CHEM 2, PT 1 PDF - Chemical Symbols and Formulae

1 CHEMICAL SYMBOLS AND FORMULAE A chemical symbol is an abbreviated form of the name of an element. An element is defined as a substance which cannot be split into simpler units by...

1 CHEMICAL SYMBOLS AND FORMULAE A chemical symbol is an abbreviated form of the name of an element. An element is defined as a substance which cannot be split into simpler units by an ordinary chemical process. Elements are represented by symbols. Group A: Symbols written with the first letter of the name of the element. Element Symbol Iodine I Nitrogen N Oxygen O Fluorine F Boron B Group B: Symbols which are the first two letters of the name of the element. Element Symbol Bromine Br Krypton Kr Xenon Xe Cobalt Co Group C: Symbols which are the first letter and any other letter in the name in the middle of the name of the element. Element Symbol Zinc Zn Manganese Mn Chromium Cr Platinum Pt Cadmium Cd Group D: Symbols of elements derived from their Latin names. English Name Latin Name Symbol Sodium Natrium Na Iron Ferrous Fe Potassium Kalium K Gold Aurum Au Mercury Hydragyrum Hg Silver Argentum Ag Tin Stannum Sn Lead Plumbum Pb Antimony Stibium Sb Tungsten Wolfram W A chemical formula is the formula of a compound which shows the type and number of elements present. A chemical formula is written using the symbols of elements, valency and radicals. Valency: is the combining power of an element. Valency of elements can be determined from their atomic numbers. 1  2 Atomic Number Name of Element Symbol Electronic Structure Valency 1 Hydrogen H 1 1 2 Helium He 2 0 3 Lithium Li 2, 1 1 4 Beryllium Be 2, 2 2 5 Boron B 2, 3 3 6 Carbon C 2, 4 4 7 Nitrogen N 2, 5 3 8 Oxygen O 2, 6 2 9 Fluorine F 2, 7 1 10 Neon Ne 2, 8 0 11 Sodium Na 2, 8, 1 1 12 Magnesium Mg 2, 8, 2 2 13 Aluminum Al 2, 8, 3 3 14 Silicon Si 2, 8, 4 4 15 Phosphorous P 2, 8, 5 3 16 Sulphur S 2, 8, 6 2 17 Chlorine Cl 2, 8, 7 1 18 Argon Ar 2, 8, 8 0 19 Potassium K 2, 8, 8, 1 1 20 Calcium Ca 2, 8, 8, 2 2 How to obtain the valency of an element Hydrogen: It has only one electron, so it is unstable, for it to be stable, the first shell must contain two electrons. It needs one more electron, therefore the valency is 1. Helium: It has two electrons in the first shell, it is stable therefore it does not need any electron, hence the valency is 0. Lithium: It has three electrons, 2 electrons goes to the first shell and the third electron goes to the second shell. For the second shell to be stable, it has to contain 8 electrons, so it needs 7 more, instead of acquiring 7 more electrons it loses 1 at the second shell to look like helium. Therefore the valency is 1 because it loses an electron and that also makes it a metal and a Li+ ion. Beryllium: The second shell has two electrons which is lost so that it can be like helium, therefore the valency is 2. Boron: It has 3 electrons in the second shell, instead of gaining 5 electrons to complete it eight, it loses three electrons to become like helium therefore the valency is 3 and is a metal with a B3+ ion. Carbon: 2  3 It has 4 electrons in the second shell, so it requires 4 or loses 4 and therefore the valency is 4. It is involved in covalent combination. Nitrogen: Instead of losing 5 electrons in the second shell, it gains 3, therefore the valency is 3. It is a non-metal because it gains electrons with a N3- ion. Oxygen: Instead of losing 6 electrons in the second shell, it gains to two, therefore the valency is 2. It is a non-metal with an O2- ion. Fluorine: It has seven valence electrons, so it requires one more electron. The valency is one a non- metal with a F- ion. Neon: It has eight electrons in the valence shell, therefore it is stable and does not require any electron. The valency is zero. It does not form any ion. Sodium: It has one valence electron, which is lost to become a Na+ ion. The valency is one. Magnesium: It has two valence electrons, which are lost to become a Mg2+ ion, a metal. Aluminium: It has three valence electrons, which is lost to become a Al3+ ion, a metal. Silicon: It has four valence electrons. It neither loses nor gains electrons, but involved in covalent combination. Phosphorous: Has five valence electrons, and therefore requires three electrons, so the valency is 3, and it has a P3- ion. Sulphur: Has six valence electrons and needs two more electrons. The valency is 2, a S2- ion and a non-metal. Chlorine: Has seven valence electrons and needs one more electron, therefore the valency is one and a Cl- ion representing a non-metal. Argon: Has eight valence electrons and therefore does not require any electron, it stable and has a zero valency. Potassium: Has one valence electron which is lost to have a valency of one and a metal with a K+ ion. Calcium: Has two valence electrons which are lost to have a valency of two, a metal with a Ca2+ ion. Radical: is a group of atoms that carry electric charge. They are treated as one entity during chemical formula. 3  4 Radicals Symbol Valency Dioxonitrate (iii) ion NO2- 1 Trioxonitrate (v) ion NO3- 1 Ammonium ion NH4+ 1 Hydroxide ion OH- 1 Trioxosulphate (iv) ion SO32- 2 Tetraoxosulphate (vi) ion SO42- 2 Trioxocarbonate (iv) ion CO32- 2 Tetraoxophosphate (v) ion PO43- 3 Hydrogen trioxocarbonate (iv) ion HCO3- 1 Hydrogen tetraoxosulphate (vi) ion HSO4- 1 Hydrogen tetraoxophosphate (v) ion HPO42- 2 Dihydrogen tetraoxophosphate (v) ion H2PO4- 1 Ion: is single atom or group of atoms that carry electric charge, eg. Na+, Cl-, OH-, CO32-. There are two types of ions:- anion and cation. Anion: is a single atom or group of atoms carrying a negative charge, eg. Cl -, CO32-, F-. Cation: is a single atom or group of atoms carrying a positive charge, eg. Na+, NH4+, Ca2+. When two oppositively charged ions combine it forms a neutral compound, eg, Na+ + Cl- NaCl Valency of some ions Cation Symbol Valency + Sodium ion Na 1 3+ Aluminum ion Al 3 2+ Magnesium ion Mg 2 + Copper (i) ion Cu 1 2+ Copper (ii) ion Cu 2 2+ Iron (ii) ion Fe 2 3+ Iron (iii) ion Fe 3 + Silver ion Ag 1 2+ Lead (ii) ion Pb 2 4+ Lead(iv) ion Pb 4 + Hydrogen ion H 1 2+ Calcium ion Ca 2 + Mercury(i) ion Hg 1 2+ Mercury(ii) ion Hg 2 Anion Symbol Valency Chloride ion Cl- 1 Oxide ion O2- 2 4  5 Bromide ion Br - 1 - Iodide ion I 1 2- Sulphide ion S 2 - Fluoride ion F 1 3- Nitride ion N 3 - Hydride ion H 1 Guidelines for working out formulae from valencies 1. Valency 1 is never indicated 2. When two elements have equal valencies, they are not indicated in the formula of a compound, eg. CuO 3. In the formula of a compound, metallic elements (cations) are written first and non-metal (anion) second, eg. Al2O3 4. Where valencies of two elements have a common factor, they can be divided by that factor, eg. C2O4 is written as CO2, where the common factor is 2 Atoms combine by exchange of valency, eg. 1. Those elements and radicals with the same valencies combine by cancelling out the valencies and the formula written like that, eg; (a) Na+ Cl- (b) Ca2+ SO42- (c ) Be2+ S2- (d) H+ F- (e) Mg2+ O2- 1 1 2 2 2 2 1 1 2 2 NaCl CaSO4 BeS HF MgO Sodium chloride Calcium tetraoxosulphate (VI) Beryllium sulphide Hydrogen fluoride Magnesium oxide 2. Those elements and radicals with different valencies, they combine by exchange of their valencies, eg; (a) Al3+ O2- (b) Al3+ Cl- (c) H+ CO32- (d) Mg2+ N3- 3 2 3 1 1 2 2 3 Al2O3 AlCl3 H2CO3 Mg3N2 Aluminium oxide Aluminium chloride Trioxocarbonate (IV) acid Magnesium nitride 3. Those elements with different valencies, but have a common factor, it is divided by that factor, eg; C4+ O2- 4 2 4 2 /2 /2 2 1 CO2 Carbon (IV) oxide 4. Those radicals that assume a valency more than one, they are put in parenthesis, bracket, eg; 5  6 (a) NH4+ SO42- (b) Al3+ CO32- (c) Ca2+ HCO3- 1 2 3 2 2 1 (NH4)2SO4 Al2(CO3)3 Ca(HCO3)2 Ammonium tetraoxosulphate (VI) Aluminium trioxocarbonate (IV) Calcium hydrogen trioxocarbonate (IV) Why do elements combine? Element combine in order to attain the duplet and octet stability. Octet stability refers to the stability of atoms which is achieved by the presence of eight electrons in their outermost shells, while duplet stability is attained by the presence of two electrons in the first shell if the first shell happens to be the outermost or only shell. Some Compounds and their Chemical Formulae Compounds Chemical Formula Sodium Chloride NaCl Ammonium tetraoxosulphate (vi) (NH4)2SO4 Aluminum Hydroxide Al(OH)3 Calcium Sulphide CaS Aluminum tetraoxosulphate (vi) Al2(SO4)3 Magnesium Chloride MgCl2 Sodium trioxocarbonate (iv) Na2CO3 Sodium Sulphide Na2S Calcium trioxocarbonate (iv) CaCO3 Aluminum Sulphide Al2S3 Potassium bromide KBr Sodium Iodide NaI Aluminum bromide AlBr3 Magnesium Nitride Mg3N2 Magnesium oxide MgO How to get the charge on a radical from the name a. Trioxonitrate (v) acid = H N O3 place 5 under nitrogen H+ NO3r 5 + (-2 x 3) = r 5 -6 = r -1 = r. It means NO3 is -1, ie. valency of 1 H NO3- + 1 1 HNO3 b. Hydrogen trioxocarbonate(iv) HCO3r +1 + 4 +(-2 x 3) = r +5 – 6 = r -1 = r, therefore the valency is 1 c. Ammonium ion 6  7 NH4r -3 + (+1 x 4) = r -3 + 4 = r +1 = r, ie, the valency is 1 Test Yourself Find the valency or charge on the following: a. Hydroxide ion, OHr b. Tetraoxosulphate (VI) ion, SO42- Practice Questions Write the formulae of the following compounds: 1. Iron (ii) trioxonitrate (v) 13. Nitrogen (iv) oxide 2. Iron (ii) tetraoxosulphate (vi) 14. Sodium trioxocarbonate(iv) 3. Iron (ii) trioxosulphate (iv) 15. Iron (iii) trioxonitrate (v) 4. Calcium oxide 16. Aluminum tetraoxosulphate (vi) 5. Sodium hydride 17. Magnesium tetraoxosulphate (vi) 6. Potassium iodide 18. Sodium trioxonitrate (v) 7. Ammonium chloride 19. Barium tetraoxosulphate (vi) 8. Sodium hydroxide 20. Calcium hydrogen trioxocarbonate(iv) 9. Tetraoxosulphate (vi) acid 21. Potassium trioxocarbonate(iv) 10. Copper (i) oxide 11. Dioxochlorate (iii) acid 12. Copper (ii) oxide Empirical and Molecular Formula Empirical Formula: is the simplest formula of a compound which indicates each kind of atom and their ratios in one molecule of a substance. Calculation 1. Calculate the empirical formula of a compound which contains 52.2% carbon, 13.0% hydrogen and 34.8% oxygen. (C = 12, O = 16, H = 1) Solution: C H O % composition: 52.2 13.0 34.8 52.2 Divide by atomic mass: /12 13.0/1 34.8/16 Mole: 4.35 13.0 2.175 4.35 Divide by the smallest: /2.175 / 2.175 2.175/2.175 13.0 Ratio 2 5.9 ≈ 6 1 Empirical Formula: C2H6O 2. Find the empirical formula of the following compounds with the composition below: a. Ca = 40%; C = 12%; O = 45% (Ca = 40, C = 12, O = 16) b. H = 11.1%, O = 88.9%, (H = 1, O = 16) c. Na = 37.1%, C = 9.7%, O = 38.7%, H2O = 14.5% (Na = 23, C = 12, O = 16, H2O=18) 7  8 Molecular Formula: is the formula which expresses the actual number of atoms of an element present in one mole of a compound. Calculation 1. A certain gas contains 82.8% carbon and 17.2% hydrogen by mass. The vapour density of the gas is 29. Determine its molecular formula. (C = 12, H = 1) Solution: C H 82.8 17.2 82.8 /12 17.2/1 6.9 17.2 6.9 17.2 /6.9 /6.9 1 2.5 ≈ 3 Empirical formula = CH3 (empirical formula)n = molecular mass (CH3)n = 2 x vapour density (12 + 1x3)n = 2 x 29 15n = 58 n = 58/15 = 3.9 ≈ 4 Molecular formula = (CH3)4 = C4H12 2. A certain organic acid contains 26.7% carbon, 2.2% hydrogen and 71.1% oxygen by mass. The molecular mass of the acid is 90. Calculate the molecular formula of the acid. (C = 12, H = 1, O = 16) 3. 1.008g of an organic acid contains 0.016g hydrogen, 0.192g carbon, 0.512g oxygen and 0.288g of water of crystallization. Calculate the empirical formula (H = 1; C = 12; O = 16; H2O = 18) Solution: C H O H2O 0.192 0.016 0.512 0.288 /1.008 x 100 /1.008 x 100 /1.008 x 100 /1.008 x 100 19.0% 1.6% 50.8% 28.6% 19.0 1.6 50.8 28.6 /12 /1 /16 /18 1.6 1.6 3.2 1.6 1.6 1.6 3.2 1.6 /1.6 /1.6 /1.6 /1.6 1 1 2 1 Empirical Formula = CHO2.H2O or HCO2. H2O 4. An organic compound X contains 40% carbon, 6.67% hydrogen, the rest being oxygen. If X has a relative molecular mass of 60. Determine its: a. empirical formula b. molecular formula (H= 1, C = 12, O = 16) 5. A hydrocarbon whose vapour density is 39 consists of 92.3% carbon. Determine its molecular formula (H = 1, C = 12) 6. A compound contains 40.4% carbon, 6.7% hydrogen and 53.3% oxygen. Determine its molecular formula if its molar mass is 180. (C = 12; H = 1; O = 16) 7. TWO elements Q and Z have electronic configuration of 1s2 2s2 2p6 3s2 3p4 and 1s2 2s2 2p4. Write the formula of the compound formed by Q and Z when they react together. 8  9 Oxidation Number Oxidation number is the charge an atom would have in a neutral molecule or in an ion. It is referred to as the oxidation state of the molecule or ion. Rules for Assigning Oxidation Number 1. Oxidation number of elements in uncombined state whether mono atomic or poly atomic is zero. O2 = 0; P4 = 0; S8 = 0; Na = 0 2. Oxidation number of oxygen in a combined state is -2 except for peroxides, eg. Na2O2, H2O2, BaO2, where it is -1. 3. Oxidation number of hydrogen is +1 except in metallic hydrides where it is -1,eg. NaH, CaH2. 4. Oxidation number of ions equals to the charge of the ion, eg. Ca2+ = +2, O2- = -2, etc. 5. In the combined state, the oxidation number of elements equal to the valency of the element, but carrying positive charge for metals, negative charge for non- metals. 6. In a neutral compound, the sum of the oxidation numbers of all the elements equal to zero, eg. CaCO3 = 0, HCl = 0 7. In radicals, the sum of the oxidation numbers of all the elements equal to the charge it carries, eg. NH4+ = +1, SO42- = -2 Calculation 1. What is the oxidation number of nitrogen in HNO3? Let N be x Solution; H N O3 = 0 +1+x+(-2x3) = 0 1+x–6=0 x–5=0 x = +5; tri oxo nitrate (V) acid 2. What is the oxidation number of carbon in HCO3 - ? let C be x Solution: H C O3- = -1 +1+x+(-2x3) = -1 1 + x – 6 = -1 x – 5 = -1 x = +5 – 1 x = +4 the sign must be written to score 3. Calculate the oxidation number of the underlined elements: a. CuCl b. CuCl2 c. HCl d. NH3 e. Na2O f. Na2O2 g. PO43- h. CaH2 9  10 IUPAC Nomenclature IUPAC means International Union of Pure and Applied Chemistry. This system was employed to make or set down the guiding principles of naming compounds in order to be universal. Rules for Naming Inorganic Compounds 1a. Binary compounds are compounds that contain two different elements only. Their names end with –ide, eg. Calcium oxide CaO, Magnesium chloride MgCl2 b. The less electronegative element or metallic elements/cations are named first, while the more electronegative element or non-metals/anions are named last and end with –ide. However, there are exceptions like water H2O, ammonia NH3, phosphine PH3. c. The radicals are treated like a single element, eg. KCN = potassium cyanide, NaOH = sodium hydroxide, ammonium chloride NH4Cl. d. The oxidation number of elements having more than one oxidation number is represented in Roman numerals in brackets, eg. Copper (i) oxide Cu2O, copper (ii) oxide CuO, FeCl2 = iron (ii) chloride and FeCl3 = iron (iii) chloride. When an element has a fixed or one oxidation number, it is omitted, eg. Aluminum chloride AlCl3. 2. Tertiary and Quaternary compounds. These are compounds containing more than two elements in their molecules. a. Oxo acids: These are acids having oxygen in their molecules. They are named as follows: i. Oxygen is named first as –oxo with the number of oxygen atoms indicated using the Greek /Latin prefix. 1 – mono 6 – hexa 2 – di 7 – hepta 3 – tri 8 – octa 4 – tetra 9 – nona 5 – penta 10 – deca ii. The central atom, a metal or non-metal is named, but ending with –ate followed by the oxidation number in Roman numeral in bracket, eg. 1=i 2 – ii 3 – iii 7 – vii 4 – iv 8 – viii 5–v 9 – ix 6 – vi 10 – x iii. The word acid ends the name, eg. HNO3 = Trioxonitrate (v) acid HClO3 = Trioxochlorate (v) acid HClO = monoxochlorate (i) acid or oxochlorate (i) acid b. Acid radicals: These are formed by the partial or complete removal of hydrogen atoms from acids. Their names are the names of the acids except that the acid is replaced by ion, eg. 10  11 SO42- = tetraoxosulphate (vi) ion HPO4- = hydrogen tetraoxophosphate (vi) ion H2PO42- = dihydrogen tetraoxophosphate (iv) ion PO43- = tetraoxophosphate (v) ion S2O32- = thiosulphate ion (an exception), eg, Na2S2O3 = sodium thiosulphate MnO4- = tetraoxomanganate(vii) ion c. Salts (anhydrous): Anhydrous salts are salts that do not contain water of crystallization, eg. Acid salts and normal salts. In naming them, the metallic ion or radical is named first, if it has two or more oxidation numbers, it is indicated by a Roman numeral in bracket, followed by the name of the radical, but without the ion, eg. KNO3 = Potassium trioxonitrate (v) Na2SO4 = Sodium tetraoxosulphate (vi) FeSO4 = Iron (ii) tetraoxosulphate (vi) Fe2(SO4)3 = Iron (iii) tetraoxosulphate (vi) CuO = Copper (ii) oxide Cu2O = Copper (i) oxide Acid salts include; NaHCO3 = Sodium hydrogen trioxocarbonate (iv) KHSO4 = Potassium hydrogen tetraoxosulphate (vi) NaH2PO4 = Sodium dihydrogen tetraoxophosphate (v) d. Hydrated Salts: These are salts with water molecules attached to them. In naming them, the name of the anhydrous salt is written first followed by the number of water molecules written in Latin words, eg. Na2CO3.10H2O = Sodium trioxocarbonate (iv) decahydrate CuSO4.5H2O = Copper (ii) tetraoxosulphate (vi) pentahydrate CoCl2.6H2O = Cobalt (ii) chloride hexahydrate e. Bases and Basic Salts: A base is an oxide or hydroxide of a metal. Bases are named as binary compounds, while basic salts are named as double salts, eg. Fe2O3 = Iron (iii) oxide, FeO = Iron (ii) oxide NaOH = Sodium hydroxide Zn(OH)Cl = Zinc chloride hydroxide f. Complex Salts and Complex ions: Complex salt dissolves in water to give complex ions. A complex ion usually contains a central metal atom, surrounded by a neutral molecule or ion, called ligands. A ligand is an electron-rich particle that may be a neutral molecule such as NH3, H2O, CO, NO, and ions such as Cl-, Br-, I-, OH-, or H-. Rules for naming Complexes i. The cation is named before the anion just like in other salts. ii. The ligand is named before the central ion. iii. Names of negative ligands end with the suffix ‘o’ eg. Br- = Bromo F- = Fluoro Cl- = Chloro I- = Iodo 11  12 NO2- = Nitro H- bounded to nitrogen = Nitro CN- = Cyano H- bounded to oxygen = Nitrito OH- = Hydroxo H- bounded to C2O42- = Oxalato H- = Hydrido H- bounded to SCN = Thiocyano Names of Neutral Ligands 12  13 H2O = Aquo NH3 = Ammine CO = Carbonyl NO = Nitrosyl iv. Where there are more than one particular ligand, it is named di, tri, tetra, etc. v. In an ammonic complex, the name of the central metal is modified to –ate. vi. When there are more than one ligand of different types, the ligands are named alphabetically, eg. Anions, add –ate, start naming from left to right. [Fe(CN)6]3- = hexacyanoferrate (iii) ion [Zn (OH)4]2- = Tetra hydroxozincate (ii) ion [Cr(NH3)Cl2Br3]- = Tribromo dichloro ammine chromate (iii) ion Cations, do not add –ate [Cu(NH3)4]2+ = Tetra ammine copper (ii) ion [Fe(H2O)6]2+ = Hexa aquo iron (iii) ion [Pt(NH3)2(NO2)Cl2]+ = Dichloro di ammine nitro platinium (iv) ion Neutral Complexes [Pt(NH3)2Cl4] = Tetrachloro diammine platinium (iv) [Co(NH3)3Cl3] = Trichloro triammine cobalt (iii) MOLAR MASS This is also known as formula mass. It is the mass of one molecule of a correctly written compound, obtained by adding together the appropriate relative atomic masses of all the atoms of the element present in the molecule. The unit is gram per mole (g/mol or gmol-1) Example: 1. Calculate the molar mass of calcium trioxonitrate (v) (Ca = 40, O = 16, N = 14) Solution Ca 2+ NO3 → Ca(NO3)2 - Molar mass = 40 + (14 + 16x3)2 = 40 + (62)2 = 40 + (62x2) = 40 + 124 = 164g/mol 2. Calculate the molar mass of the following: a. Sodium hydroxide (Na = 23, O =16, H = 1) molar mass = NaOH = 23 + 16 + 1 = 40g/mol b. Magnesium tetraoxosulphate (vi) heptahydrate crystals. (Mg = 24, S = 32, O = 16) Molar mass = MgSO4.7H2O = 24 + 32 + (16x4) + 7(1x2 + 16) = 24 + 32 + 64 + (7x18) = 246g/mol c. Copper (ii) tetraoxosulphate (vi) pentahydrate. (Cu = 64, S = 32, O = 16, H = 1) Formula = CuSO4.5H2O Molar mass = 64 + 32 + (16x4) + 5(1x2 + 16) = 64 + 32 + 64 + 5x18 = 64 + 32 + 64 + 90 = 250g/mol d. Lead (ii) dilead (iv) oxide or Lead (ii) trilead tetraoxide. (Pb = 207, O = 16) 13  14 Formula = Pb3O4 = (207x3) + (16x4) = 621 + 64 = 685g/mol Percentage Composition Percentage composition = mass of atom x 100 Molar mass of compound 1 1. Calculate the percentage composition of sodium in sodium hydroxide. (Na = 23, O = 16, H = 1) Formula = NaOH = 23 + 16 + 1 = 40g/mol, mass of Na = 23 % composition of Na = 23/40 x 100 = 57.5% 2. Calculate the percentage composition of water of crystallization in magnesium tetraoxosulphate (vi) heptahydrate [ Mg = 24, S = 32, O = 16, H = 1] Formula = MgSO4.7H2O = 24 + 32 + (16x4) + 7(1x2 + 16) = 24 + 32 + 64 + 7(2 + 16) = 24 + 32 + 64 + 7x18 = 24 + 32 + 64 + 126 = 246g/mol Mass of water = 126 % composition of water = 126/246 x 100 = 51.2% 3. Mass of oxygen = (16x4) + (7x16) = 64 + 112 = 176g % of oxygen = 176/246 x 100 = 71.5% Questions Calculate the percentage composition of 1. Iron in iron (ii) tetraoxosulphate(vi) {Fe = 56, S = 32, O = 16} 2. Nitrogen in Magnesium trioxonitate(v) { Mg = 24, N = 14, O = 16 } 3. Oxygen in ammonium tetraoxosulphate (vi) {N = 14, H = 1, S = 32, O = 16} 4. Oxygen in sodium trioxocarbonate (IV) monohydrate {O = 16, Na = 23, C = 12, H = 1} MOLE CONCEPT The mole is the amount of a substance which contains as many elementary unit as they are atoms in 12g of carbon – 12. This unit is also known as Avogadro’s number and is equal to 6.023 x 1023 particles. It is calculated thus with the following formulae; n1 = m/M; n2 = V/Vm; n3 = N/NA; n4 = CV(cm3)/1000; n5 = CV(dm3) where n = number of moles m = mass of substance M = atomic/molar/molecular mass V = volume of gas Vm = molar volume of gas at s.t.p.= 22.4dm3 = 22400cm3 N = number of atoms/ions/molecules/particle NA = Avogadro’s number = 6.02 x 1023 C = concentration of solution in moldm3 V = volume of solution in cm3 or dm3 Example: 14  15 1. Calculate the mole of 10.65g of atomic chlorine (Cl = 35.5) n = m/M where n = ?, m = 10.65g, M = 35.5 n = 10.65/35.5 = 0.300 mol 2. Calculate the number of moles of 8.96dm3 of oxygen. (O = 16, Molar vol. = 22.4dm3) n = V/Vm where n = ?, v = 8.96dm3, Vm = 22.4dm3 n = 8.96/22.4 = 0.400mole 3. Calculate the number of moles of hydroxide ions (OH-) in 60.23 x 1022 atoms (Avogadro’s no. = 6.02 x 1023) n = N/NA where n = ?, N = 60.23 x 1022, NA = 6.02 x 1023 n = 60.23 x 1022/6.02 x 1023 = 1.00 mole Diatomic Gases 1. Hydrogen, H2; Molar mass, M = 1 x 2 = 2 g/mol 2. Oxygen, O2; Molar mass, M = 2 x 16 = 32 g/mol 3. Nitrogen, N2; Molar mass, M = 2 x 14 = 28 g/mol 4. Fluorine, F2 5. Bromine, Br2 6. Chlorine, Cl2 7. Iodine, I2 WRITING AND BALANCING OF CHEMICAL EQUATION Writing and balancing of chemical equation. A chemical equation is a condensed statement of facts about a chemical reaction. Procedure for writing a chemical equation 1. Write the formulae of the reactants at the left hand side LHS and the formulae of the products the right hand side RHS then an arrow pointing from the reactants to products. 2. Match the number of atoms of each element to the left with its number to the right. 3. Balance these numbers by writing integer coefficients INFRONT of these formulae deficient in certain atoms. 4. Again match the numbers of atoms of the same elements on both sides of the arrow. To do this, multiply the coefficient by the number of each atom in a formula to get the total number of atoms of that element. Information from Chemical Equation 1. The ratio of masses of reactants and products. 2. The relative volumes of reactants and products, if gaseous. 3. Which reactants are in excess in the mixture of known amounts of reactants and by how much. 4. The state of reactants and products, eg, (s) = solid, (l) = liquid, (g) gas, (aq) = aqueous Information not provided by the equation It does not tell us the following: 1. The speed of the reaction. 2. The heat change during the reaction. 3. The colours of the reactants and products. 15  16 Example: 1. Dilute hydrochloric acid reacts with sodium hydroxide solution to form sodium chloride and water. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Element Reactant Product H 2 2 Cl 1 1 Na 1 1 O 1 1 It is balanced. 2. When hydrogen gas is passed over copper (ii) oxide, copper metal and water are produced. H2(g) + CuO(s) → Cu(s) + H2O(l) element reactant Product H 2 2 Cu 1 1 O 1 1 It is balanced 3. Hydrogen gas reacts with oxygen gas to produce steam. H2(g) + O2(g) → H2O(g) Element Reactant Product H 2 2 O 2 1 This equation is not balanced. It is balanced by writing 2 INFRONT of H2 and H2O. 2H2(g) + O2(g) → 2H2O(g) Element Reactant Product H 4 4 O 2 2 4. Copper (ii) trioxonitrate (v) crystals on strong heating decomposes to copper (ii) oxide, nitrogen (iv) Oxide and oxygen gas. Cu(NO3)2(s) → CuO(s) + NO2(g) + O2(g) Element Reactant Product Cu 1 1 N 2 1 O 6 5 The equation is not balanced, to balance it 2 is written INFRONT of the reactant, 2 infront of CuO and 4 infront of NO2 2Cu(NO3)2(s) → 2CuO(s) + 4NO2(g) + O2(g) Element Reactant Product Cu 2 2 N 4 4 O 12 12 5. Sodium hydroxide reacts with tetraoxosulphate (vi) acid to produce sodium 16  17 tetraoxosulphate (vi) and water. NaOH + H2SO4 → Na2SO4 + H2O Element Reactant Product Na 1 2 O 5 5 H 3 2 S 1 1 It is not balanced. 2NaOH + H2SO4 → Na2SO4 + 2H2O Element Reactant Product Na 2 2 O 6 6 H 4 4 Questions 1. Balance the following chemical equations: a. C + H2 → C2H4 b. Hg + O2 → HgO c. KCl + O2 → KClO3 d. H2S + SO2 → H2O + S 2. Write and balance the following chemical reactions: a. Nitrogen gas reacts with hydrogen gas to produce ammonia. b. Sulphur (iv) oxide reacts with oxygen to produce sulphur (vi) oxide. c. Iron (ii) sulphide reacts with hydrochloric acid to produce iron (ii) chloride and hydrogen sulphide. CALCULATIONS FROM CHEMICAL EQUATION 1. 2g of zinc reacted with excess hydrochloric acid to liberate hydrogen gas and zinc chloride. Calculate a. the number of moles of hydrogen gas liberated b. the volume of hydrogen liberated Zn + 2HCl → ZnCl2 + H2 (Zn = 65.4, Molar volume of any gas at s.t.p = 22.4dm3) Solution Zn + 2HCl → ZnCl2 + H2 Mole ratio 1 : 2 : 1 : 1 65.4 g of Zn …………1 mole of hydrogen 2 g of Zn…………….x. X = 2 x 1/ 65.4 = 0.0306 moles of hydrogen b. Zn + 2HCl → ZnCl2 + H2 Mole ratio 1 : 2 : 1 : 1 65.4 g of Zn …………1 x 22.4 dm3 of hydrogen 2 g of Zn …………..x X = 2 x 22.4/65.4 = 0.685 dm3 17  18 OR Zn + 2HCl → ZnCl2 + H2 Mole ratio 1 : 2 : 1 : 1 0.0306: 2x 0.0306 : 0.0306 : 0.0306 a. Mole of zinc = m/M = 2/65.4 = 0.0306 mole of hydrogen gas is produced b. one mole of any gas = 22.4dm3 0.0306 mole = x 1 x = 0.0306 x 22.4 = 0.685dm3 2. If 5.0g of potassium trioxochlorate (v) was decomposed by heat, determine the volume of oxygen produced at s.t.p. Equation for the reaction is: 2KClO3 → 2KCl + 3O2 (molar volume = 22.4dm3, K = 39, Cl = 35.5, O = 16) Solution 2KClO3 → 2KCl + 3O2 Mole ratio 2 : 2 : 3 2 x 122.5g of KClO3 gave 3 x 22.4dm3 of oxygen Therefore 5g of KClO3 will produce x 2 x 122.5x = 5 x 3 x 22.4 x = 5 x 3 x 22.4 = 1.37dm3 2 x 122.5 Molar mass of KClO3 = 39 + 35.5 + 16x3 = 39 + 35.5 + 48 = 122.5g/mol 3. Calculate the mass of calcium chloride that can be obtained from 25g of limestone in the presence of excess hydrogen chloride CaCO3 + 2HCl → CaCl2 + H2O + CO2 [Ca = 40, C = 12, O = 16, H = 1, Cl = 35.5] Solution Eqn: CaCO3 + 2HCl → CaCl2 + H2O + CO2 Mole ratio: 1 : 2 : 1 : 1 : 1 1x100 g of CaCO3 produced 1x111 g of CaCl2 25g of CaCO3 will produced x 100x = 25 x 111 x = 25 x 111/100 = 27.75g of CaCl2 is produced Molar mass of CaCO[3 = 40 + 12 + 16x3 = 40 + 12 + 48 = 100g/mol Molar mass of CaCl2 = 40 + 35.5x2 = 40 + 71 = 111g/mol 4. What mass of lead (ii) trioxonitrate (v) in solution would be required to yield 9g of lead(ii) chloride on the addition of excess sodium chloride solution. Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3 [Pb = 207, N = 14, O = 16, Cl = 35.5] Solution Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3 18  19 Mole ratio 1 : 2 : 1 : 2 1 x 331g of Pb(NO3)2 gave 1 x 278g of PbCl2 x……………………………..9g 278x = 331 x 9 x = 331 x 9 /278 = 10.7g Molar mass of Pb(NO3)2 = 207 + (14 + 16x3)2 = 207 + (14 + 48)2 = 207 + (62)2 = 207 + 124 = 331g/mol Molar mass of PbCl2 = 207 + 35.5x2 = 207 + 71 = 278g/mol Molar mass of CaCO3 = 40 + 12 + 16x3 = 40 + 12 + 48 = 100g/mol Molar mass of CaO = 40 + 16 = 56g/mol 7. Iron completely reacted with dilute hydrochloric acid: a. Write an equation for the reaction b. if 3.08g of iron completely reacted with 50cm3 of 2.20moldm-3 hydrochloric acid, calculate the relative atomic mass of the metal. Solution a. Fe + 2HCl → FeCl2 + H2 b. Fe + 2HCl → FeCl2 + H2 Mole ratio 1 : 2 : 1 : 1 X 0.11 2x = 1 x 0.11 x = 1 x 0.11/2 = 0.055mol Mole of iron = m/M = 3.08/M incomplete Mole of HCl = CV/1000 = 2.20 x 50/1000 = 0.11 mol Mole of iron = m/M 0.055 = 3.08/M M = 3.08/0.055 = 56 Questions 1. What mass of sodium trioxonitrate (v) would be obtained by reacting 10.7g of lead (ii) trioxonitrate (v) in the presence of excess sodium chloride. Equation for the reaction is: Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3 [Na =23, O =16, N= 14, Pb =207, Cl =35.5] (ans = 5.5g) 2. Calculate the mass of tetraoxosulphate (vi) acid that would be required to yield 31.9g of anhydrous copper (ii) tetraoxosulphate (vi) in the presence of excess metallic copper, given that the other two products of the reaction are water and sulphur (iv) oxide. Equation for the reaction is: 2H2SO4 + Cu → CuSO4 + 2H2O + SO2 19  20 [Cu = 63.5, S = 32, O = 16] (ans = 39.2) WATER Water is commonly used in our daily life. It is available as tap water, spring, river, sea, rain and borehole. It is a colourless and odourless liquid. It covers 70% of the earth's surface, 65% of body fluid. It is present in cells, blood, tissues and taken into the body through the water we drink and in the food we eat. Water is used for washing and bathing, in dyeing process, motor engines to generate electricity. Water is rarely found in its pure form in nature. Water is regarded as a universal solvent as it dissolves many substances because it is polar in nature. Some substances form suspensions in water, while others form colloids and some substances do not dissolve in water, eg, petrol, oil, kerosene, marble, iodine etc. Composition of waters Water is composed of one atom of oxygen and two atoms of hydrogen which combines covalently to give a V-shape, due to the repulsion between the 168O and 21H Oxygen atom is very electronegative, it tends to attract electrons from the hydrogen atom to form a partial negative ion, while hydrogen atom is a partial positive ion, leading to hydrogen bond formation. The presence of hydrogen bond makes water to have unusual property such as high boiling point, high melting point and low volatility, makes ice less dense than water and floats because of air spaces created by H-bond. Oceans do not freeze completely as water is available for ice to float thus preserving aquatic life. Water is formed as a result of the burning of hydrogen in oxygen 2H2(g) + O2(g) → 2H2O(g) However, chemically using Hoffman's voltameter water can be decomposed to obtain the elements hydrogen and oxygen in the electrolysis of acidified water. To obtain the volume and composition of the elements in water use an Eudiometer. Natural Sources of Water 3 Water covers about /4 of the earth's crust. The main natural sources are rains, springs, seas, rivers and streams. Rain water is the purest form of water in nature. Sea water is impure as it contains dissolved impurities. In the laboratory, water can be distilled or deionized to remove some of the impurities. Distilled or deionized water is not good for drinking as it does not contain certain essential elements that are required for growth. Physical Properties of Water 1. Pure water is a colourless liquid and tasteless. 2. It is neutral to litmus paper. 3. It boils at 100oC and freezes at 0oC. 4. It has a density of 1g/cm3 Chemical Properties of Water 1. In the activity series, water reacts with metals differently. K react with cold water vigorously with a hissing sound, giving fumes of Na colourless and odourless gas H2 which makes pop sound with a lighted splint, and 20