1.1 Atoms and molecules 1.O MATTER
LEARNING OUTCOMES At the end of this topic, students should be able to: a) Write isotopic notation. (C1) b) Interpret mass spectrum. (C2, C3) c) Calculate the average atomic mass of an element given d) The relative abundances of isotopes or a mass spectrum. (C3, C4) * Include calculatation relative atomic mass based on C-12
Anything that has mass and volume MATTER 3
Compounds Mixtures Pure Substances MATTER Elements Homogeneous Heterogeneous Physical process Chemical process Atoms molecules ions molecules protons neutrons electrons 4
❑ Two or more atoms of the same element having same proton number but different nucleon number. ISOTOPE ❑ Two or more atoms of the same element having same number of protons but different number of neutrons. @ 5
H 1 1 H (D) 2 1 H (T) 3 1 U 235 92 U 238 92 Protium Deutrium Tritium EXAMPLE 1 : 6
Isotopes Notation @ Atomic Symbol U 235 92 H 1 1 Example: The symbol for an atom: X A Z Nucleon Number Proton Number Element Symbol 7
❑ Number of protons in the nucleus of an atom of an element. ❑ Also called Atomic Number. PROTON NUMBER (Z) All carbon atoms (Z = 6) have 6 protons All oxygen atoms (Z = 8) have 8 protons All uranium atoms (Z = 92) have 92 protons EXAMPLE: 8
Nucleon Number =Number of protons + number of neutrons Number of Neutrons = Nucleon Number – Proton Number (A) (Z) NUCLEON NUMBER (A) ❑ Nucleon number also called mass number ❑ Total number of protons and neutrons in the nucleus of an atom of an element 9
10 Cl 35 17 Number of neutrons = A – Z =35 – 17 = 18 Nucleon number of chlorine, A = 35 Proton number of chlorine, Z = 17 EXAMPLE 2 :
➢ ion with a positive charge ➢ If a neutral atom loses one or more electrons it becomes a cation. Na 11 protons 11 electrons Na+ 11 protons 10 electrons CATION Loses electron 11 Cation
➢ ion with a negative charge ➢ If a neutral atom gains one or more electrons it becomes an anion. Cl 17 protons 17 electrons Cl17 protons 18 electrons Gains electron Anion 12 ANION
NOTES !! ❑The proton number, Z, is the nuclear charge and also the number of electrons in a neutral atom of the element. 0 charge +ve charge neutral cation (atom gained electrons) -ve charge anion 13 ❑No. of proton = no. of electron ❑No. of proton > no. of electron (atom lost electrons) ❑No. of proton > no. of electron
The number of neutrons = A – Z = 202 – 80 = 122 14 Total charge on the ion Nucleon number of mercury, A = 202 Proton number of mercury, Z = 80
Total charge on the ion 15 ➢ Calculate the number of electrons of ion mercury, Hg2+ . Tips: Charge positive: (+ve) Cation- atom lost electron No.of proton > no.of electron The number of electrons = =
Write the appropriate notation for each of the following species : 16 EXAMPLE 3 : Species Number of : Isotope Notation Proton Neutron Electron A 2 2 2 B 1 2 0 C 1 1 1 D 7 7 10
INTERPRET MASS SPECTROMETER 0.33 2.19 5.82 91.66 Abundance (%) mass/charge 54 56 57 58 Mass spectrum for Ferum ➢ The height is proportional to the amount of each isotope present. 17
❑height of each line = abundance of each isotope. ❑numbers of peaks = types of isotopes. ❑ratio of m/e for each species is found from the value of the accelerating voltage associated with a particular peak. ❑height of the peak is directly proportional to its abundance Information Of A Mass Spectrum 18
❑ The average of mass of its naturally occurring isotopes weighted according to their abundances AVERAGE ATOMIC MASS @ 19 = ∑( × ) ∑ = ∑(% × ) ∑ %
Calculate the average atomic mass of Mg. Atomic mass (amu) 23 24 25 26 Relative Intensity 63.0 8.1 9.1 20 EXAMPLE 4:
= 24.33 amu Answer 21 = ∑( × ) ∑ = 63.0 × 24 + (8.1 × 25) + +(9.1 × 26) 63.0 + 8.1 + 9.1
Nitrogen (N, Z = 7) has two naturally occurring isotopes. Calculate the percent abundances of 14 and 15 from the following: atomic mass (average) of N = 14.0067 amu; isotopic mass of 14 = 14.0031 amu; isotopic mass of 15 = 15.0001 amu. 22 EXAMPLE 5:
Average atomic mass of nitrogen = ( % of 14N x isotopic mass of 14N ) + (% of 15N x isotopic mass of 15N ) Answer Let : abundance of 14N = % , abundance of 15N = 100% - % 23 14.0067 = 100 × 14.0031 + 100− 100 × 15.0001 1400.67 = 14.0031 + 1500.01 – 15.0001 = 99.64 abundance of 15N = 100 – = 100 – 99.64 = 0.36 % Thus, = % abundance of 14N = 99.64 % &
Relative Mass ❑The mass of an atom relative to another atom can be determined experimentally using mass spectrometer. ❑Currently, carbon-12 isotope is used as standard to measure relative atomic mass. ❑Relative atomic mass is dimensionless. 24
RELATIVE ATOMIC MASS, ❑ A mass of one atom of an element compared to one twelfth mass of one atom of carbon-12 atom Relative atomic mass = mass of one atom of an element (amu) × mass of one 12C atom (amu) ❑Relatives atomic mass is a ratio, hence, it has no UNIT 25
Magnesium Carbon-12 1 atom Mg have the same weight as 2 atoms of 12C of Mg = 2 x 12 = 24 amu 26 How do they compare ?
Relative Atomic Mass of Mg 27 = 1 12 × 12 , = 24 1 12 × 12 = 24
28 Determine the relative atomic mass of an element Y if the ratio of the atomic mass of Y to carbon-12 atom is 0.75. Since the atomic mass ratio is 0.75 mass of one atom Y mass of one atom 12 C = 0.75 mass of one atom Y = 0.75 x 12.00 amu = 9.0 amu relative atomic mass, Y = 9.0 1 12 ×9.0 = 9.0 EXAMPLE 6: Answer
RELATIVE MOLECULAR MASS, ❑ A mass of one molecule of a compound compared to one twelfth mass of one atom of carbon-12 atom Relative molecular mass, = mass of one molecule of a compound (amu) × mass of one 12C atom (amu) ❑ Relatives molecular mass is a ratio, hence, it has no UNIT 29
9.2% 20Ne+ 0.3% Abundance (%) mass/charge 20 21 22 100 80 60 40 20 90.5% 21Ne+ 22Ne+ Calculate the relative atomic mass of neon from the mass spectrum. 30 EXAMPLE 7:
31 Answer NOTE: If the atomic mass is not given, the mass number can be used as atomic mass. = ∑(% × ) ∑ % = 90.5 × 20 + 0.3 × 21 + (9.2 × 22 ) 90.5 + 0.3 + 9.2 = 20.2 u = () × () = 20.2 1 12 × 12.0 = 20.2
Copper occurs naturally as mixture of 69.09% of 63Cu and 30.91% of 65Cu. The isotopic masses of 63Cu and 65Cu are 62.93 u and 64.93 u respectively. Calculate the relative atomic mass of copper. 32 Relative atomic mass Cu = 63.55 EXERCISE 1:
Naturally occurring iridium, Ir is composed of two isotopes, 191Ir and 193Ir in the ratio of 5:8. The relative isotopic mass of 191Ir and 193Ir are 191.021 u and 193.025 u respectively. Calculate the relative atomic mass of Iridium. 33 Relative atomic mass Ir = 192.254 EXERCISE 2:
The atomic masses of 3 6 and 3 7 are 6.0151 a.m.u. and 7.0160 a.m.u. respectively. What is the relative abundance of each isotope if the relative atomic mass of lithium is 6.941? % of 3 6 = 7.49 % % of 3 7 = 92.51 % 34 EXERCISE 3:
Edited By : NZSH,NAAN ……………………… Revised By : MPJ ……………………… Approved By : ZA ………………………… 35
1.2 MOLE CONCEPT
LEARNING OUTCOMES At the end of this topic, students should be able to: a) Define the terms empirical and molecular formulae. (C1) b) Determine empirical and molecular formulae from mass composition or combustion data. (C3) c) Determine the empirical formula (formula unit) from experiment (C3) d) Define each of the following concentration measurements:(C1) i. Molarity (M) ii. Molality (m) iii. Mole fraction (X) iv. Percentage by mass (%w/w) v. Percentage by volume (%v/v)
LEARNING OUTCOMES At the end of this topic, students should be able to: e) Calculate each of the following concentration measurements: (C3, C4) i. Molarity (M) ii. Molality (m) iii. Mole fraction (X) iv. Percentage by mass (%w/w) v. Percentage by volume (%v/v)
CHEMICAL FORMULA ❑ formula that shows the actual number of atoms of each element in a molecule ❑ formula that shows the simplest ratio of all elements in a molecule. Molecular Formula Empirical Formula 39
Molecular formula = n (Empirical formula) H2O H2O Empirical Formula Molecular Formula CH2O C6H12O6 NH2 N2H4 ➢ The relationship between empirical formula and molecular formula is : 40 EXAMPLE 1:
A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56 g mol-1 . Determine the empirical formula and molecular formula of the compound. 41 EXAMPLE 2:
= 7.1417 = 14.3 = 1.0 = 2.0 Assume: mass of hydrocarbon = 100 g Element C H Mass (g) 85.7 14.3 Number of mole (mol) Simplest ratio 42 Answer = = = 85.7 12.0 = 14.3 1.0 = 14.3 7.1417 = 7.1417 7.1417 Empirical Formula CH2
Molecular Formula = C4H8 Molecular formula = n (Empirical formula) Molecular formula = 4 (CH2 ) 43 () = [ . + . = = 56 14.0 = 4 ➢ How to find molecular formula when given molar mass and empirical formula?
A white solid was analysed and found to contain 40.0% Carbon, 6.7% Hydrogen and 53.3% oxygen by mass. What is the empirical formula of the substance? 44 EXAMPLE 3:
Total % by mass = 40.0% + 6.7% + 53.3 % = 100 % Element Empirical Formula 40.0 6.7 53.3 = 3.333 = 6.700 = 3.331 = 1 = 2 = 1 Mass (g) Number of mole (mol) Simplest mole ratio C H O 45 Answer Assume: mass of hydrocarbon = 100 g = = 3.333 3.331 = = = 6.7 1.0 = 53.3 16.0 = 40.0 12.0 = 6.700 3.331 = 3.331 3.331 CH2O
Combustion of 2.30 g of an organic sample, X, yields 3.30 g CO2 and 1.80 g of H2O. Determine the empirical formula of X. Organic X contains elements: Mass of organic X = 2.30 g Strategy : Relate the information given Mass of C + H + O = 2.30 g Mass of C = ? Mass of H = ? Mass of O = ? C, H Mass of CO2 = 3.30 g Mass of H2O = 1.80 g 46 EXAMPLE 4: Empirical Formula By Combustion Data & O
Mole of CO2 1 mol of CO2 contains 1 mol of C 0.075 mol of CO2 contain 0.075 mol of C Mass of carbon = no. of mole C x molar mass C = 0.075 x 12.0 mass of CO2 47 Answer mol of CO2 mol of C mass of C = . . − = 0.075 mol CO2 Mass of carbon = 0.90 g C Mass of C = ?
Mass of O = 2.30 – 0.90 – 0.20 Mole of H2O 1 mol of H2O contains 2 mol of H 0.10 mol of H2O contain 0.20 mol of H Mass of hydrogen = 0.20 x 1.0 = 0.20 g H Mass of C + H + O = 2.30 g Mass of O = 1.20 g 48 mass of H2O mol of H2O mol of H mass of H = . . − = 0.10 mol Mass of H = ? Mass of O = ?
Element C H O Mass (g) Number of mole Simplest mole ratio 0.90 0.20 1.20 0.90 / 12.0 = 0.075 0.20 / 1.0 = 0.200 1.20 / 16.0 = 0.075 / 0.075 = 1 / 0.075 = 2.667 / 0.075 = 1 0.075 0.200 0.075 C3 H8O3 1 x 3 = 3 2.667 x 3 =8 1 x 3 = 3 Never round off values close to whole number in order to get a simple ratio, but multiply the value by a factor until you get a whole number. 49 The empirical formula is
Another method of calculation Combustion of 2.30g of an organic sample, X, yields 3.30 g CO2 and 1.80 g of H2O. Determine the empirical formula of X. Mass of C = Mass of C = 0.90 g C 12.0 gmol−1 44.0 gmol−1 × 3.30 g 50 Answer FROM EXAMPLE 4 Mass of C = ?