Chapter 3. Molecules, Compounds, and Chemical Composition
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1 Chapter 3 Molecules, Compounds, and Chemical Composition
2 Elements and Compounds Elements combine together to make an almost limitless number of compounds. The properties of the compound are totally different from the constituent elements.
3
4 Chemical Bonds Compounds are made of atoms held together by bonds. Chemical bonds are forces of attraction between atoms. The bonding attraction comes from attractions between protons and electrons.
5 Bond Types Two general types of bonding between atoms found in compounds, ionic and covalent. Ionic bonds result when electrons have been transferred between atoms, resulting in oppositely charged ions that attract each other. Generally found when metal atoms bond to nonmetal atoms Covalent bonds result when two atoms share some of their electrons. Generally found when nonmetal atoms bond together
6 Formation of an Ionic Compound Na 0 Na + Cl - Charged Ions } Cl 0 Neutral Atoms Undergo Electron Transfer An Orderly Aggregate Called an Ionic Crystal
7 Formation of an Covalent Compound
8 Formation of an Covalent Compound
9 Representative Covalent Compounds
10 Chemical Formulas Describe Compounds A compound is a distinct substance that is composed of atoms of two or more elements. We describe the compound by describing the number and type of each atom in the simplest unit of the compound. Each element is represented by its letter symbol. The number of atoms of each element is written to the right of the element as a subscript. Polyatomic ions are placed in parentheses. (if more than one is present)
11 Types of Formula: Empirical Formula An empirical formula gives the relative number of atoms of each element in a compound. It does not describe how many atoms, the order of attachment, or the shape. For example: 1) The empirical formula for the ionic compound fluorspar is CaCl2. This means that there is 1 Ca 2+ ion for every 2 Cl ions in the compound. 2) The empirical formula for the molecular compound oxalic acid is CHO2. This means that there is 1 C atom and 1 H atom for every 2 O atoms in the molecule.
12 Types of Formula: Molecular Formula A molecular formula gives the actual number of atoms of each element in a molecule of a compound. It does not describe the order of attachment, or the shape. The empirical formula for the molecular compound oxalic acid is CHO2. The actual molecular formula is C2H2O4
13 Types of Formula: Structural Formula A structural formula uses lines to represent covalent bonds and shows how atoms in a molecule are connected or bonded to each other. Structural Formulas of Oxalic Acid O O H O C C O H
14 Molecular Models
15 Practice Find the empirical formula for each of the following The ionic compound that has two aluminum ions for every three oxide ions Al2O3 arabinose, C5H10O5 CH2O pyrimidine H C H C N C2H2N H C N C H ethylene glycol CH3O
16 Classifying Elements & Compounds Atomic elements elements whose particles are single atoms Molecular elements elements whose particles are multi-atom molecules Molecular compounds compounds whose particles are molecules Ionic compounds compounds whose particles are cations and anions
17 Molecular View of Elements and Compounds
18 Elements MOST ELEMENTS Single atoms are the constituent particles. The atoms may be physically attracted to each other, but are not chemically bonded together. A FEW ELEMENTS Molecules are the constituent particles. The molecules are made of two or more atoms chemically bonded together by covalent bonds.
19 Molecular Elements H 2 7 7A N 2 O 2 F 2 Cl2 Br 2 I 2 20
20
21 Compounds SOME COMPOUNDS Composed of ions arranged in a 3-dimensional pattern These are called ionic compounds. OTHER COMPOUNDS Composed of individual molecule units Each molecule contains atoms of different elements chemically attached by covalent bonds These are called molecular compounds.
22 Ionic vs. Molecular Compounds Propane contains individual C3H8 molecules Table salt contains an array of Na + ions and Cl - ions
23 Classify Each of the Following as Either an Atomic Element, Molecular Element, Molecular Compound, or Ionic Compound Aluminum, Al Aluminum chloride, AlCl3 Chlorine, Cl2 Acetone, C3H6O Carbon monoxide, CO Cobalt, Co atomic element ionic compound molecular element molecular compound atomic element molecular compound
24 Formula Units vs. Molecules
25 Compound must have no total charge, therefore we must balance the numbers of cations and anions in a compound to get 0 charge. Practice What are the formulas for compounds made from the following ions? Potassium ion with a nitride ion K + with N 3 K3N potassium nitride Calcium ion with a bromide ion Ca 2+ with Br CaBr2 calcium bromide Aluminum ion with a sulfide ion Al 3+ with S 2 Al2S3 aluminum sulfide
26 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 mass of 1 molecule of H2O = 2(1.01 amu H) amu O = amu mass of 1 formula unit of MgCl2 = 2(35.45 amu Cl) amu Mg = amu
27 Molar Mass
28 Molar Mass of Compounds The relative masses of molecules can be calculated from atomic masses. Formula Mass = 1 molecule of H2O = 2(1.01 amu H) amu O = amu 1 mole of H2O contains 2 moles of H and 1 mole of O. molar mass = 1 mole H2O = 2(1.01 g H) g O = g so the Molar Mass of H2O is g/mole
29 Practice How many moles are in 50.0 g of PbO2? (Pb = 207.2, O = g PbO 2 mol PbO 2 Pb = 1 x = O = 2 x = PbO2 = g/mol
30 Example: Find the number of CO2 molecules in 10.8 g of dry ice g CO 2 mol CO 2 molec CO 2 1 mol CO2 = g, 1 mol = x 10 23
31 Practice How many formula units are in 50.0 g of PbO2? (PbO2 = 239.2) g PbO 2 mol PbO 2 units PbO 2 1 mol PbO2 = g,1 mol = x 10 23
32 Practice What is the mass of 4.78 x NO2 molecules? molecules mol NO 2 g NO 2 1 mol NO2 = g, 1 mol = x 10 23
33 Percent Composition
34 Percent Composition Percentage of each element in a compound by mass Can be determined from 1. the formula of the compound 2. the experimental mass analysis of the compound
35 Find the mass percent of Cl in C2Cl4F2
36 Practice Determine the mass percent composition of the following CaCl2
37 Mass Percent as a Conversion Factor If NaCl is 39% sodium, find the mass of table salt containing 2.4 g of Na. g Na g NaCl
38 Find the mass of sodium in 6.2 g of NaCl g NaCl mol NaCl mol Na g Na g NaCl
39 Empirical Formulas
40 Empirical Formula Simplest, whole-number ratio of the atoms of elements in a compound Can be determined from elemental analysis
41 Finding an Empirical Formula from % Composition 1. Convert the percentages to grams 2. Convert grams to moles 3. Write a pseudoformula using moles as subscripts 4. Divide all by smallest number of moles 5. Multiply all mole ratios by number to make all whole numbers
42 Example: Find the empirical formula of aspirin with the given mass percent composition 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"
43 g C g H g O mol C mol H mol O pseudoformula CxHyOz empirical formula CxHyOz Manipulate subscripts to obtain whole-number ratio
44 Calculate the moles of each element Write a pseudoformula C4.996H4.44O2.220
45 C4.996H4.44O2.220 Find the mole ratio C2.25H2.00O1.00 Multiply subscripts by factor to give whole number C9H8O4 (x 4)
46 Practice Determine the empirical formula of stannous fluoride, which contains 75.7% Sn ( g/mol) and the rest fluorine (19.00 g/mol) Given: 75.7% Sn, ( ) = 24.3% F g Sn g F mol Sn mol F pseudoformula mole ratio whole number ratio empirical formula
47 Practice Determine the empirical formula of stannous fluoride, which contains 75.7% Sn ( g/mol) and the rest fluorine (19.00 g/mol) Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles Sn 75.7g X 1mol 118.7g F 24.3g X 1mol 19.00g SnF2
48 Practice Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Given: 72.4% Fe, ( ) = 27.6% O g Fe g O mol Fe mol O pseudoformula mole ratio whole number ratio empirical formula
49 Practice Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Element Ratio in Grams Molar Mass Ratio in Moles Ratio in Moles Ratio in Moles Fe 72.4g X 1mol 55.85g O 27.6g X 1mol 16.00g Fe3O4
50 Molecular Formulas
51 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.
52 Find the molecular formula of butanedione if its empirical formula is C2H3O and its molar mass (MM) is g/mol. Factor of 2
53 Practice Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01) C5 =" 5(12.01 g) = g H3 =" 3(1.01 g) = 3.03 g C5H3 " = g? 252 Molecular formula = {C5H3} x 4 = C20H12
54 Combustion Analysis
55 Combustion Analysis (Generally used for organic compounds containing C, H, O) A known mass of compound is burned in oxygen and the masses of the products formed (CO2 and H2O) are determined. By knowing the masses of the products and composition of constituent elements in the product, the original amount of constituent elements can be determined. It is assumed that all of the carbon in the original sample is converted to carbon dioxide and all of the hydrogen in the sample is converted to water.
56 Combustion Analysis
57 Example of Combustion Analysis Combustion of a g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO2 = g H2O = g This came from C. This came from H.! Determine the empirical formula of the compound. g CO 2, H 2 O mol CO 2, H 2 O mol C, H g C, H g O mol O mol C, H, O pseudo formula mol ratio empirical formula
58 1 mole CO2 = g CO2 1 mole H2O = g H2O 1 mole C = g C 1 mole H = g H 1 mole O = g O 1 mole CO2 = 1 mole C 1 mole H2O = 2 mole H In the original sample
59 In the original sample In the original sample
60 Pseudo formula C H O Empirical formula
61 Combustion of g of caproic acid produced g of H2O and 1.92 g of CO2. If the molar mass of caproic acid is g/mol, what is the molecular formula of caproic acid?
62 C H O g moles
63 Molecular formula = {C3H6O} x 2 = C6H12O2
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