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Chemistry 221 Chapter 2 Atoms and Molecules. CLASSIFICATION OF MATTER: MATTER: anything that has mass & takes up space. ELEMENT = simplest substance cannot be decomposed into simpler substances. Read Interchapter A on elemental etymology. . CLASSIFICATION OF MATTER:
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Chemistry 221 Chapter 2 Atoms and Molecules
CLASSIFICATION OF MATTER: MATTER: anything that has mass & takes up space. ELEMENT = simplest substance cannot be decomposed into simpler substances. Read Interchapter A on elemental etymology.
CLASSIFICATION OF MATTER: Element symbols e.g. in Table 2.3 to 2.5.
CLASSIFICATION OF MATTER: COMPOUND = a pure substance that can be broken down into two or more elements (e.g. sodium chloride can be decomposed to make Na metal and Cl2gas)
Mixture= substance that can be physically separated into different parts Components are not combined chemically More than one chem formula: e.g.
Fig. 2.6 thru 2.10: physmethods of separating mixtures
Law of Conservation of Mass “In a chemrxtn, matter is neither created nor destroyed” total mass of reactants = Antoine Lavoisier 1743-1794 7
Rxtn of Sodium with Chlorine to Make NaCl • mass of Na & Cl determined by # of atoms that combine • only whole atoms combine & atoms not D’d or destroyed • mass of NaCl made must = total mass of Na & Cl atoms that combine 7.7 g Na + 11.9 g Cl2 19.6 g NaCl 8
Law of Definite Proportions(Law of Constant Composition) Joseph Proust 1754-1826 All samples of a given compound, have same proportions of their elements (regardless of source of compound or how compound was prepared). 9
Proportions in Sodium Chloride (NaCl) A 100.0 g of NaClcontains 39.3 g of Na & 60.7 g of Cl A 200.0 g of NaCl contains 78.6 g of Na & 121.4 g of Cl A 58.44 g of NaCl contains 22.99 g of Na & 35.44 g of Cl 10
Try this: If a 10.0 g sample of calcite contains 4.0 g of calcium, how much calcite contains 0.24 g of calcium? 11
How much calcite contains 0.24 g of calcium? Given: Find: Sample 1: 4.0 g Ca and 10.0 g calcite Sample 2: 0.24 g Ca mass calcite in Sample 2, g Conceptual Plan: Relationships: 12
How much calcite contains 0.24 g of calcium? Solution: Sig Figs & Round: 13
Intensive Property: Mass Percent Mass %of an element in a compound = (mass of element in compound) x100 (mass of compound)
Law of Multiple Proportions John Dalton 1766-1844 When two elements (A and B) form two different compounds, the masses of B that combine with 1 g of A can be expressed as a ratio of small, whole numbers 16
Oxides of Carbon C combines with O to form 2 different compounds: carbon monoxide = CO & carbon dioxide = CO2 CO contains 1.33 g of O for every 1.00 g of C CO2 contains 2.67 g of O for every 1.00 g of C 17
Oxides of Carbon There are 2x as many O atoms per C atom in CO2 as in CO, the oxygen mass ratio should be Mass O comb. w/ 1g of C in CO2 = Mass O comb. w/ 1g of C in CO 18
two oxides of nitrogen (NO2 & N2O): consistent with the Law of Multiple Proportions? Given: Find: nitrogen dioxide: 2.28 g O per 1 g N dinitrogen monoxide: 0.570 g O per 1 g N O in nitrogen dioxide:O in dinitrogen monoxide Conceptual Plan: Relationships: 19
two oxides of nitrogen (NO2 & N2O): consistent with the Law of Multiple Proportions? Solution: because compounds have O:O ratio = a small whole #, results are consistent with Law of Multiple Proportions. Check: 20
Practice: Hematite contains 2.327 g of Fe for every 1.00 g of oxygen. Wüsite contains 3.490 g of Fe per 1.00 g of oxygen. Show these results are consistent with the Law of Multiple Proportions. 21
g Fein wüsite g Fein hematite Fe1:Fe2 Given: Find: hematite: 2.327 g Fe per 1 g O wüsite: 3.490 g Fe per 1 g O Fe in wüsite:Fe in hematite Conceptual Plan: Relationships: samples of diff compounds that have same elements show proportions by mass that are small whole # ratios Solution: Check: because compounds have Fe:Fe ratio = a small whole #, results are consistent with Law of Multiple Proportions 22
Dalton’s Atomic Theory Dalton proposed a theory of matter based on indivisible particles to explain the conservation of matter and the proportion laws 23
Dalton’s Atomic Theory • Each element is composed of tiny, indestructible particles called ________ 2. Atoms of the same element are 3. Atoms of different elements are 24
Dalton’s Atomic Theory • Atoms combine in simple, __________________ to form molecules or compounds. • In a chemrxtn, atoms of one element cannot D into atoms of another element, 25
Evidence for Dalton’s Theory: Law of Conservation of Mass – works if matter is composed of small, indestructible particles rearranging in chemical reactions
Evidence for Dalton’s Theory: 2. Law of Constant Composition – consistent with idea of indestructible particles (atoms) combining in __________________ H2O is always 11.2% H &89.8% O, no matter where you collect your sample of water Different proportions yields a different compound
Evidence for Dalton’s Theory: 3. Law of Multiple Proportions – water &hydrogen peroxide: mass ratio of O:H in H2O is 8.02 : 1. mass ratio of O:H in H2O2is 16.0 : 1.
Evidence for Dalton’s Theory: 3. Law of Multiple Proportions – compare amount of O that combines with a fixed quantity of H in two H & O compounds = a ratio of small, whole #’s:
Copper atoms can combine with zinc atoms to make gold atoms 31
Water is composed of many identical molecules that have one O atom and two H atoms 32
Because mass ratio of Fe:O in wüsite is 1.5 x > the Fe:O ratio in hematite, there must be 1.5 Fe atoms in a unit of wüsite and 1 Fe atom in a unit of hematite 34
Dalton: atoms must combine in small whole # ratios. If you could combine fractions of atoms, that would mean the atom is breakable in normal chemical reactions, and Dalton’s first premise would be incorrect. You can get the Fe:Fe mass ratio to be 1.5 if: 35
Small whole # ratios supports idea that small, indivisible particles are being rearranged in chemrxtns = fundamental basis for all matter. Dalton’s Atomic Theory allowed a system of _______________________________ to be established.
relative atomic masses Calcium oxide contains 64.5 g Ca for every 25.7 g O. If the formula of calcium oxide is CaO, then (mass of a Ca atom) / (mass of a oxygen atom) = 64.5g / 25.7g = 2.51 so a Caatom is 2.51 xas heavy as an O atom.
relative atomic masses Relative atomic weights are unitless, but if we assign H (the lightest element) to be ___________________________ then the rest of the atomic weights can be expressed in ________as well.
relative atomic masses Molecular weight (MW) is calculated by adding up atomic weights of all atoms in amolecule. MW can be expressed in amu. Let’s try one on the board! Glucose: C6H12O6
Nomenclature of Compounds Periodic Table shows metals on the left and center Non-metals on the upper right Binary compounds are either: Two non-metals,or a non-metal + a metal.
Nomenclature of Compounds METAL-NONMETAL: name metal 1st, then name nonmetal (“-ide” suffix) e.g. calcium chloride NONMETAL-NONMETAL: name lower-leftelement 1st, then upper-right element
Nomenclature of Compounds # atoms using Greek prefixes (many combine to form more than 1compound)
Nomenclature of Compounds Hydrogenis unusual and unique: can act as a metal or nonmetal if it comes 1stin formula, treat it as a metal HCl, HF, H2S, etc. if it comes last in formula, treat it as a nonmetal KH, CaH2, etc
Nomenclature of Compounds • Memorize older, commonly used names for • Water: H2O • AMMONIA: NH3 • METHANE: CH4
Notes on Charge Two kinds : (+) and (–) Opposite charges attract e.g. (+) attracted to (–) Like charges repel e.g. (+) repels (+) and (–) repels (–) To be neutral = 47
Cathode Ray Tube • Glass tube containing metal electrodes • almost all air has been evacuated • Connected to high voltage power supply, glowing area is seen emanating from 48
My man, “J.J.” Thomson Believed cathode ray was composed of tiny particles with an elec charge Expmnt demonstrated there were particles by measuring amount of force needed to deflect their path 49
Thomson’s Experiment Investigate effect of placing an elecfield around tube 1. charged matter is attracted to an electric field 2. light’s path is not deflected by an electric field +++++++++++ Anode Cathode (+) (-) ------------- - + Power Supply 50