Presentation on theme: "Stoichiometry Atomic Mass The Mole concept Molar Mass Percent Composition of Compounds Determination of Formula of Compounds Writing and Balancing Chemical."— Presentation transcript:
Stoichiometry Atomic Mass The Mole concept Molar Mass Percent Composition of Compounds Determination of Formula of Compounds Writing and Balancing Chemical Equations Interpreting balance equations, and Reaction Stoichiometry and Calculations
Atomic Masses Absolute 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;
Calculation of Relative Atomic Masses Example-1: An atomic mass spectrum gives atomic mass ratio of oxygen atom to carbon-12 as 1.3329:1. If the atomic mass of carbon-12 is exactly 12 u, what is the atomic mass of oxygen? Atomic mass of oxygen = 1.3329 x 12 u = 15.995 u
Calculation of Average Atomic Masses Example-2: Chlorine is composed of two stable naturally occurring isotopes: chlorine-35 (75.76%; 34.9689 u) and chlorine-37 (24.24%; 36.9659 u). What is the average atomic mass of copper? Atomic mass of chlorine = (0.7576 x 34.9689 u) + (0.2424 x 36.9659 u) = 35.45 u (as given in the periodic table)
Calculation of Average Atomic Masses Example-3: Copper is composed of two naturally occurring isotopes: copper-63 (69.09%; 62.93 u) and copper-65 (30.91%; 64.93 u). What is the average atomic mass of copper? Atomic mass of copper = (0.6909 x 62.93 u) + (0.3091 x 64.93 u) = 63.55 u (as given in the periodic table)
Exercise #1: Relative Atomic Mass A mass spectrometer computed the atomic mass ratio of fluorine to carbon-12 as 1.5832-to-1. If the atomic mass of carbon-12 is 12 u (exactly), what is the atomic mass of fluorine in u? (Answer: 18.998 u)
Exercise #2: Average Atomic Mass Natural boron is composed of two isotopes: 19.78% boron-10 (atomic mass = 10.0129 amu) and 80.22% boron-11 (atomic mass = 11.0093 amu). What is the average atomic mass of naturally occurring boron? (Answer: 10.81 u)
Molar Quantity The Mole: A quantity that contains the Avogadro’s number of items; Avogadro’s number = 6.022 x 10 23 12.01 g of carbon contains the Avogadro’s number of carbon atoms. 1 mole of carbon = 12.01 g 1 carbon atom = 12.01 u (or amu)
Gram-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 = 16.00 u Mass of 1 mole of oxygen = 16.00 g Gram-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.
Atomic Mass & Gram-Atomic Mass Examples: Element Atomic massGram-atomic mass Carbon 12.01 u 12.01 g/mol Oxygen16.00 u 16.00 g/mol Aluminum26.98 u 26.98 g/mol Silicon 28.09 u 28.09 g/mol Gold 197.0 u 197.0 g/mol
Molecular 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 MassMolar Mass N 2 28.02 u 28.02 g/mol H 2 O 18.02 u 18.02 g/mol C 8 H 18 114.22 u114.22 g/mol
Calculating Molar Mass Calculating the molar mass of sucrose, C 12 H 22 O 11 : (12 x 12.01 g) + (22 x 1.008 g) + (11 x 16.00 g) = 342.3 g/mole Molar mass of ammonium hydrogen phosphate, (NH 4 ) 2 HPO 4 : (2 x 14.01 g) + (9 x 1.008 g) + (1 x 30.97 g) + (4 x 16.00 g) = 132.06 g/mole
Percent Composition of a Compound Composition of aluminum sulfate, Al 2 (SO 4 ) 3 : Molar mass of Al 2 (SO 4 ) 3 = (2 x 26.98 g) + (3 x 32.06 g) + (12 x 16.00 g) = 342.14 g/mole Mass percent of Al = (53.96 g/342.14 g) x 100% = 15.77% Mass percent of S = (96.18 g/342.14 g) x 100% = 28.11% Mass percent of O = (192.0 g/342.14 g) x 100% = 56.12%
Formula of Compounds Empirical Formula A chemical formula that represents a simple whole number ratio of the number of moles of elements in the compound. Examples: MgO, Cu 2 S, CH 2 O, etc. Molecular Formula A formula that shows the actual number of atoms of each type in a molecule. Examples: C 4 H 10, C 6 H 6, C 6 H 12 O 6.
Empirical 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:O Mole of C = 68.12 g x (1 mol C/12.01 g) = 5.672 mol C Mole of H = 13.73 g x (1 mol H/1.008 g) = 13.62 mol H Mole of O = 18.15 g x (1 mol O/16.00 g) = 1.134 mol O Divide all moles by mole of O (smallest value) to get simple ratio: 5.672 mol C/1.134 mol O = 5; 13.62 mol H/1.134 mol O = 12, and 1.134 mol O/1.134 mol O = 1; Mole ratio: 5C:12H:1O Empirical formula = C 5 H 12 O
Empirical Formula-2 Empirical formula from mass of elements in a sample of compound Example: 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) = 0.0633 mol P; Mole of O = (4.49 g – 1.96 g) x (1 mol/16.00 g) = 0.158 mol O; Divide by mole of P (smaller value) to get a simple mole ratio: 0.0633 mol P/0.0633 = 1 mol P; 0.158 mol O/0.0633 = 2.5 mol O Mole ratio: 1 mol P to 2.5 mol O, OR 2 mol P to 5 mol O Empirical formula = P 2 O 5
Empirical Formula-3 Empirical 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 CO 2 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 CO 2 x (12.01 g C/44.01 g CO 2 ) = 1.44 g Mass % of C = (1.44 g C/2.32 g sample) x 100% = 62.1% Mass of H = 2.16 g H 2 O x (2 x 1.008 g/18.02 g H 2 O) = 0.24 g Mass % 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 g Mass % of O = 100 – 62.1% C – 10.4% H = 27.5% Derive empirical formula from these mass percent composition (next slide)
Empirical Formula-3 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 mol Mole of H = 0.242 g x (1 mol/1.008 g) = 0.24 mol Mole of O = 0.64 g x (1 mol/16.00 g) = 0.04 mol Divide 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 O Simple molar ratio: 3 mol C : 6 mol H : 1 mol O Empirical formula: C 3 H 6 O
Molecular Formula Molecular formula is derived from empirical formula and molecular mass, which is obtained independently Empirical formula = C x H y O z; molecular formula = (C x H y O z ) n, where n = (molecular mass/empirical formula mass) Example: A compound has an empirical formula C 3 H 6 O and its molecular formula is 116.2 u. What is the molecular formula? Solution: Empirical formula mass = (2 x 12.01 u) + (6 x 1.008 u) + 16.00 u = 58.1 u Molecular formula = (C 3 H 6 O) n ; where n = (116.2 u/58.1 u) = 2 Incorrect molecular formula = (C 3 H 6 O) 2 ; Correct molecular formula = C 6 H 12 O 2
Exercise #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: P 2 O 5 ; P 4 O 10 ; 4P + 5 O 2 P 4 O 10 )
Chemical Equation #1 Description of reaction: Iron reacts with oxygen gas and forms solid iron(III) oxide: Identity: reactants = iron (Fe) and oxygen gas (O 2 ); product = iron(III) oxide Chemical equation: Fe (s) + O 2 (g) Fe 2 O 3 (s) Balanced equation: 4Fe (s) + 3 O 2 (g) 2Fe 2 O 3 (s)
Chemical Equation #2 Description of reaction: Phosphorus reacts with oxygen gas to form solid tetraphosphorus decoxide. Equation: P (s) + O 2 (g) P 4 O 10 (s) Balanced eqn.: 4P (s) + 5 O 2 (g) P 4 O 10 (s)
Chemical Equation #3 Description of reaction: Propane gas (C 3 H 8 ) is burned in air (excess of oxygen) to form carbon dioxide gas and water vapor; Identity: reactants = C 3 H 8 (g) and O 2 (g) ; products = CO 2 (g) and H 2 O (g) ; Equation: C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O (g) ; Balanced equation: C 3 H 8 (g) + 5 O 2 (g) 3CO 2 (g) + 4H 2 O (g)
Chemical Equation 4 Description of reaction: Ammonia gas (NH 3 ) reacts with oxygen gas to form nitrogen monoxide gas and water vapor; Equation: NH 3 (g) + O 2 (g) NO (g) + H 2 O (g); Balancing the equation: 2NH 3 (g) + 5 / 2 O 2 (g) 2NO (g) + 3H 2 O (g) ; Multiply throughout by 2 to get rid of the fraction: 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6H 2 O (g) ;
Balancing Chemical Equations Rules for balancing equations: 1.Use smallest integer coefficients in front of each reactants and products as necessary; coefficient “1” need not be indicated; 2.The formula of the substances in the equation MUST NOT be changed. Helpful steps in balancing equations: 1.Begin with the compound that contains the most atoms or types of atoms. 2.Balance elements that appear only once on each side of the arrow. 3.Next balance elements that appear more than once on either side. 4.Balance free elements last. 5.Finally, check that smallest whole number coefficients are used.
Stoichiometry Stoichiometry = the quantitative relationships between one reactant to another, or between a reactant and products in a chemical reaction. Interpreting balanced equations: Example: C 3 H 8 (g) + 5 O 2 (g) 3CO 2 (g) + 4H 2 O (g) ; The equation implies that: 1 C 3 H 8 molecule reacts with 5 O 2 molecules to produce 3 CO 2 molecules and 4 H 2 O molecules; OR 1 mole of C 3 H 8 reacts with 5 moles of O 2 to produce 3 moles of CO 2 and 4 moles of H 2 O.
Stoichiometric Calculations Mole-to-mole relationship: Example: In the following reaction, if 6.0 moles of octane, C 8 H 18, is completely combusted in excess of oxygen gas, how many moles of CO 2 and H 2 O, respectively, will be formed? How many moles of O 2 does it consumed? Reaction: 2C 8 H 18 (l) + 25 O 2 (g) 16CO 2 (g) + 18H 2 O(g) Calculations: Mole CO 2 formed = 6.0 mol C 8 H 18 x (16 mol CO 2 /2mol C 8 H 18 ) = 48 moles Mole H 2 O formed = 6.0 mol C 8 H 18 x (18 mol H 2 O /2mol C 8 H 18 ) = 54 moles Mole O 2 consumed = 6.0 mol C 8 H 18 x (25 mol O 2 /2mol C 8 H 18 ) = 75 moles
Stoichiometric Calculations Mass-to-mole-to-mole-to-mass relationship: Example-1: In the following reaction, if 690 g of octane, C 8 H 18, is completely combusted in excess of oxygen gas, how many grams of CO 2 are formed? Reaction: 2C 8 H 18 (l) + 25 O 2 (g) 16CO 2 (g) + 18H 2 O (g) Calculation-1: Moles C 8 H 18 reacted = 690 g C 8 H 18 x (1 mol/114.2 g) = 6.0 moles Moles CO 2 formed = 6.0 mol C 8 H 18 x (16 mol CO 2 /2 mol C 8 H 18 ) = 48 moles CO 2 Mass of CO 2 formed = 48 mol CO 2 x (44.01 g/mol) = 2.1 x 10 3 g
Stoichiometric Calculations Mass-to-mole-to-mole-to-mass relationship: Example-2: In the following reaction, if 690 g of octane, C 8 H 18, is completely combusted in excess of oxygen gas, how many grams of H 2 O are formed? Reaction: 2C 8 H 18 (l) + 25 O 2 (g) 16CO 2 (g) + 18H 2 O (g) Calculation-2: Moles C 8 H 18 reacted = 690 g C 8 H 18 x (1 mol/114.2 g) = 6.0 moles Moles H 2 O formed = 6.0 mol C 8 H 18 x (18 mol H 2 O/2 mol C 8 H 18 ) = 54 moles CO 2 Mass of H 2 O formed = 54 mol H 2 O x (18.02 g/mol) = 970 g
Stoichiometric Calculations Mass-to-mole-to-mole-to-mass relationship: Example-3: In the following reaction, if 690 g of octane, C 8 H 18, is completely combusted in excess of oxygen gas, how many grams of oxygen gas are consumed? Reaction: 2C 8 H 18 (l) + 25 O 2 (g) 16CO 2 (g) + 18H 2 O (g) Calculation-3: Moles C 8 H 18 reacted = 690 g C 8 H 18 x (1 mol/114.2 g) = 6.0 moles Moles O 2 consumed = 6.0 mol C 8 H 18 x (25 mol O 2 /2 mol C 8 H 18 ) = 75 moles O 2 Mass of H 2 O formed = 75 mol O 2 x (32.00 g/mol) = 2.4 x 10 3 g g
Stoichiometry Involving Limiting Reactant Limiting reactant one that got completely consumed in a chemical reaction before the other reactants. Product yields depend on the amount of limiting reactant
A Reaction Stoichiometry Example: In the reaction: 2Cu (s) + S (s) Cu 2 S (s), 2 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.
Exercise #4 : Stoichiometry Calculations Ammonia is produced from the reaction with hydrogen according to the following equation: N 2 (g) + 3H 2 (g) 2NH 3 (g) If 25 N 2 molecules are reacted with 60 H 2 molecules in a sealed container, which molecules will be completely consumed? How many NH 3 molecules are formed? (Answer: H 2 ; 40 NH 3 molecules)
Exercise #5: Limiting Reactants and Reaction Yields Ammonia is produced in the following reaction: N 2 (g) + 3H 2 (g) 2NH 3 (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) N 2 ; (b) 6.0 g; (c) 143.4 g of NH 3 )
Theoretical, Actual and Percent Yields Theoretical 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 experiment Percent Yield = (Actual yield/Theoretical yield) x 100%
Exercise #6: Limiting Reactant & Yields In an ammonia production, the reactor is charged with N 2 and H 2 gases at flow rates of 805 g and 195 g per minute, respectively, at 227 o C, and the reaction is as follows: N 2 (g) + 3H 2 (g) 3 NH 3 (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 NH 3 per minute, calculate the percentage yield of the reaction. (Answer: (a) 978.5 g/min; (b) Yield = 93.5%)