Molecular and Chemical Formulas
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1 Molecular and Chemical Formulas Atoms combine (bond) to form molecules. Each molecule has a chemical formula. The chemical formula indicates which atoms are found in the molecule, and in what proportion they are found. MJ Bojan Chem 110 FA
2 CHEMICAL FORMULA Formula that gives the TOTAL number of elements in a molecule or formula unit. EMPIRICAL FORMULA Chemical formula with the smallest integer subscripts for a given composition. MJ Bojan Chem 110 FA
3 To get the molecular formula, we need more information. Molecular weight Need: molecular weight. MOLAR MASS or FORMULA WEIGHT sum of atomic weights The connection between experimental property (mass) and moles: C 8 H 18 8 C H MJ Bojan Chem 110 FA
4 PERCENT COMPOSITION % of element = What is the mass percent of C in CO 2? C = amu O = 16.0 amu FW = (16.0) = amu MJ Bojan Chem 110 FA
5 PERCENT COMPOSITION % of element = I have 2g of a sample that is 54.2% C by mass. How many grams of C are in the sample? HINT: If % C in an unknown substance is 54.2%, a 100g sample of that substance contains 54.2g of C. MJ Bojan Chem 110 FA
6 Empirical Formula from Analysis How do we determine the empirical formula? EXPERIMENT! Find mass of each component in a sample. Use to find mole ratio of atoms. Mole ratio gives empirical formula Molecular weight (obtained from a separate experiment) can be used to get MJ Bojan Chem 110 FA
7 Combustion Analysis Empirical formulas are determined by combustion analysis: 4 g sample of an alcohol produces 7.65 g of CO 2 and 4.70 g of H 2 O upon combustion. What is the empirical formula of the alcohol? MJ Bojan Chem 110 FA
8 Empirical Formula from % Composition An analysis of nicotine, a poisonous compound found in tobacco leaves, shows that it is 74.0%C, 8.65% H and 17.35% N. Its molar mass is 162 g/mol. What are the empirical and molecular formula of nicotine? MJ Bojan Chem 110 FA
9 What does a formula tell you? How many atoms are in the formula Mg 3 (PO 4 ) 2 a) 3 b) 5 c) 10 d) 13 How many moles of Mg 3 (PO 4 ) 2 are in a 10.0 g sample? a) moles b) moles c) moles How many atoms are in 10.0 g of Mg 3 (PO 4 ) 2? a) atoms b) atoms c) atoms d) atoms MJ Bojan Chem 110 FA
10 Structural Formulas Empirical and Molecular formulas tell us identity and quantity of atoms. But how are these atoms connected? NEXT: Connectivity (Lewis Structures) Deduce shape from connectivity MJ Bojan Chem 110 FA
11 Rules of Valence To aid in deducing the structure of molecules given the molecular formula, we use the rules of valence. Valence of atom is the number of covalent bond that it typically forms Rules of valence Element # Bonds H 1 C 4 N 3 O,S 2 F, Cl, Br 1 MJ Bojan Chem 110 FA
12 Using the rules of valence, structural formulas for simple molecules can be deduced. MJ Bojan Chem 110 FA
13 Organic Chemistry The chemistry of compounds containing C bonded to H. Millions of organic compounds are known. Often contains O, N, S, and halogens They are the main constituents of living matter: DNA, RNA, proteins, enzymes, MJ Bojan Chem 110 FA
14 Organic Chemistry Reasons for large numbers of organic compounds: 1. Carbon atoms can form strong, stable bonds to other C atoms (thus forming rings, chains, etc.) and to atoms such as H, O, N, S, halogens. (small size) 2. Carbon atoms form up to 4 bonds simultaneously: (valence of 4) molecules can be branched. 3. Carbon atoms form multiple bonds with C or with O, N, S: further structural variations are possible. (Small size and valence of 4) STRUCTURE AFFECTS FUNCTION MJ Bojan Chem 110 FA
15 Classes of Organic Compounds Classification is necessary to manage the large number of compounds HYDROCARBONS Simplest organic compounds Contain only C and H Classes of Hydrocarbons 1) Alkanes all single bonds 2) Alkenes one or more double bonds 3) Alkynes one or more triple bonds 4) aromatic alternating single, double bonds in a ring MJ Bojan Chem 110 FA
16 ALKANES Name Molecular Condensed molecular Lewis formula formula Structures methane CH 4 ethane C 2 H 6 CH 3 -CH 3 propane C 3 H 8 CH 3 CH 2 CH 3 butane C 4 H 10 CH 3 CH 2 CH 2 CH 3 pentane C 5 H 12 CH 3 CH 2 CH 2 CH 2 CH 3 In general: C n H 2n+2 MJ Bojan Chem 110 FA
17 Structure of alkanes All C atoms have tetrahedral geometry: the bonds point at the vertices of a tetrahedron. Bond angle = We can represent the chains as zig-zags or Both of these are octane (C 8 H 18 ): an 8 carbon chain MJ Bojan Chem 110 FA
18 What is the molecular formula? n-octane Iso-octane MJ Bojan Chem 110 FA
19 ISOMERS Isomers are compounds that have the same molecular formula, but different structures Simplest example: butane (C 4 H 10 ) 2 isomers CH 3 CH 2 CH 2 CH 3 n-butane iso-butane MJ Bojan Chem 110 FA
20 ISOMERS Isomers are compounds that have the same molecular formula, but different structures pentane has 3 isomers CH 3 CH 2 CH 2 CH 2 CH 3 n-pentane iso-pentane Neo-pentane MJ Bojan Chem 110 FA
21 CYCLIC ALKANES Alkanes can form rings: C n H 2n Examples: Cyclohexane C 6 H 12 Cyclopentane C 5 H 10 1,1-dimethylcyclobutane MJ Bojan Chem 110 FA
22 Properties and reactions of alkanes Properties: Alkanes are flexible: free rotation about each C C bond Boiling point increases as number of carbons increases Reactions: Alkanes are relatively inert C C and C H bonds are very strong Energy needed to initiate reactions Substitution reaction + Cl 2 + HCl Combustion reaction: alkanes are fuels In complete combustion: water and CO 2 are products. CH 3 CH 2 CH 3 + 5O 2 3CO 2 + 4H 2 O MJ Bojan Chem 110 FA
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Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten Chapter 3 : Calculations with Chemical Formulas and Equations John D. Bookstaver St. Charles Community
Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations
Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten Chapter 3 : Calculations with Chemical Formulas and Equations John D. Bookstaver St. Charles Community