3
at room temperature and pressure (r.t.p), standard temperature = 00Cor 273k and standard
pressure = 760 mmHg or 1.01×105 Nm-2
NOTE: 1 mole of any gaseous substance = molar volume of a gas at s.t.p. i.e.
1 mole of 02 (32g) = 22.4dm3 at s.t.p
1 mole of C02 (44g) = 22.4dm3 at s.t.p
1 mole of N2 (28g) = 22.4dm3 at s.t.p
1 mole of S02 (64g) = 22.4dm3 at s.t.p
1 mole of C12 (71) 22.4dm3 at s.t.p
Examples:
1. Calculate the volume occupied by 5 moles of carbon dioxide (carbon (iv) oxide) at s.t.p
SOLUTION:
1 mole of gas at s.t.p = 22.4dm3
1 mole of C02 = 22.4dm3 at s..t.p
5 mole of CO2 =5 x 22.4 at s.t.p
= 112.0dm3 at s.t.p
2. Determine the number of mole present in 11.2dm3 of nitrogen (IV) oxide (nitrogen dioxide) = N02 (g) at s.t.p
Mole = volume
G.m.v
1 mole of N02 (g) = 22.4dm3 at s.t.p
:. 22.4dm of N0 (g) = 1 mole of N02 at s.t.p
1 dm3 of N02 (g) =
1 ofN02 (g) at s.t.p
22.4dm3
:.11.2dm3 of N02(g)
= 1 x 11.2dm3 of N02 (g) at s.t.p
22.4dm3 = 11.22m3 of N02 (g) at s.t.p
= 0.5mole of N02 at s.t.p
How many grammes of gas are present in 5600cm3 of chlorine gas at s.t.p? (Cl=35.5)
SOLUTION:
1 mole of Cl2 (g) = Rmm of Cl2 = molar volume of gas at s.t.p.
Rmm of Cl20 = 35.5 x 2 = 71g/mol
1 mole of Cl2= 71gl/mol = 22400cm3 of Cl2 at s.t.p.= 71 g
22400
:. 5600cm3 of Cl2= 71 x 5600
22400 = 17.56g of Cl2
Calculate the number of molecules of hydrogen gas present in 2.24dm3 of the gas at s.t.p
SOLUTION:
1 mole H2 (g) = Avogadro’s No of molecule =
Molar volume of H2 at s.t.p
1 mole H2(g) = 6.023 x 10 23 molecules =
: .22.4dm3 of H2 (g) at s.t.p
Molecules of H2 (g) at s.t.p
1dm3 = 6.023 x 10 23
22.4
: .2.24dm3 = 6.023 x 10 23 x 2.24
22.4 1
6.02 x 1022 molecules of H2(g)
5. Calculate the volume at s.t.p which would occupy 2.5.6g of 58vapour (S = 32)
SOLUTION
Mole of substance = massing
G.m.m
G.m.m. of S8 = 32 x 8
Mole of S8
= 2.56g
256g/mole
= 0.01mole
Volume (dm3) = mole x molar volume (22.4dm3)
1mole of S8vapour = 22.4dm3 at s.t.p
1mole of S8vapour
= 0.01 x 22.4dm3
= 0.224dm3
SUMMARY
i. 1mole of any Rmm of = Na.GMV
Substance substance of
Gaseous gas at s.t.p
This summary holds for a gaseous substance only
ii. 1 mole of = Rmm of = Avogadro’s
Any substance substance number
Solid/liquid (NA)
This summary holds for solid and liquid substances
1. A volume of a gas was found to weigh 5.6g and when corrected to s.t.p measured 4.48dm3. Calculate the G.mm. of the gas
2. Calculate the number of:
a. atoms in 2.5mole of Na (sodium)
b. ions present in 0.5 moles of copper (II) ions (Cu2+)
3. A volume of a gas Z was found to weigh 6.5g and when corrected to s.t.p it measured 4.84dm3. Calculate the G.m.m of the gas Z
(H =1, S=32, 0=16)
EMPIRICAL AND MOLECULAR FORMULAE
Empirical formula: The formula which shows the simplest ratio of the atoms of the elements that make up a compound.
Molecular formula: The formula which shows the actual number of atoms present in one molecule of the element or compound.
EXERCISE :
EFthanoic Acid Mol. Formula = CH3COOH or C2 H4 O2
2 atoms of carbon, 4 atoms of H2 and 2 atoms of O2
Empirical formula = CH2O Ratio 1: 2: 1
Note: the 3rd type of formula is structural formula
CH3COOH = H-C-C=O-H
OH
CALCULATIONS INVOLVING E.F AND M.F
Exercise 1: A compound has the following % composition by mass, C= 40%, H= 6.67% and O = 53.3% calculate the E.F of the compound. If its ml. mass is 180, find its mol. Formula (C=12, H= 1, O= 16)
Answer: E.F = CH2O
M.F = C6H12O6
Note: M.F = (E.F)n = Rmm
Rmm = 2 x v.d
Where mass is given instead of % composition, it can still be used i.e. mass = % composition.
Exercise 2: the analysis of a compound gave the Hg results:
5.2g of the compound contained 1.935g of carbon,
3.2g of the compound contained o.46g of hydrogen,
1.2g of the compound contained 0.6g of oxygen.
Calculate the E.F (C = 12, O= 16, H=1)
Solution: C H O
% Composition 1.935 x 100% 0.46 x 100% 0.6 x 100%
52 3.60 1.2
37.2% 12.8% 50%
37.2%12.8%50%
12 1 16
Note: if 5.2g has been for all the components i.e. no 3.6g and 1.2g, then you solve directly.
Exercise 3: 6g of metal x reacts completely with 23.66g of chlorine to form 29.66g of the metallic chloride.
Find the E.F of the metallic chloride
If the v.d of the compound is 133.5
Find its mol. Formula (x = 27, Cl = 35.5)
Solution: x Cl
Mass composition 6 23.66
6/2723.66/35.5
0.22 0.67
0.220.67
0.22 0.22
1 3
E.F = XCl3
mol. Mass
v.d = 133.5
mol mass = 133.5 x 2 = 267
(XCl3)n = 267
27 + [35.5 x 3]n = 267
133.5n = 267
n= 267 =2
133.5
Mol. Formula = (XCl3)2 = X2Cl6
Exercise 4: A hydrocarbon on combustion given 0.704 g of CO2 and 0.216g of H2O. If the relative mol. Mass of the compound is 54, calculate E.F. and M.F.
Solution: Hydrocarbon contains carbon and hydrogen only.
Rmm of CO2 = 44g, RAM of Carbon = 12
Rmm of H2O = 18g, RAM of Hydrogen = 2
4 4g of CO2 → 12g of C
Therefore 0.704g CO2 → x
X = 12/44 x 0.704 = 0.192 of C
18g of H2O → 2g of H
Therefore 0.216g → x
X = 0.216 x 2 = 0.024g of hydrogen
18
C H
Mass composition 0.192 0.024
RAM 12 1
0.1920.024
12 1
0.0160.024
0.016 0.016
= 1 : 1.5
2 3
E.F = C2H3
M.F = (E.F)n = mm
(C2H3)n = 54
27n = 54
n = 2
Therefore M.F = C4H6
Note: Where the V.D and RMM are not given use this formula to get V.D.
v.d = mass of a certain vol. of a gas
mass of an equal vol. of H2
PERCENTAGE BY MASS OF ELEMENT IN A COMPOUND.
Calculate the percentage by mass of nitrogen in HNO3 (H = 1, N = 14, O = 16)
Solution :
Percentage by mass of N = mass of N /molar mass of HNO3 x 100%
Molar mass of HNO3
= 1 + 14 + (16 x 3) = 63
Mass of Nitrogen, N = 14
= 14/ 63 x 100% = 22.2%
Calculate the percentage by mass of all the component elements in NaNO3 ( Na = 23, N= 14, O= 16)
Solution:
%by mass of Na = 23/85 x 100% = 27%
% by mass of N = 14/85 x 100% = 16.5%
% by mass of O = (3 x 16) / 85 x 100% = 56.5%
Evaluation :
ANSWER THE FOLLOWING QUESTIONS
1. A volume of a gas Z was found to weigh 6.5g and when corrected to s.t.p it measured 4.84dm3. Calculate the G.m.m of the gas Z
(H =1, S=32, 0=16)
2. Calculate the number of:
a. atoms in 2.5mole of Na (sodium)
b. ions present in 0.5 moles of copper (II) ions (Cu2+)
3. A compound contains 40 C, 6.66% H and a certain % of O. Calculate the E.F. If its mol. mass is 180. Calculate M.F.
4. 5.05g of a compound was found to contain 4g of Ca and 0.35g of S and 0.70g of Oxygen calculate its E.F (Ca = 40, S= 32, O=16)
5. A gaseous hydrocarbon contains 92.3% C, and 7.7% hydrogen by mass. 300cm3 of the hydrocarbon weighs 0.301g and under the same conditions of temperature and pressure;
6. 300cm3 of H weighs 0.023g.
Find the E.F of the hydrocarbon
Determine the M.F of the hydrocarbon (C = 12, H =1).
CONCLUSION : At the end of the lesson learners were able to answer the questions.
ASSIGNMENT :
1. The relative atomic mass of calcium atom is 40. This means that
(A) the mass of calcium is 40g
(B) the calcium is 40 times heavier than that of 1 atom of hydrogen
(C) calcium is 40 times that of 1g of hydrogen
(D) calcium is related to hydrogen through 40 digits
2. The relative molecular mass of lead (II ) trioxonitrate (v) is ________ (Pb=108,N=14,O=16).
(A).170 (B).222 (C).232 (D).132
3. Which is heavier? 1 mole of PbCl2 , 1 mole of H2 and 1 mole of pb(NO3)2?
(A) .PbCl2 (B) H2 (C) None of them (D) Pb(NO3)2
4. How many toms are contained I mole of hydrogen molecule.
(A) 18.09 x 10 23 atoms (B) 12.06 x 10 23 atoms
(C) 6.02 x 10 23 atoms (D) 6.02 x 10 23 molecules
6.The percentage of oxygen in sulphur (iv)oxide is (S = 32, O = 16)
(A) 5% (B) 50% (C) 500% (D) 25%
7. The empirical formula of C6H6 is
(A) CH (B) C3H3 (C) C6H6 (D) 3CH
8. If the relative molecular mass of CH2O is 60, calculate the empirical formula. (C = 12, H = 1, O = 6 )
(A) 4 (B) 1 (C) 2 (D) 3