1.1 Introduction to the Particulate Nature of Matter and Chemical Change MATTER. Homogeneous (SOLUTIONS)

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1 TOPIC 1: STOICHIOMETRIC RELATIONS 1.1 Introduction to the Particulate Nature of Matter and Chemical Change MATTER Mass Volume Particles Particles in constant motion MATTER Pure Matters Mixtures ELEMENTS COMPOUNDS Homogeneous (SOLUTIONS) Heterogeneous Suspension Emulsion Aerosol ELEMENT COMPOUND MIXTURE Pure Pure Not pure Contain 1 type of atom Contain 1 type of molecule Contain different types of with different atoms particles Shown by symbols Shown by formula No specific figures - Constant ratio btw No fixed ratio btw components components Homogeneous Homogeneous Homogeneous or Heterogeneous Fixed phys. properties Fixed phys. properties Not fixed properties - Formed/separated by chemical methods Formed/separated physical methods by Exercise 4, Test yourself pg 7

2 STATES OF MATTER Endothermic + heat Sublimation + heat + heat Melting Boiling Solid Liquid Gas Freezing - heat Condensation - heat - heat Deposition Exothermic Fixed volume Fixed shape Fixed volume No shape No fixed volume No shape Solid + heat Liquid + heat Gas From solid to liquid Volume generally increases Density generally decreases Inter-particular forces get weaker Total energy increases. Randomness increases.

3 CHANGES OF STATE Heating Graph of a pure substance Temperature(t, C) t 2 L G t 1 S L t 0 S I II III IV 0 Q 1 Q 2 Q 3 Q 4 Heat (Q, j) CHEMICAL EQUATIONS A (s) + B (l) C (aq) + D (g) Reactants Products Equality between reactants and products (s): Solid (l): Liquid (g): Gas (aq): Aqueous solution Balancing Equations...NaOH +...H 2SO 4...Na 2SO H 2O In chemical reactions, Mass is conserved. Types of atoms are conserved. Numbers of each atom is conserved. Total electrical charge is conserved. Exercises 1, 2, 3, 5 Test yourself pg 6

4 HOMEWORK Balance given chemical equations. N2(g) + H2(g) NH3(g) N2O4(g) NO2(g) C2H4 + O2 CO2 + H2O C6H12O6 + O2 CO2 + H2O Li + H2O LiOH + H2 Ca + H2O Ca(OH)2 + H2 H2SO4 + NaOH Na2SO4 + H2O HCl + NH3 NH4Cl H2(g) + O2(g) H2O(l) Fe(s) + O2(g) Fe2O3(s) C2H5OH + O2 CO2 + H2O CS2 + O2 CO2 + SO2 Na + H2O NaOH + H2

5 Fe + H2O Fe(OH)2 + H2 H3PO4 + NaOH Na3PO4 + H2O CaCO3(s) CaO(s) + CO2(g) KClO3(s) KCl(s) + O2 (g) KCl (s) + O2(g) KClO3(s) Na(s) + HCl(aq) NaCl(s) + H2O(l) AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) Ca(s) + ZnCl2(aq) CaCl2(aq) + Zn(s) BaCl2(aq) Na2SO4(aq) BaSO4(s) + NaCl(aq) AlCl3(aq) + KOH(aq) Al(OH)3(aq) + KCl(aq) K2SO4(aq) + FeBr3(aq) Fe2(SO4)3(s) + KBr(aq)

6 HOMEWORK 2 1) Balance following chemical equations with the smallest integer numbers. a) SO 2 + O 2 SO 3 b) CS 2 + O 2 CO 2 +SO 2 c) Fe + H 2O Fe 3O 4 + H 2 d) C 7H 6O 2 + O 2 CO 2 +H 2O e) NH 3 + F 2 N 2F 4 + HF f) Al 4C 3 + HBr AlBr 3 +CH 4 g) C 3H 8 + O 2 CO 2 + H 2O h) C 3H 7COOH + O 2 CO 2 +H 2O i) Al + H 2SO 4 Al 2(SO 4) 3 + H 2 j) H 3PO 4 +Ca(OH) 2 Ca 3(PO 4) 2 + H 2O 2) Find the substances shown by X/Y/Z in the following equations. a) CaC 2 + 2H 2O Ca(OH) 2 + X b) 4X + 5O 2 4NO + 6H 2O c) X N 2 +4H 2O +Cr 2O 7 d) 9Fe 2O 3 +2NH 3 6X +N 2 +3H 2O e) 2KMnO 4 +4X +O 2 2KMnO 4 +2H 2O f) 4Zn +10 HNO 3 4X +NH 4NO 3 +3H 2O g) Bi +4X +3H 2O Bi(NO 3) 3 +5H 2O +NO h) CaO +3C X + CO i) X+2H 2O Ca(OH) 2 +Y j) xc 2H 6O +yo 2 zco 2 +th 2O

1.2 The Mole Concept Number of moles: n Number of particles: N N = n * L L: Avogadro s constant WE 1, Exercises 13, Test yourself pg 12 Relative Atomic Mass; A r The mass of an individual atom in a

7 1.2 The Mole Concept Number of moles: n Number of particles: N N = n * L L: Avogadro s constant WE 1, Exercises 13, Test yourself pg 12 Relative Atomic Mass; A r The mass of an individual atom in a sample of an element is taken as a weighted average of its different (isotope) masses. A r has no unit. Weighted average of one atom of the element A r = 1 mass of one atom of C Mass spectrometer is an instrument to measure the mass of individual atoms. The mass of a H atom = 1.67*10-24 g The mass of a C atom = 1.99*10-23 g C/H = The mass needs to be recorded relative to some agreed standard. As C is a very common element which is easy to transport and store because it is a solid. The mass of standard 12 C isotope is 12, but relative atomic mass of C element is , because 12 C has 13 C and 14 C isotopes. Relative Average Mass Q: The mass spectrum of Zirconium is shown as below: So, calculate the relative average mass of Zirconium

8 Q: Chlorine has 2 isotopes as 35 Cl and 37 Cl. If the abundance of 35 Cl is %75, what is the average weight of chlorine? Relative Formula Mass; M r The mass of a compound (or molecular element) calculated by the relative atomic masses of its atoms: (No unit again) For example: H 2 = 2*1.01 =2.02 H 2O = 2* = Exercises 29, Test yourself pg 9 Molar Mass; M The mass of 1 mole substance. Has unit g.mol -1 For example: H 2 = 2*1.01 =2.02 g.mol -1 H 2O = 2* = g.mol -1 Exercises 16 23, Test yourself pg 11 Empirical Formula of a Compound The formula which is the simplest whole-number ratio of the elements in a compound. For example: The empirical formula of C 6H 6 is CH. WE 2, 3, 4, Exercises 24, 25, 26, 27, 34, TY q.13, 16, 17, 18, 19 Molecular Formula of a Compound The formula which shows all atoms present in a molecule of a compound. For example: C 6H 6 is the molecular formula of benzene. WE 6, 7, Exercises 31, 32, 33, TY q.14, 15

9 Percentage Composition by Mass Let s see CaCO 3 (A r; Ca: C:12.01 O: 16.00) Exercises 28, 30, TY pg 14 SL & HL Questions on the mole and Avogadro s constant SL & HL Questions on Empirical and molecular formulas

10 1.3 Reacting Masses and Volumes Limiting reactant determines the quantity of product. Excess is reactant which is not fully used. Q: Zinc metal reacts with copper (II) sulfate solution according to the following equation: Zn (s) + CuSO 4(aq) ZnSO 4(aq) + Cu (s) Determine the maximum mass of copper that can be deposited when 1.20 g of zinc is added to 50.0 cm 3 of 2.00 x 10-1 mol dm -3 copper (II) sulfate solution. Q: Calculate the mass of carbon dioxide produced when 150 cm 3 of 1.00 mol dm -3 hydrochloric acid, HCl (aq), is added to 10.0 g of calcium carbonate, CaCO 3. Theoretical Yield = Maximum amount of product obtainable by assuming 100% of the limiting reactant is converted to product. Experimental (Actual) Yield Percentage Yield =.100 Theoretical Yield Q: A student prepared ethene by dehydrating ethanol. C 2H 5OH (l) C 2H 4(g) + H 2O (l) She started with 9.36 g of ethanol and made 2.12 g of ethene. Calculate the percentage yield she obtained for this reaction. Pearson Exercises pg 32

11 ATOM ECONOMY Atom economy is the mass of product you want as a % of the mass of all the products you make. % Atom economy = mass of desired product. 100 mass of all products Q: Find the atom economy for these 2 methods of extracting copper: RAM; Cu: 64, O: 16, C: 12, S: Heat copper oxide with carbon 2CuO + C 2Cu + CO 2 2. Heat copper sulphide with oxygen CuS + O 2 Cu + SO 2 Q: Find the atom economy to get MgO with the following rxn: (Mg: 24, O: 16) 2Mg + O 2 2MgO Q: Find the atom economy to get Fe with the following rxn: (Fe: 56, C: 12, O: 16) Fe 2O 3 + 3CO 2Fe + 3CO 2

12 GASES Ideal & Real Gases Ideal gas: The volume of gas particles is negligibly small compared to the space between gas particles. The interactions between gas particles are negligibly weak. Real Gas: The type of gases whose volume cannot be neglected and interactions between gas particles are not too weak to ignore. Gases deviate most from ideal behaviour at high P and low T. Avogadro s law At the same conditions equal volume of all gases contain equal number of particles. At standard temperature and pressure (STP: 100 kpa, 273 K), 1 mole of any gas has a volume of 2, m 3 mol -1 (22,7 dm 3 mol -1 ) At STP n = V(dm3 ) 22.7 Q: The molecular formula of a gaseous hydrocarbon can be determined by combusting it completely in excess oxygen and then passing it through potassium hydroxide solution to absorb the carbon dioxide produced. In an experiment 200 cm 3 of a hydrocarbon was reacted with 1500 cm 3 of oxygen. After the hydrocarbon had combusted completely 1000 cm 3 of gas remained. This volume was reduced to 200 cm 3 after the gas had been passed through a solution of potassium hydroxide. All volumes were measured under the same conditions of temperature and pressure. Deduce the formula of the hydrocarbon. Pearson Exercises pg 36, Cambridge Test Yourself pg 31, 32

13 Boyle s Law At constant T & n, the volume of an ideal is inversely proportional to its pressure. Pα 1 V P*V = k P P PV V 1/V P Charles Law At constant P & n, the volume of an ideal gas is directly proportional to its temperature. T α V V T =k V V T ( ) t ( ) Gay-Lussac s Law At constant T & n, the pressure of an ideal gas is directly proportional to its temperature. P P α V or P T =k P T ( ) t ( )

14 The Overall Gas Equation BOYLE CHARLES GAY-LUSSAC P V = k V T =k P T =k P V P 1 V 1 T =k = P 2 V 2 T 1 T 2 Q: A sample of gas occupies 67.2 cm 3 at a temperature of 22 C and a pressure of 9.38x10 4 Pa. Calculate the volume the gas will occupy if the temperature is increased to 29 C and the pressure increased to 1.06 x 10 5 Pa. Ideal Gas Equation P: Nm -2 or Pa V: m 3 PV = nrt T: K R=8.31 JK -1 mol -1 Q: 2.50 dm 3 of gas at a temperature of 19 C and a pressure of 1.01 x 10 5 Pa has a mass of 4.59 g. Determine the molar mass of the gas. Test yourself pg 39

15 SOLUTIONS Solution = Solute + Solvent Concentration: The amount of solute dissolved in a unit volume of solution. Molar Concentration; M M = n (mol) V (dm 3 ) Molar Concentrations of Ions Q: Calculate molar concentrations of each ion in 2.50*10-3 moles of 50 cm 3 FeCl 3 solution. Dilution (Reduce concentration) n 1 V 1 Pure + water V 2 = n 1 V last V last = V 1 + V 2 M 1 = n 1 V 1 M=0 n=0 M last = n 1 V last M 1* V 1= M last* V last Q: Determine the final concentration of a 73 cm 3 solution of HCl with concentration 0.40 mol.dm -3, which is diluted to a volume of 300 cm 3.

16 Percent Concentration; c% C% = Concentrations of very dilute solutions ppm (parts per million) ppm = mass of solute mass of solution *100 mass of solute mass of solution *106 ppb (parts per billion) ppb = mass of solute mass of solution *109 Titration It is a technique which is used to determine the reacting volume precisely. 1 1 NaOH + HCl NaCl + H 2O M 1 V 1 M 2 V 2 n 1 n 2 M V 1 1 n = n 1 2 = M 2 V 2 Test yourself pg 45 Exam-Style questions pg 51 53

Stoichiometric relationships 1

Stoichiometric relationships 1 Chapter outline Describe the three states of matter. Recall that atoms of diff erent elements combine in fi xed ratios to form compounds which have diff erent properties