Elements and Compounds

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10 18 alkane hydrocarbons Chemical Bonds compounds are made of atoms held together by chemical

1 Molecules, Compounds, and Chemical Equations Elements and Compounds elements combine to give almost limitless number of compounds C n H n + approx alkane hydrocarbons Chemical Bonds compounds are made of atoms held together by chemical bonds bonds are net electrostatic forces of attraction between atoms the bonding attraction comes from attractions between protons and electrons 1

Bond Types two extreme types: ionic and covalent ionic bonds electrons transferred between atoms, resulting in oppositely charged

some of their electrons generally found when nonmetal atoms bonded together real bonds lie along a spectrum or continuum of ionic

Representing Compounds with Chemical Formula compounds are generally represented with a chemical formula amount of information

2 Bond Types two extreme types: ionic and covalent ionic bonds electrons transferred between atoms, resulting in oppositely charged ions that attract each other generally found when metal atoms bonded to nonmetal atoms covalent bonds or molecular atoms share some of their electrons generally found when nonmetal atoms bonded together real bonds lie along a spectrum or continuum of ionic and covalent types. Representing Compounds with Chemical Formula compounds are generally represented with a chemical formula amount of information about structure varies with the type of formula all formula and models convey a limited amount of information none is a perfect representations all chemical formulas tell what elements are in the compound use the letter symbol of the element

3 Types of Formula Empirical Formula Empirical Formula describe the kinds of elements and the ratio of their atoms they do not describe how many atoms, the order of attachment, or the shape the formulas for ionic compounds are empirical e.g., HO could be H O C 4 H 9 could be C 8 H 18 Types of Formula Molecular Formula Molecular Formula kinds of atoms and the numbers of their atoms they do not describe the order of attachment, or the shape Types of Formula Structural Formula Structural Formula kinds of atoms numbers of their atoms order of atom attachment kind of attachment do not directly describe the 3-dimensional shape, but an experienced chemist can make a good guess at it use lines to represent covalent bonds each line describes the number of electrons shared by the bonded atoms single line = shared electrons, a single covalent bond double line = 4 shared electrons, a double covalent bond triple line = 6 shared electrons, a triple covalent bond 3

Representing Compounds Molecular Models Models 3-dimensional structure along with all the other

atoms and sticks to represent the electron cloud attachments between them Space-Filling Models use

Hydrogen Peroxide Molecular Formula = H O Empirical Formula = HO Benzene Molecular Formula = C 6 H

4 Representing Compounds Molecular Models Models 3-dimensional structure along with all the other information given in structural formula Ball-and-Stick d S c Models suse balls to represent ese the atoms and sticks to represent the electron cloud attachments between them Space-Filling Models use interconnected spheres to show the electron clouds of atoms connecting together Chemical Formulas Hydrogen Peroxide Molecular Formula = H O Empirical Formula = HO Benzene Molecular Formula = C 6 H 6 Empirical Formula = CH Glucose Molecular Formula = C 6 H 1 O 6 Empirical Formula = CH O Types of Formula 4

Molecular View of Elements and Compounds Classifying Materials atomic elements = elements whose particles are single atoms molecular elements = elements whose particles are multi-atom molecules

5 Molecular View of Elements and Compounds Classifying Materials atomic elements = elements whose particles are single atoms molecular elements = elements whose particles are multi-atom molecules molecular compounds = compounds whose particles are molecules made of only nonmetals ionic compounds = compounds whose particles are cations and anions Molecular Elements Certain elements occur as atom molecules Rule of 7 s Other elements occur as polyatomic molecules P 4, S 8, Se 8 H 7 7A N O F Cl Br I 5

C 3 H 8 an array of Na + ions molecules and Cl - ions.

NOMENCLATURE Diatomic ionic compounds Representative/main group

6 Molecular Elements Ionic vs. Molecular Compounds Propane contains Table salt contains individual C 3 H 8 an array of Na + ions molecules and Cl - ions. NaCl is Not a molecule! What about Ag(NH 3 ) Cl huh? NOMENCLATURE Diatomic ionic compounds Representative/main group Transition Metal/d-block metals Molecular or covalent compounds Hydrates Polyatomic ions Acids and bases Formula to name and name to formula 6

Formula Mass the mass of an individual molecule or formula unit also known as molecular mass or molecular weight sum of the masses of the atoms in a

0 amu Molar Mass of Compounds the relative masses of molecules can be calculated from atomic masses Formula Mass = 1 molecule of H O = (1.

7 Formula Mass the mass of an individual molecule or formula unit also known as molecular mass or molecular weight sum of the masses of the atoms in a single molecule or formula unit whole = sum of the parts! mass of 1 molecule of H O = (1.01 amu H) amu O = 18.0 amu Molar Mass of Compounds the relative masses of molecules can be calculated from atomic masses Formula Mass = 1 molecule of H O = (1.01 amu H) amu O = 18.0 amu since 1 mole of H O contains moles of H and 1 mole of O Molar Mass = 1 mole H O = (1.01 g H) g O = 18.0 g so the Molar Mass of H O is 18.0 g/mole 7

8 Example Find the number of CO molecules in 10.8 g of dry ice Given: Find: Concept Plan: Rlti Relationships: Solution: Check: 10.8 g CO molecules CO g CO mol CO 1mol molec CO molecules 44.01g 1mol 1 mol CO = g, 1 mol = 6.0 x mol CO molecules 10.8 g CO 44.01g CO 1mol molecules CO since the given amount is much less than 1 mol CO, the number makes sense Practice - Converting Grams to Molecules How many molecules are in 50.0 g of PbO? (PbO = 39.) Practice - Converting Grams to Molecules How many molecules are in 50.0 g of PbO? Given: 50.0 g PbO Find: molecules PbO Relationships: 1 mole PbO 39. g; 1 mol 6.0 x 10 3 molec Concept Plan: gpbo mol PbO molec oecpbo 1mole PbO molec 39. g 1mole PbO Apply Solution Map: 1mole PbO molec 50.0 g PbO molec PbO 39. g 1mole PbO Check Answer: Units are correct. Number makes sense because given amount less than 1 mole 8

9 Percent Composition Percentage of each element in a compound By mass Can be determined from 1. the formula of the compound. the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding Percentage part whole 100% Example 3.13 Find the mass percent of Cl in C Cl 4 F Given: Find: Concept Plan: Rlti Relationships: C Cl 4 F % Cl by mass 4 molar mass Cl Mass % Cl 100% molar mass CCl4F mass element X in 1mol Mass % element X 100% mass1mol of compound Solution: 4 molar mass Cl 4(35.45 g/mol) g/mol molar mass CCl4F (1.01) 4(35.45) (19.00) 03.8 g/mol g/mol Mass % Cl 100% 69.58% 03.8 g/mol Check: since the percentage is less than 100 and Cl is much heavier than the other atoms, the number makes sense Practice - Determine the Mass Percent Composition of the following CaCl (Ca = 40.08, Cl = 35.45) 9

10 CaCl Practice - Determine the Percent Composition of the following Mass % Cl Mass % Ca Mass % Cl molar mass Ca 100% molar mass CaCl molar mass Cl 100% molar mass CaCl molar mass Cl (35.45 g/mol) g/mol molar mass CaCl 1(40.08) (35.45) g/mol g/mol Mass % Ca 100% 36.11% g/mol g/mol g/mol 100% 63.88% Mass Percent as a Conversion Factor the mass percent tells you the mass of a constituent element in 100 g of the compound the fact that CCl F is 58.64% Cl by mass means that 100 g of CCl F contains g Cl this can be used as a conversion factor 100 g CCl F : g Cl g Cl 100 g CCl g CClF g Cl F g Cl g CClF 100 g CClF g Cl Example 3.14 Find the mass of table salt containing.4 g of Na Given: Find: Concept Plan: Rlti Relationships: Solution: Check:.4 g Na, 39% Na g NaCl g Na 100. g NaCl : 39 g Na 100 g NaCl 39 g Na g NaCl 100 g NaCl.4 g Na 6. g NaCl 39 g Na since the mass of NaCl is more than x the mass of Na, the number makes sense 30

11 Practice Benzaldehyde is 79.% carbon. What mass of benzaldehyde contains 19.8 g of C? Practice Benzaldehyde is 79.% carbon. What mass of benzaldehyde contains 19.8 g of C? Given: Find: Concept Plan: Rlti Relationships: 19.8 g C, 79.% C g benzaldehyde g C 100 g benzaldehyde 79.g C 100. g benzaldehyde : 79. g C g benzaldehyde Solution: 100 g benzaldehyde 19.8 g C 5.0 g benzaldehyde 79. g C Check: since the mass of benzaldehyde is more than the mass of C, the number makes sense Conversion Factors in Chemical Formulas chemical formulas have inherent in them relationships between numbers of atoms and molecules or moles of atoms and molecules these relationships can be used to convert between amounts of constituent elements and molecules like percent composition 31

12 Example 3.15 Find the mass of hydrogen in 1.00 gal of water Given: Find: Concept Plan: 1.00 gal H O, d HO = 1.00 g/ml g H gal H O L H O ml H O g H O g H O mol H O mol H g H Rlti Relationships: L = 1 gal, 1 L = 1000 ml, 1.00 g H O = 1 ml, 1 mol H O = 18.0 g, 1 mol H = g, mol H : 1 mol H O Solution: L 1000 ml 1.00 g gal HO g HO 1gal 1L 1mL 3 1mol HO mol H g H g HO g H 18.0 g 1mol HO 1mol H Check: since 1 gallon weighs about 3800 g, and H is light, the number makes sense Practice - How many grams of sodium are in 6. g of NaCl? (Na =.99; Cl = 35.45) How many grams of sodium are in 6. g of NaCl? Given: 6. g NaCl Find: g Na Rel: 1 mole NaCl g; 1 mol Na 1 mol NaCl; 1 mol Na.99 g Na Concept Plan: gnacl 1 mole NaCl mol NaCl 1 mole Na mol Na.99 g Na g Na g 1mol NaCl 1mol Na Apply Concept Plan: 1mole NaCl 1mole Na.99 g Na 6. g NaCl.4 g Na g 1mole NaCl 1mole Na Check Answer: Units are correct. Number makes sense because given amount less than 1 mole NaCl. 3

13 Empirical Formula simplest, whole-number ratio of the atoms of elements in a compound can be determined from elemental analyses of various kinds. Obtain masses of elements identified or released through decomposition reaction or other kinds. E.g., combustion analysis percent composition Finding an Empirical Formula 1) convert the percentages to grams a) assume you start with 100 g of the compound b) skip if already grams ) convert grams to moles a) use molar mass of each element 3) write a pseudoformula using moles as subscripts 4) divide all by smallest number of moles a) if result is within 0.1 of whole number, round to whole number 5) multiply all mole ratios by number to make all whole numbers a) if ratio?.5, multiply all by ; if ratio?.33 or?.67, multiply all by 3; if ratio 0.5 or 0.75, multiply all by 4; etc. b) skip if already whole numbers Example 3.17 Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53% 33

14 Example: formula of aspirin with the given mass percent composition. Write down the given quantity and its units. Given: C = 60.00% H = 4.48% O = 35.53% Therefore, in 100 g of aspirin there are g C, 4.48 g H, and g O Example: formula of aspirin with the given mass percent composition. Given: g C, 4.48 g H, g O Write down the quantity to find and/or its units. Find: empirical formula, C xh yo z Example: formula of aspirin with the given mass percent composition. Given: g C, 4.48 g H, g O Write a Concept Plan: g C, H, O mol C, H, O mol ratio empirical formula 34

15 Example: formula of aspirin with the given mass percent composition. Collect Needed Relationships: 1 mole C = 1.01 g C 1 mole H = g H 1 mole O = g O Given: g C, 4.48 g H, g O CP: g C,H,O mol C,H,O mol ratio empirical formula Example: formula of aspirin with the given mass percent composition. Given: g C, 4.48 g H, g O CP: g C,H,O mol C,H,O mol ratio empirical formula Rel: 1 mol C = 1.01 g; 1 mol H = g; 1 mol O = g Apply the Concept Plan: calculate the moles of each element 1mol C g C mol C 1.01g C 1mol H 4.48 g H 4.44 mol H g H 1mol O g O.0 mol O g O Example: formula of aspirin with the given mass percent composition. Apply the Concept Plan: write a pseudoformula Given: mol C, 4.44 mol H,.0 mol O CP: g C,H,O mol C,H,O mol ratio empirical formula Rel: 1 mol C = 1.01 g; 1 mol H = g; 1 mol O = g C H 4.44 O.0 35

16 Example: formula of aspirin with the given mass percent composition. Given: C H 4.44 O.0 CP: g C,H,O mol C,H,O mol ratio empirical formula Rel: 1 mol C = 1.01 g; 1 mol H = g; 1 mol O = g Apply the Concept Plan: find the mole ratio by dividing by the smallest number of moles C H O C H O Example: formula of aspirin with the given mass percent composition. Given: C.5 H O 1 CP: g C,H,O mol C,H,O mol ratio empirical formula Rel: 1 mol C = 1.01 g; 1 mol H = g; 1 mol O = g Apply the Concept Plan: multiply subscripts by factor to give whole number {C.5 H O 1 } x 4 C 9 H 8 O 4 Practice Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) 36

17 Practice Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) Given: 75.7% Sn, ( ) = 4.3% F in 100 g stannous fluoride there are 75.7 g Sn and 4.3 g F Find: Sn x F y Rel: 1 mol Sn = g; 1 mol F = g Concept Plan: g Sn mol Sn pseudoformula g F mol F mole ratio whole number ratio empirical formula Practice Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) Apply the Concept Plan: 1mol Sn g Sn mol Sn g 1mol F 43g 4.3 F 18mol 1.8 F g Sn F 1.8 Sn 0.638F1.8 Sn1F SnF Practice Determine the empirical formula of hematite, which contains 7.4% Fe (55.85) and the rest oxygen (16.00) 37

18 Practice Determine the empirical formula of hematite, which contains 7.4% Fe (55.85) and the rest oxygen (16.00) Given: 7.4% Fe, ( ) = 7.6% O in 100 g hematite there are 7.4 g Fe and 7.6 g O Find: Fe x O y Rel: 1 mol Fe = g; 1 mol O = g Concept Plan: g Fe mol Fe pseudoformula g O mol O mole ratio whole number ratio empirical formula Practice Determine the empirical formula of hematite, which contains 7.4% Fe (55.85) and the rest oxygen (16.00) Apply the Concept Plan: 1mol Fe 7. 4 g Fe 1.30 mol Fe g 1mol O g O mol O g Fe 1.30 O 1.73 Fe O Fe O O Fe3O Fe 4 Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound Molar Massmolecular formula multiplyin g factor, n Empirical Formula Mass 38

19 Example 3.18 Find the molecular formula of butanedione emp. form. = C H 3 O; MM = g/mol Given: Find: molecular formula Concept Plan: Molecular Form. Emp. Form. n and Molar Mass n Relationships: Emp. Form. Molar Mass Solution: Molar Mass Emp. Form. (1.01g/mol) 3(1.008 g/mol) 1(16.00 g/mol) g/mol g/mol g/mol n Molecular Formula CH3O C4H6O Check: the molar mass of the calculated formula is in agreement with the given molar mass Practice Benzopyrene has a molar mass of 5 g/mol and an empirical formula of C 5 H 3. What is its molecular formula? (C = 1.01, H=1.01) Practice Benzopyrene has a molar mass of 5 g and an empirical formula of C 5 H 3. What is its molecular formula? (C = 1.01, H=1.01) C 5 = 5(1.01 g) = g H 3 = 3(1.01 g) = 3.03 g C 5 H 3 = g n 5 g/mol g/mol 4 Molecular Formula = {C 5 H 3 } x 4 = C 0 H 1 39

Combustion Analysis a common technique for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made generally used for organic compounds containing C, H, O by

20 Combustion Analysis a common technique for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made generally used for organic compounds containing C, H, O by knowing the mass of the product and composition of constituent element in the product, the original amount of constituent element can be determined all the original C forms CO, the original H forms H O, the original mass of O is found by subtraction once the masses of all the constituent elements in the original compound have been determined, the empirical formula can be found Combustion Analysis Example 3.0 Combustion of a g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO =.445 g H O = g Determine the empirical formula of the compound 40

21 Example 3.0: formula of compound with the given amounts of combustion products Write down the given quantity and its units. Given: compound = g CO =.445 g H O = g Example 3.0: formula of compound with the given amounts of combustion products Given: g compound,.445 g CO, g H Write down the quantity to find and/or its units. Find: empirical formula, C xh yo z Example 3.0: formula of compound with the given amounts of combustion products Given: g compound,.445 g CO, g H Write a Concept Plan: g mol mol g g mol CO, H O CO, H O C, H C, H O O mol C, H, O mol ratio empirical formula 41

22 Example 3.0: formula of compound with the given amounts of combustion products Given: g compound,.445 g CO, g H O CP: g CO & H O mol CO & H O mol C & H g C & H g O mol O mol ratio empirical formula Collect Needed Relationships: 1moleCO = g CO 1 mole H O = 18.0 g H O 1 mole C = 1.01 g C 1 mole H = g H 1 mole O = g O 1 mole CO = 1 mole C 1 mole H O = mole H Example 3.0: formula of compound with the given amounts of combustion products Apply the Concept Plan: calculate the moles of C and H 1mol CO 1mol C.445 g CO 44.01g CO 1mol CO Given: g compound.445 g CO, g H O CP: g CO & H O mol CO & H O mol C & H g C & H g O mol O mol ratio empirical formula Rel: MM of CO, H O, C, H, O; mol element : 1 mol compound 1mol HO mol H g H O 18.0 g H O 1mol H O mol C mol H Example 3.0: formula of compound with the given amounts of combustion products Given: g compound,.445 g CO, g H O, mol C, mol H CP: g CO & H O mol CO & H O mol C & H g C & H g O mol O mol ratio emp. formula Rel: MM of CO, H O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: calculate the grams of C and H 1.01g mol C g C 1mol C g mol H g H 1mol H 4

23 Example 3.0: formula of compound with the given amounts of combustion products Given: g compound,.445 g CO, g H O, mol C, g C, mol H, g H, CP: g CO & H O mol CO & H O mol C & H g C & H g O mol O mol ratio emp. formula Rel: MM of CO, H O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: calculate the grams and moles of O g compound - ( g C g H) g O 1mol O g O mol O g Example 3.0: formula of compound with the given amounts of combustion products Apply the Concept Plan: write a pseudoformula Given: g compound,.445 g CO, g H O, mol C, g C, mol H, g H, g O, mol O CP: g CO & H O mol CO & H O mol C & H g C & H g O mol O mol ratio emp. formula Rel: MM of CO, H O, C, H, O; mol element : 1 mol compound C H O Example 3.0: formula of compound with the given amounts of combustion products Given: g compound,.445 g CO, g H O, mol C, g C, mol H, g H, g O, mol O CP: g CO & H O mol CO & H O mol C & H g C & H g O mol O mol ratio emp. formula Rel: MM of CO, H O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: find the mole ratio by dividing by the smallest number of moles C H O C10H1O1 43

24 Example 3.0: formula of compound with the given amounts of combustion products Given: g compound,.445 g CO, g H O, mol C, g C, mol H, g H, g O, mol O CP: g CO & H O mol CO & H O mol C & H g C & H g O mol O mol ratio emp. formula Rel: MM of CO, H O, C, H, O; mol element : 1 mol compound Apply the Concept Plan: multiply subscripts by factor to give whole number, if necessary write the empirical formula C 10H1O1 The smell of dirty gym socks is caused by the compound caproic acid. Combustion of g of caproic acid produced g of H O and 1.9 g of CO. If the molar mass of caproic acid is 116. g/mol, what is the molecular formula of caproic acid? (MM C = 1.01, H = 1.008, O = 16.00) Combustion of g of caproic acid produced g of H O and 1.9 g of CO. If the molar mass of caproic acid is 116. g/mol, what is the molecular formula of caproic acid? 1mol CO 1mol C 1.9 g CO mol C 44.01g CO 1mol CO 1mol H O mol H g HO mol H 18.0 g HO 1mol HO 1.01g mol C 0.54 g C 1mol C g mol H g H 1mol H 44

25 g moles C H O g compound - (0.54 g C g H) 0.3 g O 1mol O 0.3 g O mol O g C H O C H C H O 3 6 O C H O C g g 3 H g g 6 O g g g 116. g/mol n g/mol Molecular Formula = {C 3 H 6 O} x = C 6 H 1 O Organic Structure & Nomenclature Next 45

Chapter 6 Empirical and Molecular Formulas

Chapter 6 Empirical and Molecular Formulas EMPIRICAL FORMULA A chemical formula that indicates the relative proportions of the elements in a molecule rather than the actual number of atoms of the elements.