1. Stoichiometry Relationships between reactants and products in a chemical reaction

2. General definitions  Element- a pure substance which cannot be broken down by any ordinary chemical means.  Molecule- two or more atoms chemically bonded together. Example O 2, H 2 O  Compound- two or more elements chemically bonded together. Physical and chemical properties change.  Solution: homogeneous, containing 2 or more substances. No reaction to make

3.  Mixtures: Heterogeneous, The substances are not evenly distributed. Ex. Flour in water,  Physical Change: There is no change in the chemical formula of the substance. Ex. H 2 O (s) H 2 O (l) have same formulas, chemical and some physical properties. Color, texture, density, taste, rate of evaporation, boiling and melting points Color, texture, density, taste, rate of evaporation, boiling and melting points

4.  Chemical Change: In a chemical change there is a change in the arrangement of the atoms, that is, the chemical formula changes.  Chemical properties: flammability, temp. at which something ignites, oxidation (rusting), acidity, pH, reducing ability, etc.  Some indications of chemical changes: heat produced, change in color or smell, gas or flames produced, precipitation.

5. Classification of Matter Matter Pure Substances ElementsCompounds Mixtures Homogeneous Mixtures Solutions Heterogeneous Mixtures

6. The Mole  What is it? A unit of measurement which equals 6.0 x 10 23 of something A unit of measurement which equals 6.0 x 10 23 of something Similar to 1 dozen = 12 of something Similar to 1 dozen = 12 of something

7.  Why do we use it? Other conventional quantities such as “one dozen” are too small a number to represent atoms Other conventional quantities such as “one dozen” are too small a number to represent atoms Gives us a way to “count” atoms Gives us a way to “count” atoms ex. 1g H = 1mole of H = 6.0 x 10 23 atoms

8.  How the mole was developed. The mass of each type of atom was determined using a mass spectrometer. The mass of each type of atom was determined using a mass spectrometer. ex. H = 1.66 x 10 -24 g/atom The masses of the atoms were compared and ratios were developed. The masses of the atoms were compared and ratios were developed. ex. He: 6.64 x 10 -24 = 4.0 (amu)  relative mass ratio H 1.66 x 10 -24 H 1.66 x 10 -24

9. Mass Spectrometer

10. Mass spectrometer

11. The Mole cont’d. The ratios were used to make standard masses. The ratios were used to make standard masses. Made “H” have a mass of 1g. Other elements masses are determined by multiplying 1g by the ratio.Made “H” have a mass of 1g. Other elements masses are determined by multiplying 1g by the ratio. ex. 1.0 g H x 4.0 for He (since He is 4 times heavier) x 4.0 for He (since He is 4 times heavier) 4.0g for He 4.0g for He These masses represent one mole of the element and are A.M.U. or atomic mass units. These masses represent one mole of the element and are A.M.U. or atomic mass units.

12.  6.02 x 10 23 was determined by: Atomic mass = # of atoms in mass Mass of one atom ex. 1.00g H = 6.02 x 10 23 atoms 1.66 x 10 -24 g/atom 1.66 x 10 -24 g/atom

13. Amedeo Avogadro

16. Mole conversions

18. Chemical Formulas Molecular FormulaEmpirical Formula C 6 H 12 O 6 CH 2 O -Elements are written in their # of atoms found in nature. -This is often not the lowest ratio. -Hard to determine in the lab. -Elements are written in the lowest ratio. -Easy to experimentally determine.

19. Empirical Formula  Ex. You are heating a 0.2636g sample of Ni and reacting with oxygen. After reacting the nickel oxide weighed 0.3354g. Find the empirical formula. Find the mass of each element Find the mass of each element.3354g Ni ox..3354g Ni ox. -.2636g Ni  masses 0.0718g O  mass Convert to moles. Convert to moles..2636g Ni x 1mol Ni =.004491mol Ni 58.69g 58.69g.0718g O x 1mol O =.004490mol O 16.00g 16.00g

20. Empirical Formula cont’d. Make ratio – use smallest mole value on the bottom. Make ratio – use smallest mole value on the bottom..004491mol Ni = 1mol Ni.004490mol O 1mol O Use the ratio to determine the formula. Use the ratio to determine the formula.NiO

21. Ex. An Aluminum oxide compound was found to contain 4.151g Al and 3.692g O. What is Aluminum oxides empirical formula? Determine masses.Determine masses. 4.151g Al, 3.692g O 4.151g Al x 1mol Al = 0.1539mol Al4.151g Al x 1mol Al = 0.1539mol Al 26.98g 26.98g 3.692g O x 1mol O = 0.2308mol O 1 16.00 1 16.00

22. Ratio Ratio 0.2308mol O = 1.5mol O 0.1539mol Al 1mol Al Multiply the ratio by the smallest integer in order to get whole #s. Multiply the ratio by the smallest integer in order to get whole #s. 2 x (1.5) = 3mol O 1 2mol Al 1 2mol Al Formula. Formula. Al 2 O 3

23. Empirical Formula cont’d. Ex. A Platinum compound was found to contain by mass: 65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl. Determine its empirical formula.  Find the masses. 65.02g Pt, 9.34g N, …  65.02g Pt x 1mol Pt = 0.3333m Pt 195.1g 195.1g 9.34g N x 1mol N = 0.667m N 14.0 14.0 2.02g H x 1mol H = 2.00m H 1.00 1.00

24. continued  Ratios N : 0.667m = 2 NH = 2.00m H = 6H Pt 0.333m 1 PtPt 0.3333m Pt 1 Cl = 0.6667mol C = 2.0 Cl Pt 0.3333mol Pt 1 Pt  Formula? PtN 2 H 6 Cl 2

25. Percent Mass  Calculating the % by mass of an element in a molecular formula/empirical formula.  Equation: Mass of element (in 1mole) x 100% = Molecular mass Molecular mass  Find the % mass of Chromium in the molecular formula K 2 Cr 2 O 7. Find molecular mass: Find molecular mass: K 2 Cr 2 O 7 = 294.2g Find mass of element in compound. Find mass of element in compound. Cr 2 = 104g Substitute. Substitute. Mass of element x 100% Molecular mass 104g Cr x 100 = 35.4% Cr 294.2g  % mass can be useful for determining the empirical formula of a compound.

26.  Definitions ex. 2H 2(g) + O 2(g)  2H 2 O (g) everything on the left of the arrow is a reactant, everything on the right of the arrow is a product. states of matter (g) = gas (l) = liquid (s) = solid (aq) = aqueous (dissolved in H 2 O) Chemical Equations

27.   symbol can be interpreted as makes, produces, yields, decomposes materials above and below the arrow: MnO 2 MnO 2 ex. H 2 O 2  H 2 O + O 2 chemicals = catalysts e- = electricity Δ = heat

28. Types of Chemical Reactions  Single Displacement Cu (s) + 2AgNO 3(aq)  2Ag (s) + Cu(NO 3 ) 2(aq) Cu (s) + 2AgNO 3(aq)  2Ag (s) + Cu(NO 3 ) 2(aq)  Double Displacement CoCl 2(aq) + 2NaOH (aq)  Co(OH) 2(s) + 2NaCl (aq)  Decomposition 2H 2 O 2  2H 2 O + O 2 (NH 4 ) 2 Cr 2 O 7(s)  N 2(g) + Cr 2 O 3(s) + 4H 2 O (g)  Redox (Oxidation-reduction) Fe 2 O 3 + 3CO  2Fe + 3CO 2  Synthesis 2H 2(g) + O 2(g)  2H 2 O (g)

29. Law of Conservation of Mass  This law states that matter is neither created or destroyed in a chemical reaction. The mass used in the beginning of a reaction equals the mass made at the end of a reaction.  Balancing: This is a process by which the written reaction is modified so the number of atoms on the reactant side equals the number of atoms on the product side.

30. Antoine Lavoisier

31. Percent Yield  Theoretical and actual yield Theoretical yield – maximum amount of product that can be produced. Determined by math. Theoretical yield – maximum amount of product that can be produced. Determined by math. Actual yield – the “actual” amount of product produced. Actual yield – the “actual” amount of product produced.  Percent yield Actual yield x 100 = % yield Actual yield x 100 = % yield Theoretical yield  Problems with % yield. (Either the % yield or the actual yield must be given in the problem.)