Chemistry F4 Unit 3 Formula & Equations

Unit 3: Chemical Formula and Equations(By Lim Kai Yee, SMK Tun Mamat) Chemical formulae are simply there to describe chemical reactions as denoted by the chemical equations. 1. Relative atomic mass, Ar is the atomic mass of an atom when compared to a standardatom 2. Standard atom: Hydrogen scale: hydrogen is the lightest atom of all and the mass of one hydrogen atom was assigned 1 unit. Weakness of Hydrogen scale: a)not too many elements can react readily with hydrogen, b)the reactive masses of some elements were not accurate, c)hydrogen exists as a gas at room temperature and d)has a number of isotopes with different masses. Helium scale: the second lightest atom of all and the mass of one helium atom was assigned 1 unit. Weakness of Helium scale: a)Mass of 1 helium atom = 4 times the mass of a hydrogen atom So, mass of 1 helium atom = 4 times 1/12 mass of a carbon atom b)helium exists as a gas at room temperature and c)helium is an inert gas. Oxygen scale: chose as the standard atom to compare the masses of atoms Weakness of Oxygen scale: a)the existence of three isotopes of oxygen were discovered, b)natural oxygen (containing all the three isotopes) as the standard (Chemist) and c)used the isotopes oxygen-16 as the standard (Physicists). Carbon scale: standard atom of comparison internationally. a carbon-12 atom is 12 times heavier than an atom of hydrogen, a)used as the reference standard in mass spectrometers, b)exists as a solid at room temperature, c)most abundant carbon isotope, happening about 98.89% and d)carbon-12 is close to the agreement based on oxygen. 3. Relative molecular mass, Mr of a substances is the average mass of a molecule (two or more atoms) of the substances when compared 1/12 with of the mass of a carbon-12 atom. 4. Relative formula mass, Fr is for ionic compound which is calculated by adding up therelative atomic masses of all the atoms. 5. Example: Relative atomic mass, Ar of helium = 4 Relative molecular mass, Mr of CO2 = 12 + 2(16) = 44 Relative formula mass, Fr of NaCl = 23 + 35.5 = 58.5 Relative formula mass, Na2CO3·10H2O = 2(23) + 12 + 3(16) + 10 [2(1) + 16] = 286

6. Italian physicist Amedeo Avogadro (Name at birth: Lorenzo Romano Amedeo Carlo Avogadro) Born: 9 August 1776 Birthplace: Turin, Piedmont, Italy Died: 9 July 1856 Best Known As: The guy they named Avogadro’s number after 7. Avogadro constant / Avogadro’s number is 6.02 x 1023 8. Atomic substances Elements – all the particles are atoms. Example: zinc (Zn), sodium (Na), aluminium (Al) and all noble gases, argon (Ar), helium (He) and neon (Ne). RAM (Relative Atomic Mass) of Na = 23 9. Molecular substances Covalent compounds – the particles are molecules. Example: carbon dioxide (CO2), water (H2O) and non-metal elements, iodine (I2), nitrogen (N2) and oxygen (O2). RMM (Relative Molecular Mass) of I2 = 127 + 127 = 254 10. Ionic substances Ionic compounds – the particles are ions. Example: sodium chloride (NaCl), hydrochloric acid (HCl) and potassium iodide (KI). RFM (Relative Formula Mass) of HCl = 1 + 35.5 = 36.5 11. Avogadro’s Law / Gas Law states that equal volumes of all gases contain the same numberof molecules under the same temperature and pressure. 1

Example: equal volumes of molecular hydrogen and nitrogen would contain the same number of molecules under the same temperature and pressure. 12. Volume of gas (dm3) = Number of moles of gas x Molar volume 13. Room temperature and pressure (r.t.p.) = 24 dm3 mol-1 (25°C and 1 atm) Example: What is the volume of 5.0 mol helium gas at s.t.p.? Volume of gas = Number of moles x Molar gas volume = 5.0 mol x 24 dm3 mol-1 = 120 dm3 14. Standard temperature and pressure (s.t.p.) = 22.4 dm3 mol-1 (0°C and 1 atm) Example: What is the volume of 5.0 mol helium gas at s.t.p.? Volume of gas = Number of moles x Molar gas volume = 5.0 mol x 22.4 dm3 mol-1 = 112 dm3 15. Mass (g) = Number of moles x Molar mass 16. Number of particles = Number of moles x Avogadro constant 17. Volume (dm3) = Number of moles x Molar volume Empirical and Molecular Formulae 1. Empirical (simplest ratio of atoms of each element that present in the compound) and molecular formulae (actual number of atoms of each element that are present in one molecule of the compound) indicate: the types of the elements the symbols of the elements and the ratio of atoms or moles of atoms of each element in a compound. 2. Molecular formula = (empirical formula)n n is a positive number Compound Molecular formula n Empirical formula Carbon dioxide CO2 1 (CO2) = CO2 Ethane CH3 2 (CH3)2 = C2H6 Propene CH2 3 (CH2)3 = C3H6 Glucose CH2O 6 (CH2O)6 = C6H12O6 Quinine C10H12NO 2 C20H24N2O2 3. Chemical formulae for covalent compounds. Name Chemical formula Number of each element Nitrogen gas N2 2 nitrogen atoms Oxygen gas O2 2 oxygen atoms Ammonia NH3 1 nitrogen atom and 3 hydrogen atoms Water H2O 2 hydrogen atoms and 1 oxygen atom 4. Cations are positively-charged ions. Charge Cations Formula Charge Cations Formula + +1 Ammonium ion NH4 +2 * Copper(II) ion Cu2+ + +1 * Copper(I) ion Cu +2 * Iron(II) ion Fe2+ +1 Hydrogen ion H+ +2 * Lead(II) ion Pb2+ +1 Lithium ion Li+ +2 Magnesium ion Mg2+ +1 * Nickel(I) ion Ni+ +2 * Manganese(II) ion Mn2+ + +1 Potassium ion K +2 Nickel(II) ion Ni2+ +1 Silver ion Ag+ +2 * Tin(II) ion Sn2+ +1 Sodium ion Na+ +2 Zinc ion Zn2+ 2+ +2 Barium ion Ba +3 Aluminium ion Al3+ +2 Calcium ion Ca2+ +3 * Chromium(III) ion Cr3+ +3 * Iron(III) ion Fe3+ +4 * Tin(IV) ion Sn4+ +4 * Lead(IV) ion Pb4+ * refer to the Roman numerals 2

5. Anions are negatively-charged ions. Charge Anions Formula Charge Anions Formula -1 Bromide ion Br -2 Oxide ion O2-1 Chloride ion Cl-2 Carbonate ion CO32-1 Chlorate(V) ion ClO3-2 Chromate(VI) ion CrO42-1 Ethanoate ion CH3COO-2 Dichromate(VI) ion Cr2O72-1 Fluoride ion F -2 Sulphide ion S2-1 Hydride ion H-2 Sulphate ion SO42-1 Hydroxide ion OH-2 Sulphite ion SO32-1 Iodide ion I-2 Thiosulphate ion S2O32-1 Manganate(VII) ion MnO4 -3 Nitride ion N3-1 Nitrate ion NO3-3 Phosphate ion PO43-1 Nitrite ion NO2-3 Phosphite ion PO336. Chemical formulae for ionic compounds Name Chemical formula Number of cation Number of anion Zinc chloride ZnCl2 1 Zn2+ 2 ClCopper(II) sulphate CuSO4 2 Cu2+ 2 SO423+ Aluminium sulphate Al2(SO4)3 2 Al 3 SO427. Meaning of prefixes Prefix Meaning Prefix Meaning Mono1 Hexa6 Di2 Hepta7 Tri3 Octa8 Tetra4 Nona9 Penta5 Deca10 8. Naming of chemical (non-metal) compounds with Greek numerical prefixes. Non-metal compound Chemical formula Carbon monoxide CO Carbon dioxide CO2 Sulphur dioxide SO2 Sulphur trioxide SO3 Carbon tetrachloride (tetrachloromethane) CCl4 Chemical Equation 1. Importance of chemical equation: The types of reactants; the physical conditions; the quantity of reactants and products and stated in moles. nA + nB –> pC + pD 2. Reactants are written in the left side of the reaction and products are written in the right side of the reaction. Example 1: Word equation: Sodium hydroxide + sulphuric acid –> sodium sulphate + water Chemical equation: NaOH + H2SO4 –> Na2SO4 + H2O Balancing equation: 2NaOH + H2SO4 –> Na2SO4 + 2H2O Complete chemical equation: 2NaOH + H2SO4 –> Na2SO4 + 2H2O Example 2: Word equation: Aluminium + copper(II) oxide –> aluminium(III) oxide + copper Chemical equation: Al + CuO –> Al2O3 + Cu Balancing equation: 2Al + 3CuO –> Al2O3 + 3Cu Complete chemical equation: 2Al + 3CuO –> Al2O3 + 3Cu

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Example 3: Word equation: Nitrogen + hydrogen ammonia Chemical equation: N2 + H2 NH3 Balancing equation: N2 + 3H2 2NH3 Complete chemical equation: N2 + 3H2 2NH3 3. Information obtainable from chemical equations. i) mass of reactants ii) volume of reacting gas iii) mass of products formed iv) volume of gas produced Example: 2 cm3 of lead (II) nitrate solution is added to excess of potassium iodide solution. How many molecules of potassium nitrate will be formed? [Relative atomic mass: N, 14; O, 16; K, 39; I, 127; Pb, 207; Avogadro's constant: 6.02 x 10 23mol-1] Step 1: Write a complete chemical equation. Pb(NO3)2(aq) + 2KI(aq) –> PbI2(s) + 2KNO3(aq) From the equation, 1 mole of Pb(NO3)2 reacts with 2 moles of KI formed 1 mole PbI2 of and2 moles of KNO3. Step 2: Convert to moles. No. of moles of Pb(NO3)2 = Mass of Pb(NO3)2 / Relative molecular mass = 2 / [2018/2/20147 + 2(14 + 3 x 16)] = 6.04 x 10-3 mol Step 3: Ratio of moles. Number of moles of KNO3/ Number of moles of Pb(NO3)2 = 2/1 Number of moles of KNO3 = (2 x 6.04 x 10-3) / 1 = 12.08 x 10-3 mol Step 4: Convert to the number of molecules of potassium nitrate. Number of molecules of KNO3 = 12.08 x 10-3 x 6.02 x 1023 = 7.27 x 1021

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