CHEM180/181 2004 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. Using the Law of Conservation of Mass 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). 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. Mercury + oxygen red-orange residue 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? Solution: Chemical Equations From the law of conservation of mass: Total mass of substances before reaction = Total mass of substance after reaction Therefore: That is: 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. 1 CHEM180/181 2004 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. 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) Consider the burning of natural gas: CH4 + O2 You can also indicate the conditions under which a reaction takes place, as well as the presence of a catalyst. CO2 + H2O eg. 2H2O2(aq) Pt 2H2O(l) + O2(g) 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. C3H8 + O2 CO2 + H2O C3H8 + O2 CO2 + H2O Balancing Simple Equations Balance the following equations: CS2 + O2 CO2 + SO2 C2H5OH + O2 Now have C3H8 + O2 CO2 + H2O BALANCED 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) CO2 + H2O NH3 + O2 NO + H2O NH3 + O2 NO + H2O Atomic mass unit - amu Today based on carbon-12 Mass of 12C = 12 amu 1 amu = 1 12 mass of 12C The atomic weight of an element, is the average atomic mass for the naturally occurring element, expressed in atomic mass units. 2 CHEM180/181 2004 Alternate Example Fractional Abundance The fractional abundance of an isotope is the fraction of the total number of atoms that is composed of a particular isotope An element has four naturally occurring isotopes. The mass and percentage abundance of each isotope are as follows: Percentage abundance Mass 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? Percentage Composition The mass percentages sums to 100 ie. 1.48 + 23.6 + 22.6 + 52.3 = 99.98 = 100 (SIG FIG’S) Percentage composition of a compound = the percentage by mass contributed by each element in the substance The percentage of element B Is defined as: Mass % B = Atomic weight = Therefore the element is : (Atoms of B)(AW of B) x 100% mass of the whole Remember the sum of the percentages of each element in the substance add up to 100% Example Lead(II)chromate is used as a paint pigment (chrome yellow). What is the percentage composition of lead(II)chromate? The chemical name for table sugar is sucrose, C12H22O11. How many grams of carbon are in 61.8 g of sucrose? Lead(II)chromate = PbCrO4 Ar(Pb) = 207.2 g/mol Ar(Cr) = 51.9961 g/mol Ar(O) = 15.9994 g/mol %Pb = Mr(PbCrO4) = Consider 1 mol of PbCrO4 %Cr = %O = 3 CHEM180/181 2004 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 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. Example b) methylamine = CH3NH2 Calculate the formula weight of the following compounds from their formulae: a) calcium hydroxide 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) The term mole, just like a dozen or gross, refers to a particular number of things. one dozen one gross one mole = = = 12 144 6.0221367 x 1023 Avogadro’s number (NA) 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. 4 CHEM180/181 2004 The molar mass of a substance is the mass of one mole of the substance Calculating the Mass of an Atom or Molecule What is the mass of a nitric acid molecule? nitric acid = For all substances, the molar mass in grams per mole is numerically equal to the formula weight in atomic mass units. Example Mole Calculations Converting Grams of Substance to Moles Because molar mass is the mass per mole we have the formula: Mass or Molar Mass = Mr = m Moles n 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? Remember that mass must be expressed in grams here! Example Converting Moles of Substance to Grams A sample of nitric acid contains 0.253 mol HNO3. How many grams is this? Mr(HNO3) = 63.013 g.mol-1 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? (see previous example) n = 0.253 mol 5 CHEM180/181 2004 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 Empirical formula = HO Compounds with different molecular formulae can have the same empirical formula, and such substances will have the same percentage composition. Eg. acetylene = C2H2 benzene = C6H6 both have the empirical formula = Example Empirical formula from Composition Consider the following flow-diagram: Percent composition Mass Composition 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? 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 Example Determining The Empirical Formula from Percentage Composition. (General) 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. Ratio C : H = Empirical Formula = 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. 6 CHEM180/181 2004 Example Stoichiometry Determining the Molecular Formula from the Percentage Composition and Molecular Weight. Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction 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 formula weight = Empirical Formula = 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. Consider the reaction: N2(g) + 3H2(g) Example 2NH3(g) Molecular Interpretation: 1 molecule N2 + 3 molecules H2 Molar interpretation: 1 mol N2 + 3 mol H2 Mass interpretation: 28.0 g N2 + 3 x 2.02 g H2 Amounts of Substance in a Chemical Reaction 2 molecules NH3 2 mol NH3 2 x 17.0 g NH3 Find the amount of water produced when 3.68 mol of NH3 is consumed according to the following equation: 4NH3 + 5O2 4 NO + 6H2O We assume that sufficient O2 is present. From the equation we know that : Notice that : The number of moles involved in a reaction is proportional to the coefficients in the balanced chemical equation. 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 Example Calculate the mass of sulfur dioxide (SO2) produced when 3.84 mol O2 is reacted with FeS2 according to the equation: 4FeS2 + 11O2 3.84 mol 2Fe2O3 + 8SO2 m=? 2.79 mol Use molar mass of B Mass of substance B 7 CHEM180/181 2004 Another Example One of the most spectacular reactions of aluminium, the thermite reaction, is with iron oxide, Fe2O3, by which metallic iron is made. 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: 2Al(s) + Fe2O3(s) mass of Fe mol of Fe mol of Fe2O3 Al2O3(s) + 2Fe(l) Limiting Reactants mol of Fe mol of Fe2O3 mass of Fe2O3 The Cheese Sandwich Analogy 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. 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. 8 CHEM180/181 2004 Example 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 More Challenging!! Iron can be obtained by reacting the ore heamatite (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? Thus the limiting reagent is 2Fe2O3 + 3C 4Fe + 3CO2 (b) How many megagrams of iron can you make with the materials you have? Recommended tut questions 2.17 2.18 2.26 2.27 3.1 3.5 3.21 3.27 3.33 3.35 3.59 3.63 3.96 9