Presentation on theme: "Quantities of Reactants and Products"— Presentation transcript:
1 Quantities of Reactants and Products Chapter 4Quantities of Reactants and ProductsAntoine Lavoisier“Fatherof modern chemistry.”Recognized true elements.Used quantitative measurementsin chemical reactions.1111
2 2H2 + O2 2H2O Chemical Equations Reactants Product Lavoisier: mass is conserved in a chemical reaction.Chemical equations: descriptions of chemical reactions.Two parts to an equation: reactants and products:2H2 + O22H2OReactantsProduct
4 Stoichiometric coefficients: numbers in front of the chemical formulas give numbers of molecules or atoms reacting (and numbers being produced).
5 CH4 + O2 CO2 + H2O Law of Conservation of Mass: All reactions must be balancedCH4 + O CO2 + H2Ois not balanced. (Why?)Count atoms:Reactants: Products:1 C1 C4 H2 H2 O3 O
6 Balance reactions only by changing coefficients, not by altering chemical formula
7 C3H8(g) + 5O2(g) 3CO2 (g) + 4H2O(l) Combustion is the burning of a substance in oxygen:C3H8(g) + 5O2(g) 3CO2 (g) + 4H2O(l)
8 Which is correct? (Blue=A; Red=B) a) A2 + B A2Bb) A2 + 4B AB2c) 2A + B AB2d) A + B AB2
9 Atomic and Molecular Weights Percentage Composition from FormulasPercent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:
10 Percentage Composition from Formulas What is % O in H2SO4 (by mass)?FW= (2x1) + (1x32)+ (4x16)= 98 amuFW of O in H2SO4 = 4 x 16 = 64 amu%O = x 100 = 65.3%98
11 The Mole* Experimentally, 1 mole = 6.02 x 1023 things (atoms) *MSJ Ch 3 ppThe “amu” is an “atomic mass unit.”O has a mass of 16 amu – but we can’t weigh out anything in amuIf we want to keep the number “16” for the mass of oxygen in some real units (like grams) then we are dealing with a whole bunch of atoms (in 16 g of oxygen).That bunch of atoms is called a mole.Experimentally,1 mole = 6.02 x 1023 things (atoms)This number is called Avogadro’s number.
12 The mole is defined so that one mole of a substance has a mass equal to its AW or MW in gramsBasically, you are replacing amu with grams,e.g.The mass of a P atom is 31 amu.The molar mass of P is 31 grams.The mass of a Ca atom is 40 amuThe molar mass of Ca is 40 gramsThis amount of P or Ca each contains Avogadro’s number (6.02 x 1023) of atoms of P and Ca.
13 The MoleExperimentally, 1 mole of 12C has a mass of exactly 12 g. (recall from Ch. 2)Molar MassMolar mass: mass in grams of 1 mole of substanceUnits: g/mol or g.mol-1.Mass of 1 mole of 12C = 12 g exactly
15 This photograph showsone mole of :solid NaCl (58.5 g),liquid H2O (18 g), andgaseous N2 (28 g).
16 The Mole Molar Mass What is Molar Mass of H2SO4? Molar mass: sum of the molar masses of the atoms:What is Molar Mass of H2SO4?
17 The MoleInterconverting Masses, Moles, and Numbers of Particles
18 Example 1: g of P(a) contains mol of P(b) contains atoms of PExample 2: x 1024 atoms of C(a) equals mol of C(b) has mass equal to grams.
19 Example 3: mol CO2:(a) has what mass?(b) contains how many molecules of CO2?(c) contains how many atoms of O?
20 Empirical Formulas from Analyses Start with mass % of elements (i.e. empirical data) and calculate a formula.
21 Empirical Formulas from Analyses* Example: compound of N and OGiven analysis: N: 25.9%; O: 74.1%*MSJ Ch 3 ppAssume 100g; N: 25.9 g; O: 74.1 gChange to mol:N: g x mol = 1.85 mol14 gO: g x mol = 4.63 mol16 gPreliminary emp. Formula: N1.85O4.63Clean it up: divide both by 1.85:Get N1O2.5; get rid of fractions, multiply both by 2:Get N2O5 which is the empirical (simplest) formulaWhat are some possible molecular formulas?
22 Empirical Formulas from Analyses Molecular Formula from Empirical FormulaOnce we know the empirical formula, we need the MW to find the molecular formula.Subscripts in the molecular formula are always whole-number multiples of subscripts in the empirical formula.Example: suppose compound of C and H hasempirical formula of C3H8 and a MW = 176 g/mol.What is the molecular formula?
23 Quantitative Information from Balanced Equations Balanced chemical equation gives number of molecules (or moles) that react to form products.Interpretation: balanced equation gives us the ratio of number of moles of reactant to product (or v.v.).These ratios are called stoichiometric ratios.Example:2 H2 + O H2OMolecules:Moles:Ratio of O2:H2O = 1:2 (either molecules or moles)
24 Quantitative Information from Balanced Equations The ratio of grams of reactant cannot be directly related to the grams of product.
25 Stoichiometry aluminum sulfide + water Problem:aluminum sulfide + wateraluminum hydroxide + hydrogen sulfide(a) Write balanced reaction:(b) How many g aluminum hydroxide obtained from 10.5 g of aluminum sulfide?
26 Stoichiometry Problem: 2 NaN3(s) 2Na(s) + 3 N2(g) (a) how many mol N2 produced from 2.50 mol NaN3?(b) how many g NaN3 needed to form 6.00 g N2(c) how many g NaN3 needed to produce 10.0 ft3 of N2?(1.00 ft3 = 28.3 L; density of N2 = 1.25 g/L)
27 Limiting ReactantsIf the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess).Limiting Reactant: one reactant that is consumed.O2H2O2 INXS4444
28 -------------------------------------------------------- RECIPE:3 cups flour eggs cups sugar cakeOn Hand: cups flour eggs cups sugarHow many cakes can be made from these amounts?What is the “limiting reactant” (LR)What’s left over and how much of it is left over?
29 -------------------------------------------------------- RECIPE:3 cups flour eggs cups sugar cakeOn Hand: cups flour eggs cups sugarHow many cakes can be made from these amounts?What is the “limiting reactant” (LR)What’s left over and how much of it is left over?
30 Limiting Reagent (a)Assume a reactant (any one) is LR. Calculate stoichiometric amount of product (any product) formed.(b) Pick another reactant and make it the LR.Calculate the stoichiometric amount of same product formed.(c) Whichever reactant gives the smaller amount of product is the Limiting Reagent
31 Limiting ReagentThe LR is always used up in a chemical reaction. Everything else is in excess (INXS).
32 Limiting Reagent Problem: 4 NH3 + 5 O NO + 6 H2O2.25 g NH3 mixed with 3.75 g O2 and allowed to react(a) which is LR?(b) how many grams NO formed?(c) how much of excess reactant remains?
33 Limiting Reactants Theoretical Yields The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield.The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:
34 Theoretical Yield C6H6 + Br2 C6H5Br + HBr Problem:C6H6 + Br C6H5Br + HBr(a) theoretical yield of C6H5Br when 30.0 gof C6H6 reacts with 65.0 g of Br2?(b) if actual yield of C6H5Br is 56.7 g,calculate %yield