3 chemical formulae and equations

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3 chemical formulae and equations

  1. 1. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and EquationsCHAPTER 3 : CHEMICAL FORMULAE AND EQUATIONSA RELATIVE ATOMIC MASS (RAM) AND RELATIVE MOLECULAR MASS (RMM) Learning Outcomes You should be able to: state the meaning of relative atomic mass based on carbon-12 scale, state the meaning of relative molecular mass based on carbon-12 scale, state why carbon-12 is used as a standard for determining relative atomic mass and relative molecular mass, calculate the relative molecular mass of substances.Activity 1 (refer text book pg 28 ) Relative atomic mass of an element , Ar = The average mass of an atom of the element 1/12 x the mass of an atom of carbon-12 Example: Ar of C=12 Ar of O=16 Ar of Mg=241. The Relative atomic mass of an element is ……………………………………………………………... …………………………………. when compare with 1/12 of the mass of an atom of carbon – 12.2. Carbon-12 is chosen because it is a ………………………. and can be easily handled.3. Find the relative atomic masses of these elements. Element Relative Atomic Mass Element Relative Atomic Mass Calcium, Ca Argon, Ar Sodium, Na Silver, Ag Iron, Fe Caesium, Cs Copper, Cu Lead, Pb Carbon, C Chlorine, Cl Hydrogen, H Flourine, F Potassium, K Aluminium, Al Lithium, Li Zinc, Zn Bromine, Br Helium, HeActivity 2 (refer text book pg 29 )Relative molecular mass of a substance, Mr= The Average mass of a molecule of the substance 1/12 x the mass of an atom of carbon-12 1
  2. 2. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and EquationsCalculating Relative molecular mass,MrMr= The sum of Ar of all atoms present in one molecule 2 Hydrogen Molecular atomsExample: formula Mr of Water, H2O = 2(1) + 16 = 18 Relative atomic mass Relative atomic mass for Oxygen for Hydrogen Mr of Carbon dioxide, CO2 = 12 + 2(16) = 44 All Ar, Mr and Fr have no unitFor ionic substance , Relative formula mass , Fr= The sum of Ar of all atoms present in the formulaExample:Fr of Magnesium oxide, MgO = 24 + 16 = 40Fr of Sodium chloride, NaCl = 23 + 35.5 = 58.5 1. The relative molecular mass of a molecule is ……………………………………………… ………………………………………………………. when compared with 1/12 of the mass of one atom of …………………………………………… 2. Calculate the relative molecular masses of the substances in the table below. Substance Molecular formula Relative molecular mass, Mr Hydrogen gas H2 2(1) = 2 Propane C3H8 Ethanol C2H5OH Bromine gas Br2 Methane CH4 Glucose C6H12O6 Ammonia NH3 [Relative atomic mass : H,1; C,12; O,16; Br,80 ; N,14 ] 3. Calculate the relative formula masses of the following ionic compounds in the table. Substance Compound formula Relative formula mass, Fr Potassium oxide K2O 2(39) + 16 = 94 Aluminium sulphate Al2(SO4)3 2(27)+3[32+4(16)]=342 2
  3. 3. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations Zinc nitrate Zn(NO3)2 Aluminium nitrate Al(NO3)3 Calcium carbonate CaCO3 Calcium hydroxide Ca(OH)2 Hydrated copper(II) CuSO4.5H2O 64 + 32 + 4(16) + 5[2(1) + 16]=250 sulphate Hydrated sodium Na2CO3.10H2O carbonate Sodium hydrogen NaHSO4 sulphate Aluminium chloride AlCl3 Copper(II) sulphate CuSO4 Zinc carbonate ZnCO3 Potassium K2CO3 carbonate [Relative atomic mass: O,16; C,12; H,1; K,39 ; Cu,64 ; Zn, 65; Cl, 35.5 ; Al, 27 S,32 ; Ca, 40; Na,23; N, 14]B THE MOLE AND THE NUMBER OF PARTICLES Learning Outcomes You should be able to: define a mole as the amount of matter that contains as many particles as the 12 number of atoms in 12 g of C, state the meaning of Avogadro constant, relate the number of particles in one mole of a substance with the Avogadro constant, solve numerical problems to convert the number of moles to the number of particles of a given substance and vice versa. 3
  4. 4. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and EquationsActivity 3 (refer text book pg 30 ) 1. To describe the amount of atoms, ions or molecules , mole is used. 2. A mole is an amount of substance that contains as many particles as the ……………….. …………………………………………………………….. in exactly 12g of carbon-12. 3. A mole is an amount of substance which contains a constant number of particles atoms, ions, molecules which is 6.02 x 1023 4. The number 6.02 x 1023 is called …………………………………… (NA) 5. In other words: 1 mol of atomic substance contains ……………………………. atoms 1 mol of molecular substance contains ……………………………. molecules 1 mol of ionic substance contains …… …………………………….. formula units 6. Relationship between number of moles and number of particles (atom/ion/molecules): x Avogadro Constant number of moles number of particles ∻ A vogadro Constant Number of moles Number of particles0.5 mol of carbon atoms …………………………………… atoms of carbon 0.2 moles of hydrogen gas ( H2) (i) …………………………..molecules of hydrogen gas (ii) …………………………….Atoms of hydrogen2 mol of carbon dioxide molecules ………………x 10 23 molecules of carbon dioxide gas contains : ………………. atoms of C and …………………. atoms of O 4
  5. 5. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations0.007 mol of calcium ions ……………………… calcium ions…………………………. mol of water 6.02 x 10 25 molecules of water0.4 mol of ozone gas ( O3) ………………….x 10 23 molecules of ozone, contains : ……………………… atoms of Oxygen.7. Complete these sentences . a) 1 mol of calcium contains ………………………………………….. atoms b) 2 mol of iron contains ……………………………………………….. atoms c) 2 mol of magnesium oxide, (MgO) contains ………………………………………….. ions d) 2 mol of sodium carbonate, (Na2CO3) contains ………………………………………. e) 3 mol of carbon dioxide, (CO2) contains …………………………………….. molecules f) 0.5 mol Copper (II) nitrate, Cu(NO3)2 contains ………………………………….. Cu2+ ions and …………………………………………………. NO3- ionsC NUMBER OF MOLES AND MASS OF SUBSTANCESLearning OutcomesYou should be able to: state the meaning of molar mass, relate molar mass to the Avogadro constant, relate molar mass of a substance to its relative atomic mass or relative molecular mass, solve numerical problems to convert the number of moles of a given substance to its mass and vice versa.Activity 4 (refer text book pg 33 ) 1. The molar mass of a substance = The molar mass of _________________ mole of the substance. = The mass of (NA) number of particles = The mass of ____________________ particles 5
  6. 6. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations x Molar mass Number of moles Mass in g ∻ Molar mass2. Calculating the Mass from a number of Moles Number of moles = . mass of the substance . Mass of 1 mole of the substanceTherefore : Mass of substance = Number of moles x Mass of 1 moleExample 1 : What is the mass of 2 moles of carbon ? Mass = 2 x 12 = 24gExample 2 : What is the mass of 2 moles of H2O ? Mass = 2 x [ 2(1) + 16 ] = 36g3. Calculate the masses of these substancesa) 2 moles of aluminium atoms b) 10 moles of iodine atoms Mass = Mass =c) 3 moles of lithium atoms d) 0.5 moles of oxygen gas (O2) Mass = Mass =e) 0.1 moles of sodium f) 2 moles of chlorine molecules (Cl2) Mass = Mass =g) 1 mole of carbon dioxide ( CO2) h) 3 moles of nitric acid, ( HNO3 ) Mass = Mass =i) 2 moles of calcium carbonate (CaCO3 ) j) 0.25 moles of calcium chloride (CaCl2 ) Mass = Mass = 6
  7. 7. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equationsk) 0.25 moles of sodium hydroxide (NaOH) l) 0.25 moles of sodium carbonate (Na2CO3) Mass = Mass =m) 0.5 moles of potassium manganate (VII) n) 0.25 moles of hydrated magnesium sulphate (KMnO4) (MgSO4.7H2O) Mass = Mass =Activity 54. Calculate the Number of Moles from a given MassExample : How many moles are there in 88g of CO2 Number of moles = 88 = 2 moles 44 a) 2g of helium atoms b) 6g of carbon atoms Number of moles = Number of moles = c) 16g of helium atoms d) 4g of sulphur atoms Number of moles = Number of moles = e) 4g of oxygen molecules (O2) f) 213g of chlorine molecules (Cl2) Number of moles = Number of moles = g) 0.56g of nitrogen molecules (N2) h) 254g of iodine molecules (I2) Number of moles = Number of moles = i) 88g of carbon dioxide (CO2) j) 3.1g of sulphur dioxide (SO2) Number of moles = Number of moles = k) 560g of potassium hydroxide (KOH) l) 392g of sulphuric acid (H2SO4) Number of moles = Number of moles = 7
  8. 8. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations m) 170g of ammonia (NH3) n) 120g of magnesium oxide (MgO) Number of moles = Number of moles = o) 4g of sodium hydroxide (NaOH) p) 73g of hydrogen choride (HCl) Number of moles = Number of moles = q) 15.8g of potassium manganate (VII) r) 8g of ammonium nitrate (NH4NO3) KMnO4 Number of moles = Number of moles = s) 0.78g of aluminium hydroxide Al(OH)3 t) 0.92g of ethanol (C2H5OH) Number of moles = Number of moles =Activity 65. Complete the following table. ChemicalElement/compound formulae Molar mass Calculate Copper Cu RAM= 64 (a)Mass of 1 mol = ……………g (b) Mass of 2 mol = …………. g (c)Mass of ½ mol = ………….g (d)Mass of 3.01x1023 Cu atoms = Sodium hydroxide NaOH RFM= 40 (a) Mass of 3 mol of sodium hydroxide = (b) Number of moles of sodium hydroxide in 20 g = Zinc nitrate Zn(NO3)2 RFM = a) Number of moles in 37.8 g of zinc nitrate : 8
  9. 9. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and EquationsD NUMBER OF MOLES AND VOLUME OF GAS Learning Outcomes You should be able to: state the meaning of molar volume of a gas, relate molar volume of a gas to the Avogadro constant, make generalization on the molar volume of a gas at a given temperature and pressure, calculate the volume of gases at STP or room conditions from the number of moles and vice versa, solve numerical problems involving number of particles, number of moles, mass of substances and volume of gases at STP or room conditions.Activity 7 (refer text book pg 36, 37 )1. The molar volume of a gas is defined as the …………………………………………………. …………………………………………………………….2. One mole of any gas always has the …………………………………………… under the same temperature and pressure.3. The molar volume of any gas is 24 dm3 at ……………………………………………… or 22.4 dm3 at …………………………………………….Example :1 mol of oxygen gas, 1 mol of ammonia gas, 1 mol helium gas and 1 mol sulphur dioxide gas occupiesthe same volume of 24 dm3 at room condition x 22.4 / 24 dm3 Number of moles of gas Volume of gas x 22.4/24 dm3 ∻22.4/24 dm34. Calculate the volume of gas in the following numbers of moles at STPExample : Find the volume of 1 mole of CO2 gas Volume = number of moles x 22.4 dm3 = 1 x 22.4 dm3 = 22.4 dm3 9
  10. 10. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations a) 3 moles of oxygen b) 2 moles of CH4 Volume = Volume = c) 0.3 moles of Argon d) 0.2 moles of SO3 Volume = Volume = e) 0.1 moles of N2 f) 1.5 mol of N2 Volume = Volume =5. Complete the diagram below . (Refer to Page 33,34 & 38-Chemistry textbook) Volume of gas (dm3) Mass in gram Number of moles No of particlesActivity 8Solve these numerical problems1. What is the volume of 0.3 mole of sulphur dioxide gas at STP? [Molar volume: 22.4 dm3 mol-1 at STP] (Ans: 6.72 dm3)2. Find the number of moles of oxygen gas contained in a sample of 120 cm3 of the gas at room conditions. [Molar volume: 24 dm3 mol-1 at room conditions] 10
  11. 11. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations (Ans: 0.005 mol)3. Calculate the number of water molecules in 90 g of water, H2O. [Relative atomic mass: H, 1; O, 16. Avogadro constant, NA: 6.02 x 1023 mol-1] 24 (Ans; 3.01x 10 molecules)4. What is the volume of 24 g methane ,CH4 at STP? [Relative atomic mass: H, 1; C, 12. Molar volume: 22.4 dm3 mol-1 at STP] 3 (Ans: 33.6 dm )5. How many aluminium ions are there in 20.4 g of aluminium oxide, Al2O3? [Relative atomic mass: O, 16; Al, 27. Avogadro constant, NA: 6.02 x 1023 mol- 23 (2 x 0.2 x 6.02 x10 )6. Calculate the number of hydrogen molecules contained in 6 dm3 of hydrogen gas at room conditions. [Molar volume: 24 dm3 mol-1 at room conditions Avogadro constant, NA: 6.02 x 1023 mol-1] 11
  12. 12. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations 23 (Ans: 1.505x10 molecules) 3 237. Find the volume of nitrogen in cm at STP that consists of 2.408 x 10 nitrogen molecules. [Molar volume: 22.4 dm3 mol-1 at STP. Avogadro constant, NA: 6.02 x 1023 mol-1] 3 (Ans: 8.96 dm )E CHEMICAL FORMULAELearning OutcomesYou should be able to state the meaning of chemical formula state the meaning of empirical formula state the meaning of molecular formula determine empirical and molecular formula of substances compare and contrast empirical formula with molecular formula solve numerical problems involving empirical and molecular formula. write ionic formula of ions construct chemical formulaf ionic compounds state names of chemical compounds using IUPAC nomenclature. use symbols and chemical formula for easy and systematic communication in the field of chemistry.ACTIVITY 9 (Refer text book pg 40)1) A Chemical formula - A representation of a chemical substance using letters for ……………………………………… and subscripts to show the numbers of each type of …………………….. that are present in the substance. The letter H shows Subscript shows 2 ……………. hidrogen atoms in ……………. H2 a molecule2) Complete this table Chemical subtance Chemical Notes formulae 12
  13. 13. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations Water …………….. 2 atoms of H combine with 1 atom of O ……….. NH3 ……. atoms of H combine with 1 atom of N Propane C3H8 …….. atoms of C combine with ……. atoms of H Magnesium oxide …………….. ……………………………………………. ……………….. H2SO4 ……………………………………………3). There are two types of chemical formulae. Complete the following:** Empirical Formula  The simplest ………… ……….. ratio of atoms of each ………. in the compound.** Molecular Formula  The actual …………… of atoms of each …………… that are present in a molecule of the compound Molecular formula = (Empirical formula)n Remember: Example: (i) Compound – Ethene (ii) Compound – Glucose Molecular formula - C2 H 4 Molecular formula - C6 H12 O 6 Empirical formula - ................... Empirical formula - ....................Activity 101 Find the empirical formula of a compound Example of calculation: a) When 11.95 g of metal X oxide is reduced by hydrogen, 10.35 g of metal X is produced. Find the empirical formula of metal X oxide [ RAM; X,207; O,16 ] Element X O Mass of element(g) 10.35 11.95-10.35Number of moles of atoms 10.35÷207 (11.95-10.35)÷16 Ratio of moles Simplest ratio of moles 13
  14. 14. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and EquationsEmpirical formula : …………b) A certain compound contains the following composition: Na 15.23%, Br 52.98% , O 31.79%, [ RAM : O, 16; Na, 23; Br,80] (Assume that 100g of substance is used) Element Na Br O Mass of element(g) 15.23 52.98 31.79 Number of moles atoms 15.23 ÷23 52.98÷80 31.79÷16 Ratio of moles Simplest ratio of molesEmpirical formula:: ……………………………………………….c) Complete the table below. Compound Molecular Formula Empirical formula Value of n Water H2O Carbon Dioxide CO2 CO2 Sulphuric Acid H2SO4 Ethene C2H4 CH2 Benzene C6H6 Glucose C6H12O6d) 2.52g of a hydrocarbon contains 2.16 g of carbon. The relative molecular mass of the hydrocarbon is 84. [RAM H,1; C,12] i. Find the empirical formula of the hydrocarbon ii. Find the molecular formula of the carbon. 14
  15. 15. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and EquationsActivity 11 :Chemical Formula for ionic compounds:Complete the table below : Cation Formula Anion Formula Hydrogen ion H Flouride ion F Lithium ion Chloride ion Sodium ion Bromide ion Potassium ion Iodide ion Magnesium ion Hydroxide ion Calcium ion Ca 2 Nitrate ion Barium ion Ba 2 Manganate(VII) ion Copper(II) ion Ethanoate ion CH 3COO Iron(II) ion O2 Iron (III) ion Sulphate ion Lead (II) ion Sulphide ion S2 Zinc ion Carbonate ion Chromium (III) ion Dichromate (VI) ion Cr2O7 2 Aluminium ion Al 3 PO4 3 Ammonium ion Chromate (VI) ionAvtivity 12a) Chemical formula of an ionic compound comprising of the ions Xm+ and Yn- is constructed by exchanging the charges of each element. The formula obtained will XnYm Example : Sodium oxide Copper (II) nitrate Na+ O2- Cu2+ NO3- +1 -2 +2 -1 2 1 1 2 = Na2O = .................... 15
  16. 16. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equationsb) Construct a chemical formula for each of the following ionic compounds:(i) Magnesium chloride (ii) Potassium carbonate(iii) Calcium sulphate (iv) Copper (II) oxide(v) Silver nitrate (vi) Zinc nitrate(vii) Aluminium oxide (viii) Iron(II) hydroxide(ix) Lead(II) sulphide (x) Chromium(III) sulphateCHEMICAL EQUATIONS Learning Outcomes You should be able to 1. state the meaning of chemical equation identify the reactants and products of a chemical equation 2. write and balance chemical equations 3. interpret chemical equations quantitatively and qualitatively 4. solve numerical problems using chemical equations 5. identify positive scientific attitudes and values practiced by scientist in doing research 6. justify the need to practice positive scientific attitudes and good values in doing researsh 7. use chemical equations for easy and systematic communication in the field of chemistry.Activity 13 (refer text book pg 48) Example: C (s) + O2 (g)  CO 2 (g) Reactant product1) Qualitative aspect of chemical equation: a) Arrow in the equation  the way the reaction is occurring b) Substances on the left-hand side  …………………….. c) Substances on the right-hand side  ……………………… d) State of each substance  ………: (s), ………………(l), gas ……….and aqueous solution ………………. 16
  17. 17. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations2) Quantitative aspect of chemical equations Coefficients in a balanced equation  the exact proportions of reactants and products in equation.Example: 2 H 2 (g) + O2 (g)  2 H 2 O (l)(Interpreting): 2 molecules (2 mol) of H 2 react with 1 molecule (1 mol) of O2 to produced 2 molecules(2mol) of water Complete the following word equations and write in chemical equation a) Sodium + chlorine  ………………………….. ………… + ……………  NaCl b) Carbon + ………..  Carbon dioxide ………. + …………  …………………….. c) Sulphur + oxygen  …………………………… ……….. + ………..  ………………………….. d) Zinc + oxygen  ……………………………….. ………… + O2  ………………………………..3) Write a balanced equation for each of the following reactions and interpret the equations quantitatively.(a). Carbon monoxide gas + oxygen gas  carbon dioxide gas ………………………………………………………………………………………………………Interpreting:……………………………………………………………………………………………………………(b). Hydrogen gas + nitrogen gas  ammonia gas ……………………………………………………………………………………………………….Interpreting:…………………………………………………………………………………………………………..(c). Aluminium + Iron (III) oxide  aluminium oxide + Iron ……………………………………………………………………………………………………….Interpreting:……………………………………………………………………………………………………………. 17
  18. 18. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and EquationsActivity 14** Numerical Problems Involving Chemical EquationsHydrogen peroxide decomposes according to the following equation: 2 H 2O2 (l)  2 H 2 O (l) + O2 (g)1). Calculate the volume of oxygen gas, O2 measured at STP that can be obtained from the decomposition of 34 g of hydrogen peroxide, H 2O2 . [Relative atomic mass : H, 1 ; O, 16. Molar volume : 22.4 dm3 m ol 1 at STP] 3 (Ans: 11.2 dm )2).Silver carbonate Ag2CO3 breaks down easily when heated to produce silver metal 2 Ag2CO3(l) 4 Ag (s) + 2 CO2 (g) + O2Find the mass of silver carbonate that is required to produce 10 g of silver[Relative atomic mass: C, 12 ; O, 16 ; Ag, 108] (Ans : 12.77g) 18
  19. 19. WAJA F4 Chemistry 2010 Chapter 3 : Chemical Formulae and Equations3). 16 g of copper (II) oxide, CuO is reacted with excess methane, CH 4 . Using the equation below, find the mass of copper that is produced. [Relative atomic mass : Cu, 64 ; O, 16] 4 CuO (s) + CH 4 (g)  4 Cu (s) + CO2 (g) + 2 H 2 O (l) (Ans : 12.8 g) 4). A student heats 20 g of calcium carbonate CaCO3 strongly. It decomposes according to the equation below: CaCO3 (s) CaO (s) + CO2 (g). (a). If the carbon dioxide produced is collected at room conditions, what is its volume? (b). Calculate the mass of calcium oxide, CaO produced. [Relative atomic mass: C, 12 ; O, 16; Ca, 40. Molar volume : 24 dm3 m ol 1 at room conditions] 3 (Ans : (a). 4.8 dm (b) 11.2 g) 19

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