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Week 5: Lectures 13 – 15 Chemical Formulas Lecture 13: W 9/21 Lecture 14: F 9/23 Lecture 15: M 9/26 Reading: BLB Ch 2.6, 2.9, 3.3 – 3.5; 9.1 – 9.3 Homework: BLB 3:25, 29, 35, 37, 51, 53; Supp 3:1 – 9 BLB 9:23, 25, 27, 38; Sup 9:1 – 7 Reminder: ALEKS Objective 5 due on Tue, 9/27 “No Score” from Exam 1?? Not a problem—Go see Mike Joyce in 210 Whitmore or email him at [email protected] Molecular Formula: chemical formula with the total number of atoms and their relative proportions in each molecule. Empirical Formula: chemical formula with the smallest integer subscripts for a given molecule. Structural Formula: chemical formula with structural information about connectivity and angles. Jensen Chem 110 Chap3 Page: 2 Connecting microscopic (atoms) to macroscopic (weight) using Important Connections • Avogadro!s number (NA): Connects molecules (or atoms) to moles • Molar Mass (in g/mol) connects mass to moles; connects experimentally measured property to moles (or molecules) of substance; the molar mass (in g/mol) of any substance is always numerically equal to its formula weight (in amu). • Empirical Formula: chemical formula with the smallest integer subscripts; tells relative number of moles (molar ratio) of each element in a compound; obtained from mass percent or molecular formula; • Balanced chemical equation: connects moles (molecules) of reactants with moles (molecules) of products; related to conservation of mass • Conservation of mass and energy: mass of products = mass of reactants; energy is conserved but can change its form Jensen Chem 110 Chap3 Page: 3 Coefficients and Subscripts • In a molecule, subscripts denote the number of ________ per molecule or formula unit C6H12O6 : one C6H12O6 molecule has __ C, __ O, and __ H atoms. Mg(NO3)2 : one Mg(NO3)2 formula unit has __ Mg, __ N, and __ O atoms • In a reaction, coefficients denote the relative number of ________ of each compound 2 H2 + O 2 ! 2 H 2O ! Changing the subscripts changes the _________ of the molecule. ! Changing the coefficient changes the _________ of that molecule, not the identity. Jensen Chem 110 Chap3 Page: 4 Working with Formulas Working with Formulas Example: How many moles of ammonium ions are present in 30.9 g of ammonium sulfate? A. B. C. D. E. 0.468 mol 0.288 mol 0.234 mol 2.14 mol 3.47 mol Jensen Practice Example: Which of the following has the greatest number of total atoms present? A. B. C. D. E. Chem 110 Chap3 Page: 5 6.022 x 1023 O2 molecules 0.1 mol C6H12O6 56 g Fe 0.4 mol CH4 4.4 g of H2 Jensen Chem 110 Chap3 Page: 6 Percent Composition: Mass Percent contributed by each element in compound (Note: percentage must add to 100) A) 18 g 1. Given chemical formula, % element = ! Practice Example: How many grams of oxygen are in 65 g of C 2H 2O 2? (# of element) (AW of element) "100 FW of compound B) 29 g C) 9.0 g Example: What is the mass percent of C in CO2? D) 36 g C= _____ amu; O = _____ amu; CO2 = _______ amu E) 130 g %C = 2. If we have 100 g of sample, % element = ! mass of element "100 100 g compound Example: I have 2g of a sample that is 27.3% C by mass, how many grams of C are in the sample? Jensen Chem 110 Chap3 Page: 7 Jensen Chem 110 Chap3 Page: 8 Connect Mass Percent to Chemical Formula Given one quantity, we can calculate the other: • Start with a chemical formula and calculate the mass % elements; or • Start with mass % of elements (i.e. empirical data) and calculate a chemical formula. Jensen Chem 110 Chap3 Page: 9 Practice Example: Butyric acid, which has the smell of rancid butter, is 54.53% carbon by mass and 9.15% hydrogen by mass, and is composed only of carbon, hydrogen, and oxygen. It has a molar mass of 88.10 g/mol. What is the molecular formula of butyric acid? A. B. C. D. E. C4H 8O 2 C2H 4O C5H 4O C 4H 2O 2 C4H16O Jensen Chem 110 Chap3 Page: 10 Combustion Analysis Solve: 1. Assume 100 g of sample C = _______ g H = _______ g O = ______ g 2. Find moles of each element C = _____________________________________ mol H = _____________________________________ mol O = _____________________________________ mol 3. Find ratio of moles of elements (divide by Fuel C, H, O + O2 CO2 + H2O Given grams of sample in excess O2, grams of CO2 and H2O produced; may have N or O in formula (non-C,H atoms); determine empirical formula or molecular formula of the sample. Combustion Analysis Procedures smallest) 1. Use grams of CO2 and H2O to find moles of Cand H- elements in the products and in the sample; Empirical Formula: 4. Get molecular formula from empirical formula 2. Use massnon-C,H = masssample - massC - massH to find mass and moles of non-C,H elements in the sample; 3. Calculate the ratio of C, H, and non-C,H elements (divide by smallest) in the sample; 4. Write empirical formula using ratios; and convert to molecular formula using molar mass of sample (if known) Jensen Chem 110 Chap3 Page: 11 Jensen Chem 110 Chap3 Page: 12 Practice Example: 11.0 g sample of a compound containing only C, H, and O was burned completely in air to produce 15.6 g of CO2 and 9.60 g of water. What is the empirical formula of this compound? Solve: a) C4H9O 2. Calculate mass of C, H, and O elements and moles of O elements in the sample b) C2H6O 1. Use grams of CO2 and H2O to calculate moles of C, H elements in products and in the sample C = _____________________________________ mol H = _____________________________________ mol C = ______________________________________ g c) C3H6O2 d) CH3O H = ______________________________________ g e) CH2O O = ______________________________________ g O = _____________________________________ mol 3. Find ratio of moles of elements (divide by smallest) Empirical Formula: Jensen Chem 110 Chap3 Page: 13 Jensen Chem 110 Chap3 Page: 14 What’s Next? VSEPR Model Chemical Formulas tell us how atoms are connected to each other Lewis structures tell us where the electrons are. Electron Domain Geometry (EDG) Electron pairs " Electron domains One lone pair = one electron domain Lewis Structures (2D) One bond (single, double or triple bond) = one electron domain. — Bonding e pairs — Lone e pairs Molecular Shape (3D): Arrangement of Atoms in the Molecule Valence Shell Electron Pair Repulsion (VSEPR): Predict molecular shape (spatial arrangement of atoms) from Lewis structure VSEPR: Electron domains around a central atom repel each other. They must stay near nucleus but otherwise try to get as far apart from each other as possible Jensen Chem 110 Chap 9 Page: 1 Electron Domain Geometries (EDG) Determine Electron Domain Geometry (EDG) and Molecular Geometry (MG) 1. Draw the Lewis structure of the molecule 2. Count the number of electron domains in the Lewis structure: one bond (single, double, or triple) counts as ______ domain! And one lone pair counts as ______ domain!) 3. The electron domain geometry corresponds to the number of electron domains 4. The molecular geometry is defined by the positions of only ________________ in the molecules, _______ the nonbonding pairs 5. Some electron domains have more than one molecular geometry Jensen Chem 110 Chap 9 Page: 2 Jensen Chem 110 Chap 9 Page: 3 1. Linear EDG (2 electron domains) Two possible MG for trigonal planar EDG: # If all bonding electron domains, MG = _________ # If there is a nonbonding (lone) pair, MG = ______ 3. Tetrahedral EDG (4 electron domains) # In this domain, only one possible MG: linear. # NOTE: If there are only two atoms in the molecule, the molecule will be linear no matter what the electron domain is. 2. Trigonal planar EDG (3 electron domains) Three possible MG for tetrahedral EDG: # If all are bonding pairs, MG = ________________ # If there is one lone pair, MG= _________________ # If there are two lone pairs, MG= _______________ Jensen Chem 110 Chap 9 Page: 4 Jensen Chem 110 Chap 9 Page: 5 The Trigonal Bipyramidal Case Lone Pairs and Bond Angles 4. Trigonal bipyramidal EDG (5 electron domains) ! Nonbonding pairs are physically larger Two types of sites ________ – _____ and ______ • Two types of• sites –_______and • Angles between sitesbetween are different • Angles sitesfor are different for these two types. these two types. than bonding pairs ! Therefore, their repulsions are greater; this tends to _________ bond angels 900 on the other side of the molecule Eq Ax Eq 1200 A Eq 900 Ax • Lone pairs prefer to be equatorial. Why? Multiple Bond and Bond Angles ! greater electron density on one side of the central atom ! Therefore, bond angles involving multiple bond are _____________, while angles on other side of Lone pairs prefer to be equatorial. the molecule are ___________ CH110 FA11 SAS Jensen Chem 110 Chap 9 Page: 6 Jensen Chem 110 Chap 9 Page: 7 4. Trigonal bipyramidal EDG (5 electron domains) 5. Octahedral EDG (6 electron domains) • All positions are equivalent • Two lone pairs go on ___________ sides of the Four possible MG for trigonal bipyramidal EDG: ! ! ! ! Trigonal bipyramidal Seesaw T-shaped Linear Jensen Chem 110 Chap 9 molecule Three possible MG for octahedral EDG: ! Octahedral ! Square pyramidal ! Square planar Page: 8 Jensen Chem 110 Chap 9 Page: 9 Determine the Geometry of a Molecule 1. Draw the Lewis structure of the molecule. 2. Determine the electron domain geometry (EDG: count electron domains: multiple bonds count as one domain). 3. Focus on bonding electron pairs ONLY to determine the molecular geometry (MG) Example: What is the molecular geometry of IF5? Practice Example The molecular geometry of which molecule is square planar? A. B. C. D. E. CCl4 XeF4 PH3 XeF2 BrF3 A. see-saw B. octahedral C. square pyramidal D. square bipyramidal E. square planar Jensen Chem 110 Chap 9 Page: 10 Jensen Chem 110 Chap 9 Page: 11 VSEPR can be applied to Larger Molecules ! Describe geometry around Determine Approximate Bond Angel Practice Example: Determine the approximate bond angles indicated. a particular atom Example: What is the approximate C—C—O bond angel as indicated? A. 90° B. 105° O 1 NCH2CH2CH2C 2 H OH H Angle #1 Angle #2 A. 109° 109° B. 180° 120° C. 120° 90° D. 109° 120° E. 120° 109° C. 109° D. 120° E. 180° Jensen Chem 110 Chap 9 Page: 12 Jensen Chem 110 Chap 9 Page: 13 Molecular Geometries: Summary # of electron domains 2 # of bonded atoms 2 3 molecular geometry Example linear CO2 3 trigonal planar BF3 3 2 bent O3 4 4 tetrahedral CH4 4 3 4 2 5 5 5 4 trigonal bipyramidal seesaw 5 3 T-shaped BrF3 5 2 linear XeF2 6 6 octahedral SF6 6 5 6 4 Jensen trigonal pyramidal bent square pyramidal square planar Chem 110 Chap 9 NH3 Molecular Polarity Molecular geometry and relative electronegativity of elements determine many properties of molecules • How does microwave cooking work? • Why do water and methane differ so much in their boiling point (BP)? H 2O BP of H2O = 100 ˚C; BP of CH4 = -161 ˚C PCl5 • Why do alcohols and water mix but oil and water don!t? SF4 BrF5 XeF4 Page: 14 Jensen Chem 110 Chap 9 Page: 15 Bond Polarity Molecular Polarity The polarity of individual bond is determined • Just because a molecule has polar bonds does not mean the molecule as a whole will be polar. by electronegativity difference between elements; The greater the difference, the Polarity of entire molecule: Dipole Moment (µ) more polar the bond. For a diatomic molecule: µ = Q r H r Cl units = debye (D) •%%%% • • Most polar: ionic compound, e.g. NaF +Q • Polar covalent: (two different atoms) &Q • A molecule is polar if there is a NET charge separation between two “ends” of the molecule: a negative “end” and a positive “end” A molecule is polar (has a net dipole) if it has: a. Polar bonds b. Geometry where bond dipoles do not cancel out $+ and $— are partial charges for H and F To determine the polarity of a molecule that has more than two atoms: • Nonpolar covalent: (two identical atoms); e.g. H2, F2, Cl2, O2, 1. Find molecular shape using VSEPR 2. Find “bond” dipoles 3. Use vector analysis to find NET molecular dipole Jensen Jensen Chem 110 Chap 9 Page: 16 Chem 110 Chap 9 Page: 17 Examples: Examples: CO2 NH3 (electronegativity: N > H) (electronegativity: O > C) EDG: tetrahedral EDG: Linear MG: trigonal pyrimidal MG: Linear Bond dipoles? Bond dipoles? Net dipole moment? N H H H Net dipole moment? CF4 (electronegativity: F > C) H 2O (electronegativity: O > H) EDG: tetrahedral EDG: tetrahedral MG: tetrahedral MG: bent Bond dipoles? Bond dipoles? Net dipole moment? F Net dipole moment? Jensen Chem 110 Chap 9 Page: 18 Jensen F Chem 110 Chap 9 C F F Page: 19 Example: Which molecule below is polar? Molecular Geometry & Molecular Polarity Do not memorize this table!! Draw the generic structures and think about them to verify! Practice Example: Which molecule shown below has non-zero dipole moment? linear bent trigonal pyramidal T-shaped seesaw square pyramidal A. XeF2 B. BF3 C. SF4 D. IF2– E. BeCl2 Jensen MG linear trigonal planar tetrahedral trigonal bipyramidal octahedral square planar Formula Example Polar? AX2 CO2 AX3 BF3 AX4 CCl4 AX5 PCl5 AX6 SF6 AX4 XeF4 AX AX2 AX3 AX3 AX4 AX5 HF, HCl H2O, SO2 NH3 BrF3 SF4 BrF5 NOTE: X is the same atom on all positions Chem 110 Chap 9 Page: 20 Jensen Chem 110 Chap 9 Page: 21