1Chem 1A Chapter 3 Lecture Outlines StoichiometryAtomic MassThe Mole conceptMolar MassPercent Composition of CompoundsDetermination of Formula of CompoundsWriting and Balancing Chemical EquationsInterpreting balance equations, andReaction Stoichiometry and Calculations
2Atomic MassesAbsolute masses of atoms cannot be obtained – too small to measure the mass directly;Relative atomic masses are used instead – masses relative to a chosen standard or reference.Carbon-12 is used as atomic mass reference – it is assigned an atomic mass of 12 u exactly;Other atoms are assigned masses relative to that of carbon-12;Relative atomic masses are determined using mass spectrometer;
7Calculation of Relative Atomic Masses Example-1:An atomic mass spectrum gives atomic mass ratio of oxygen atom to carbon-12 as :1. If the atomic mass of carbon-12 is exactly 12 u, what is the atomic mass of oxygen?Atomic mass of oxygen = x 12 u= u
8Calculation of Average Atomic Masses Example-2:Chlorine is composed of two stable naturally occurring isotopes: chlorine-35 (75.76%; u) and chlorine-37 (24.24%; u). What is the average atomic mass of copper?Atomic mass of chlorine= ( x u) + ( x u)= u (as given in the periodic table)
9Calculation of Average Atomic Masses Example-3:Copper is composed of two naturally occurring isotopes: copper-63 (69.09%; u) and copper-65 (30.91%; u). What is the average atomic mass of copper?Atomic mass of copper= ( x u) + ( x u)= u (as given in the periodic table)
10Exercise #1: Relative Atomic Mass A mass spectrometer computed the atomic mass ratio of fluorine to carbon-12 as to-1. If the atomic mass of carbon-12 is 12 u (exactly), what is the atomic mass of fluorine in u?(Answer: u)
11Exercise #2: Average Atomic Mass Natural boron is composed of two isotopes: 19.78% boron-10 (atomic mass = amu) and 80.22% boron-11 (atomic mass = amu). What is the average atomic mass of naturally occurring boron?(Answer: u)
12Molar Quantity The Mole: A quantity that contains the Avogadro’s number of items;Avogadro’s number = x 102312.01 g of carbon contains the Avogadro’s number of carbon atoms.1 mole of carbon = g1 carbon atom = u (or amu)
13Gram-Atomic Mass Mass of 1 carbon-12 atom = 12 u (exactly); Mass of 1 mole of carbon-12 = 12 g;Mass of 1 oxygen atom = uMass of 1 mole of oxygen = gGram-atomic mass = mass (in grams) of 1 mole of an element – that is, the mass (in grams) that contains the Avogadro’s number of atoms of that element.gram-atomic mass is the molar mass of an element in grams.
14Atomic Mass & Gram-Atomic Mass Examples:Element Atomic mass Gram-atomic massCarbon u g/molOxygen u g/molAluminum u g/molSilicon u g/molGold u g/mol
15Molecular Mass and Molar Mass Molecular mass = the mass of a molecule in u;Molar mass = the mass of one mole of an element or a compound, expressed in grams.Examples:Molecular Mass Molar MassN u g/molH2O u g/molC8H u g/mol
16Calculating Molar Mass Calculating the molar mass of sucrose, C12H22O11:(12 x g) + (22 x g) + (11 x g)= g/moleMolar mass of ammonium hydrogen phosphate, (NH4)2HPO4:(2 x g) + (9 x g) + (1 x g) + (4 x g)= g/mole
17Percent Composition of a Compound Composition of aluminum sulfate, Al2(SO4)3:Molar mass of Al2(SO4)3 =(2 x g) + (3 x g) + (12 x g) = g/moleMass percent of Al = (53.96 g/ g) x 100% = 15.77%Mass percent of S = (96.18 g/ g) x 100% = 28.11%Mass percent of O = (192.0 g/ g) x 100% = %
18Formula of Compounds Empirical Formula Molecular Formula A chemical formula that represents a simple whole number ratio of the number of moles of elements in the compound. Examples: MgO, Cu2S, CH2O, etc.Molecular FormulaA formula that shows the actual number of atoms of each type in a molecule.Examples: C4H10, C6H6, C6H12O6.
19Empirical Formula-1 Empirical formula from composition: Example: A compound containing carbon, hydrogen, and oxygen has the following composition (by mass percent): 68.12% C, 13.73% H, and 18.15% O, Determine its empirical formula.Solution:Use mass percent to calculate mole and mole ratio of C:H:OMole of C = g x (1 mol C/12.01 g) = mol CMole of H = g x (1 mol H/1.008 g) = mol HMole of O = g x (1 mol O/16.00 g) = mol ODivide all moles by mole of O (smallest value) to get simple ratio:5.672 mol C/1.134 mol O = 5; mol H/1.134 mol O = 12, and1.134 mol O/1.134 mol O = 1;Mole ratio: 5C:12H:1O Empirical formula = C5H12O
20Empirical Formula-2 Empirical formula = P2O5 Empirical formula from mass of elements in a sample of compoundExample: When 1.96 g of phosphorus is burned, 4.49 g of a phosphorus oxide is obtained. Calculate the empirical formula of the phosphorus oxide.Solution:Calculate moles of P and O in sample and obtain a simple mole ratio;Mole of P = 1.96 g P x (1 mol/30.97 g) = mol P;Mole of O = (4.49 g – 1.96 g) x (1 mol/16.00 g) = mol O;Divide by mole of P (smaller value) to get a simple mole ratio:mol P/ = 1 mol P; mol O/ = 2.5 mol OMole ratio: 1 mol P to 2.5 mol O, OR 2 mol P to 5 mol OEmpirical formula = P2O5
21Empirical Formula-3Empirical formula from data of combustion reaction:Example: A compound is composed of carbon, hydrogen, and oxygen. When 2.32 g of this compound is burned in excess of oxygen, it produces 5.28 g of CO2 gas and 2.16 g of water. Calculate the composition (in mass percent) of the compound and determine its empirical formula.Solution:Find mass of C, H, and O in the sample and then calculate their mass percent:Mass of C = 5.28 g CO2 x (12.01 g C/44.01 g CO2) = 1.44 gMass % of C = (1.44 g C/2.32 g sample) x 100% = 62.1%Mass of H = 2.16 g H2O x (2 x g/18.02 g H2O) = 0.24 gMass % of H = (0.242 g H/2.32 g sample) x 100% = 10.4%Mass of O = 2.32 g sample – 1.44 g C – 0.24 g H = 0.64 gMass % of O = 100 – 62.1% C – 10.4% H = 27.5%Derive empirical formula from these mass percent composition(next slide)
22Empirical Formula-3 Empirical formula: C3H6O Empirical formula from data of combustion (continued):Calculate mole and simple mole ratio from calculated mass of each element:Mole of C = 1.44 g C x (1 mol/12.01 g) = 0.12 molMole of H = g x (1 mol/1.008 g) = 0.24 molMole of O = 0.64 g x (1 mol/16.00 g) = 0.04 molDivide all moles by mole of O (smallest mole) to obtain a simple ratio:0.12 mol C/0.04 = 3 mol C; 0.24 mol H/0.04 = 6 mol H;0.04 mol O/0.04 = 1 mol OSimple molar ratio: 3 mol C : 6 mol H : 1 mol OEmpirical formula: C3H6O
23Molecular Formula Correct molecular formula = C6H12O2 Molecular formula is derived from empirical formula and molecular mass, which is obtained independentlyEmpirical formula = CxHyOz; molecular formula = (CxHyOz)n,where n = (molecular mass/empirical formula mass)Example:A compound has an empirical formula C3H6O and its molecular formula is u. What is the molecular formula?Solution:Empirical formula mass = (2 x u) + (6 x u) u= 58.1 uMolecular formula = (C3H6O)n; where n = (116.2 u/58.1 u) = 2Incorrect molecular formula = (C3H6O)2;Correct molecular formula = C6H12O2
24Exercise #3: Determination of Formulas A 2.00-gram sample of phosphorus is completely reacted with oxygen gas, which yields 4.58 g of product that is composed of only phosphorus and oxygen. In separate analyses, the compound is found to have molar mass of about 284 g/mol.(a) Determine the empirical and molecular formulas of the compound. (b) Write an equation for the reaction of phosphorus with oxygen gas.(Answer: P2O5; P4O10; 4P + 5 O2 P4O10)
25Chemical Equation #1 4Fe(s) + 3 O2(g) 2Fe2O3(s) Description of reaction:Iron reacts with oxygen gas and forms solid iron(III) oxide:Identity: reactants = iron (Fe) and oxygen gas (O2); product = iron(III) oxideChemical equation: Fe(s) + O2(g) Fe2O3(s)Balanced equation:4Fe(s) + 3 O2(g) 2Fe2O3(s)
26Chemical Equation #2 Description of reaction: Phosphorus reacts with oxygen gas to form solid tetraphosphorus decoxide.Equation: P(s) + O2(g) P4O10(s)Balanced eqn.: 4P(s) + 5 O2(g) P4O10(s)
27Chemical Equation #3 Description of reaction: Propane gas (C3H8) is burned in air (excess of oxygen) to form carbon dioxide gas and water vapor;Identity: reactants = C3H8(g) and O2(g);products = CO2(g) and H2O(g);Equation: C3H8(g) + O2(g) CO2(g) + H2O(g);Balanced equation:C3H8(g) + 5 O2(g) 3CO2(g) + 4H2O(g)
28Chemical Equation 4 Description of reaction: Ammonia gas (NH3) reacts with oxygen gas to form nitrogen monoxide gas and water vapor;Equation: NH3(g) + O2(g) NO(g) + H2O(g);Balancing the equation:2NH3(g) + 5/2 O2(g) 2NO(g) + 3H2O(g);Multiply throughout by 2 to get rid of the fraction:4NH3(g) + 5 O2(g) 4NO(g) + 6H2O(g);
29Balancing Chemical Equations Rules for balancing equations:Use smallest integer coefficients in front of each reactants and products as necessary; coefficient “1” need not be indicated;The formula of the substances in the equation MUST NOT be changed.Helpful steps in balancing equations:Begin with the compound that contains the most atoms or types of atoms.Balance elements that appear only once on each side of the arrow.Next balance elements that appear more than once on either side.Balance free elements last.Finally, check that smallest whole number coefficients are used.
30StoichiometryStoichiometry = the quantitative relationships between one reactant to another, or between a reactant and products in a chemical reaction.Interpreting balanced equations:Example: C3H8(g) + 5 O2(g) 3CO2(g) + 4H2O(g);The equation implies that:1 C3H8 molecule reacts with 5 O2 molecules to produce 3 CO2 molecules and 4 H2O molecules; OR1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O.
31Stoichiometric Calculations Mole-to-mole relationship:Example: In the following reaction, if 6.0 moles of octane, C8H18, is completely combusted in excess of oxygen gas, how many moles of CO2 and H2O, respectively, will be formed? How many moles of O2 does it consumed?Reaction: 2C8H18(l) + 25 O2(g) 16CO2(g) + 18H2O(g)Calculations:Mole CO2 formed = 6.0 mol C8H18 x (16 mol CO2/2mol C8H18) = 48 molesMole H2O formed = 6.0 mol C8H18 x (18 mol H2O/2mol C8H18) = 54 molesMole O2 consumed = 6.0 mol C8H18 x (25 mol O2/2mol C8H18) = 75 moles
32Stoichiometric Calculations Mass-to-mole-to-mole-to-mass relationship:Example-1: In the following reaction, if 690 g of octane, C8H18, is completely combusted in excess of oxygen gas, how many grams of CO2 are formed?Reaction: 2C8H18(l) + 25 O2(g) 16CO2(g) + 18H2O(g)Calculation-1:Moles C8H18 reacted = 690 g C8H18 x (1 mol/114.2 g) = 6.0 molesMoles CO2 formed = 6.0 mol C8H18 x (16 mol CO2/2 mol C8H18)= 48 moles CO2Mass of CO2 formed = 48 mol CO2 x (44.01 g/mol) = 2.1 x 103 g
33Stoichiometric Calculations Mass-to-mole-to-mole-to-mass relationship:Example-2: In the following reaction, if 690 g of octane, C8H18, is completely combusted in excess of oxygen gas, how many grams of H2O are formed?Reaction: 2C8H18(l) + 25 O2(g) 16CO2(g) + 18H2O(g)Calculation-2:Moles C8H18 reacted = 690 g C8H18 x (1 mol/114.2 g) = 6.0 molesMoles H2O formed = 6.0 mol C8H18 x (18 mol H2O/2 mol C8H18)= 54 moles CO2Mass of H2O formed = 54 mol H2O x (18.02 g/mol) = 970 g
34Stoichiometric Calculations Mass-to-mole-to-mole-to-mass relationship:Example-3: In the following reaction, if 690 g of octane, C8H18, is completely combusted in excess of oxygen gas, how many grams of oxygen gas are consumed?Reaction: 2C8H18(l) + 25 O2(g) 16CO2(g) + 18H2O(g)Calculation-3:Moles C8H18 reacted = 690 g C8H18 x (1 mol/114.2 g) = 6.0 molesMoles O2 consumed = 6.0 mol C8H18 x (25 mol O2/2 mol C8H18)= 75 moles O2Mass of H2O formed = 75 mol O2 x (32.00 g/mol) = 2.4 x 103 g g
35Stoichiometry Involving Limiting Reactant one that got completely consumed in a chemical reaction before the other reactants.Product yields depend on the amount of limiting reactant
36A Reaction Stoichiometry Example:In the reaction: 2Cu(s) + S(s) Cu2S(s), moles of copper are required to react completely with 1 mole of sulfur, which will produce 1 mole of copper(I) sulfide.If a reaction is carried out using 1 mole of copper and 1 mole of sulfur, then copper will be the limiting reactant and sulfur is in excess. Only 0.5 mole of copper(I) sulfide is obtained.
37Exercise #4: Stoichiometry Calculations Ammonia is produced from the reaction with hydrogen according to the following equation:N2(g) + 3H2(g) 2NH3(g)If 25 N2 molecules are reacted with 60 H2 molecules in a sealed container, which molecules will be completely consumed? How many NH3 molecules are formed?(Answer: H2; 40 NH3 molecules)
38Exercise #5: Limiting Reactants and Reaction Yields Ammonia is produced in the following reaction:N2(g) + 3H2(g) 2NH3(g)(a) If 118 g of nitrogen gas is reacted with 31.5 g of hydrogen gas, which reactant will be completely consumed at the end of the reaction? (b) How many grams of the excess reactant will remain (unreacted)? (c) How many grams ammonia will be produced when the limiting reactant is completely reacted and the yield is 100%?(Answer: (a) N2; (b) 6.0 g; (c) g of NH3)
39Theoretical, Actual and Percent Yields Chem 1A Chapter 3 Lecture OutlinesTheoretical, Actual and Percent YieldsTheoretical yield:yield of product calculated based on the stoichiometry of balanced equation and amount of limiting reactant (assuming the reaction goes to completion and the limiting reactant is completely consumed).Actual Yield:Yield of product actually obtained from experimentPercent Yield = (Actual yield/Theoretical yield) x 100%
40Exercise #6: Limiting Reactant & Yields In an ammonia production, the reactor is charged with N2 and H2 gases at flow rates of 805 g and 195 g per minute, respectively, at 227oC, and the reaction is as follows:N2(g) + 3H2(g) 3 NH3(g)(a) What is the rate (in g/min) that ammonia is produced if the yield is 100%? (b) If the reaction produces 915 g of NH3 per minute, calculate the percentage yield of the reaction.(Answer: (a) g/min; (b) Yield = 93.5%)