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STOICHIOMETRY

STOICHIOMETRY. Chapter 3. Law of conservation of mass. The balance became an important instrument in the eighteen century. Balances measure mass, which is the quantity of matter in a material. You should note the distinction between the terms mass and weight in precise usage.

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STOICHIOMETRY

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  1. STOICHIOMETRY Chapter 3

  2. Law of conservation of mass • The balance became an important instrument in the eighteen century. • Balances measure mass, which is the quantity of matter in a material. You should note the distinction between the terms mass and weight in precise usage.

  3. By weighing substances before and after chemical change, Lavoisier (1743-1794) demonstrated the law of conservation of mass which states that: • the total mass remains constant during a chemical change (chemical reaction).

  4. Mercury + oxygen red-orange residue Using the Law of Conservation of Mass You heat 2.53 g of metallic mercury in air, which produces 2.73 g of a red-orange residue. Assume that the chemical change is the reaction of the metal with oxygen in air. What is the mass of oxygen that reacts? When you strongly heat the red-orange residue, it decomposes to give back mercury and releases the oxygen, which you collect. What is the mass of oxygen you collect?

  5. Solution: From the law of conservation of mass: Total mass of substances before reaction =Total mass of substance after reaction

  6. Chemical Equations A chemical equation is a symbolic representation of a chemical reaction in terms of chemical formulas. eg. 2Na + Cl2 2NaCl A reactant is a starting substance in a chemical reaction A product is a substance that results from a reaction.

  7. CH4 + O2 CO2 + H2O Balancing Chemical Equations A chemical equation is said to be balanced, when the numbers of atoms of each element are equal on either side of the arrow. Consider the burning of natural gas:

  8. You can also indicate the conditions under which a reaction takes place, as well as the presence of a catalyst. eg. 2H2O2(aq) Pt 2H2O(l) + O2(g) The state or phase of a substance may be indicated with one of the following labels: (g) ; (I) ; (s) ; (aq) eg. 2Na(s) + Cl2(g)  2NaCl(s)

  9. Consider the burning or propane gas C3H8 + O2 CO2 + H2O To balance the equation, you select coefficients that will make the numbers of atoms of each element equal on both sides of the equation.

  10. Balancing Simple Equations Balance the following equations: CS2 + O2CO2 + SO2 C2H5OH + O2CO2 + H2O NH3 + O2NO + H2O

  11. Atomic Weights & Molecular Weights Central to Dalton’s atomic theory was the idea that each atom had a characteristic mass. We now know that each isotope has its own characteristic mass. Dalton actually calculated the average atomic mass, based on the relative abundance of each isotope. Dalton could not weigh individual atoms, instead he could find the average mass of one atom relative to the average mass of another. (using H as a base)

  12. Today based on carbon-12 Mass of 12C = 12 amu 1 amu = 12 mass of 12C 1 Atomic mass unit - amu The atomic weight of an element, is the average atomic mass for the naturally occurring element, expressed in atomic mass units.

  13. Fractional Abundance The fractional abundance of an isotope is the fraction of the total number of atoms that is composed of a particular isotope

  14. Alternate Example An element has four naturally occurring isotopes. The mass and percentage abundance of each isotope are as follows: Percentage abundanceMass 1.50 203.973 23.6 205.9745 22.6 206.9745 52.3 207.9766 What is the atomic weight and the name of the element?

  15. The mass percentages sums to 100 ie. 1.48 + 23.6 + 22.6 + 52.3 = 99.98 = 100 (SIG FIG’S)

  16. (Atoms of B)(AW of B) Mass % B = x 100% mass of the whole Percentage Composition Percentage composition of a compound = the percentage by mass contributed by each element in the substance The percentage of element B Is defined as: Remember the sum of the percentages of each element in the substance add up to 100%

  17. Ar(Pb) = 207.2 g/mol Ar(Cr) = 51.9961 g/mol Ar(O) = 15.9994 g/mol Example Lead(II)chromate is used as a paint pigment (chrome yellow). What is the percentage composition of lead(II)chromate? Lead(II)chromate= PbCrO4

  18. Example The chemical name for table sugar is sucrose, C12H22O11. How many grams of carbon are in 61.8 g of sucrose?

  19. Mass and Moles of Substance • We have seen so far that atoms / molecules react in simple ratios. • Atoms and molecules are too small to count. • We need find another way of expressing such large numbers In pursuit of this end we are going to consider the following concepts: • Molecular weight • Formula weight • the Chemical Mole

  20. Formula and Molecular Weight The formula weight of a substance is the sum of the atomic weights of all atoms in a formula unit of the compound. The molecular weight of a substance is the sum of the atomic weights of all the atoms in a molecule of the substance. Note: the formula weight is applicable to both molecular and ionic compounds.

  21. Example Calculate the formula weight of the following compounds from their formulae: a) calcium hydroxide b) methylamine = CH3NH2

  22. The Mole Chemists have adopted the mole concept as a convenient way to deal with the enormous numbers of atoms, molecules or ions in the samples they work with. Definition of Mole: A mole (symbol mol) = the amount of matter that contains as many Objects (atoms, molecules etc) as the number of atoms in exactly 12 grams of carbon-12. The number of atoms in a 12 gram sample of carbon-12 is called Avogadro’s number (NA)

  23. Avogadro’s number (NA) The term mole, just like a dozen or gross, refers to a particular number of things. one dozen = 12 one gross = 144 one mole = 6.0221367 x 1023 When using the term mole it is important to specify what is being referred to! eg : 1 mol of CH3COOH molecules contains 2 moles of C atoms etc.

  24. The molar mass of a substance is the mass of one mole of the substance For all substances, the molar mass in grams per mole is numerically equal to the formula weight in atomic mass units.

  25. Calculating the Mass of an Atom or Molecule What is the mass of a nitric acid molecule?

  26. m Mass Mr = Molar Mass = n Moles Mole Calculations Because molar mass is the mass per mole we have the formula: or Remember that mass must be expressed in grams here!

  27. Example Converting Grams of Substance to Moles Calcite is a mineral composed of calcium carbonate, CaCO3. A sample of calcite composed of pure CaCO3 weighs 23.6 g. How many moles of calcium carbonate is this?

  28. Mr(HNO3) = 63.013 g.mol-1 (see previous example) Example Converting Moles of Substance to Grams A sample of nitric acid contains 0.253 mol HNO3. How many grams is this?

  29. Example Calculating the Number of Molecules in a Given Mass The daily requirement of chromium in the human diet is 1.0x10-6 g. How many atoms of chromium does this represent?

  30. Determining formulae The percentage composition of a compound leads directly to its empirical formula. Recall:An empirical formula for a compound is the formula of a substance written with the smallest integer subscripts. Eg. Consider hydrogen peroxide: Molecular formula = H2O2 Empiricalformula =

  31. Compounds with different molecular formulae can have the same empirical formula, and such substances will have the same percentage composition. Eg. acetylene = C2H2benzene = C6H6 both have the empirical formula =

  32. Empirical formula from Composition Consider the following flow-diagram: Percent composition Mass Composition Number of moles of each element Divide by smallest number of moles to find the molar ratios Multiply by appropriate number to get whole number subscripts

  33. Example Determining the Empirical Formula from the Masses of Elements (Binary Compound) We have determined the percentage composition of benzene: 92.3% C and 7.7% H. What is the empirical formula of benzene?

  34. Example Determining The Empirical Formula from Percentage Composition. Sodium pyrophosphate is used in detergent preparations. The mass percentages of the elements in this compound are 34.6% Na, 23.3% P and 42.1%O. What is the empirical formula of sodium pyrophosphate? Hint: Consider a 100 g sample.

  35. Molecular Formula from Empirical Formula The molecular formula of a compound is a multiple of its empirical formula. Molecular weight = n x empirical formula weight where n = number of empirical formula units in the molecule.

  36. Example Determining the Molecular Formula from the Percentage Composition and Molecular Weight. We have already determined the mass composition and empirical formula of benzene. In a separate experiment, the molecular weight of benzene was determined to be 78.1. What is the molecular formula of benzene

  37. Stoichiometry Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction It is based on the chemical equation and the relationship between mass and moles. We can interpret a chemical equation in terms of number of molecules (or ions or formula units) or in terms of number of moles of molecules etc. depending on our needs. Remember that because moles can be converted to mass, we can also produce a mass interpretation of a chemical equation.

  38. Consider the reaction: N2(g) + 3H2(g) 2NH3(g) 1molecule N2 + 3 molecules H22 molecules NH3 1 mol N2 + 3 mol H22 mol NH3 28.0 g N2 + 3 x 2.02g H22 x 17.0 g NH3 Molecular Interpretation: Molar interpretation: Mass interpretation: Notice that : The number of moles involved in a reaction is proportional to the coefficients in the balanced chemical equation.

  39. Example Amounts of Substance in a Chemical Reaction Find the amount of water produced when 3.68 mol of NH3 is consumed according to the following equation: 4NH3 + 5O2 4NO + 6H2O We assume that sufficient O2 is present.

  40. The Steps in a Stoichiometric Calculation Mass of substance A Use molar mass of A Moles of substance A Use coefficients of A & B in balanced eqn Moles of substance B Use molar mass of B Mass of substance B

  41. 4FeS2 + 11O2 2Fe2O3 + 8SO2 Example Calculate the mass of sulfur dioxide (SO2) produced when 3.84 mol O2 is reacted with FeS2 according to the equation:

  42. Another Example One of the most spectacular reactions of aluminium, the thermite reaction, is with iron oxide, Fe2O3, by which metallic iron is made.

  43. The equation is : 2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(l) A certain welding operation, requires that at least 86.0 g of Fe be produced. What is the minimum mass in grams of Fe2O3 that must be used for the operation? Calculate also how many grams of aluminium are needed. Strategy:

  44. Limiting Reactants The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when a reaction goes to completion. The moles of product are always determined by the starting moles of the limiting reactant.

  45. The Cheese Sandwich Analogy

  46. Which is the limiting reactant? Strategy: • Use the relationships from the balanced chemical equation • You take each reactant in turn and ask how much product would be obtain, if each were totally consumed. • The reactant that gives the smaller amount of product is the limiting reactant.

  47. Example When 100.0 g mercury is reacted with 100.0 g bromine to form mercuric bromide, which is the limiting reagent? Hg + Br2 HgBr2

  48. More Challenging!! 2Fe2O3 + 3C 4Fe + 3CO2 Example Iron can be obtained by reacting the ore haematite (Fe2O3) with coke (C). The latter is converted to CO2. As manager of a blast furnace you are told that you have 20.5 Mg of Fe2O3 and 2.84 Mg of coke on hand. (a) Which should you order first - another shipment of iron ore or coke? (b) How many megagrams of iron can you make with the materials you have?

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