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Stoichiometry

Stoichiometry

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Stoichiometry

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  1. Stoichiometry

  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 O2 , H2O • 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. H2O (s) H2O (l) have same formulas, chemical and some physical properties. • 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

  6. The Mole • What is it? • A unit of measurement which equals 6.0 x 1023 of something • Similar to 1 dozen = 12 of something • Why do we use it? • Other conventional quantities such as “one dozen” are too small a number to represent atoms • Gives us a way to “count” atoms ex. 1g H = 1mole of H = 6.0 x 1023 atoms

  7. How the mole was developed. • The mass of each type of atom was determined using a mass spectrometer. ex. H = 1.66 x 10-24g/atom • 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

  8. The Mole cont’d. • 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. ex. 1.0 g H x 4.0 for He (since He is 4 times heavier) 4.0g for He • These masses represent one mole of the element and are A.M.U. or atomic mass units.

  9. 6.02 x 1023 was determined by: Standard mass = # of atoms in mass Mass of one atom ex. 1.00g H = 6.02 x 1023atoms 1.66 x 10-24g/atom

  10. Mass spectrometer

  11. Mass spectrometer

  12. Chemical Formulas Molecular Formula Empirical Formula -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. C6H12O6 CH2O

  13. 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 .3354g Ni ox. • .2636g Ni masses 0.0718g O  mass • Convert to moles. .2636g Ni x 1mol Ni = .00491mol Ni 58.69g .0718g O x 1mol O = .0049mol O 16.00g

  14. Empirical Formula cont’d. • Make ratio – use smallest mole value on the bottom. .004491mol Ni = 1mol Ni .00449mol O 1mol O • Use the ratio to determine the formula. NiO

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

  16. 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. 2 x (1.5) = 3mol O 1 2mol Al • Formula. Al2O3

  17. 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 9.34g N x 1mol N = 0.667m N 14.0 2.02g H x 1mol H = 2.00m H 1.00

  18. continued • Ratios N : 0.667m = 2 NH = 2.00m H = 6H Pt 0.333m 1 Pt Pt 0.3333m Pt 1 Cl = 0.6667mol C = 2.0 Cl Pt 0.3333mol Pt 1 Pt • Formula? PtN2H6Cl2

  19. 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 • Find the % mass of Chromium in the molecular formula K2Cr2O7. • Find molecular mass: K2Cr2O7 = 294.2g • Find mass of element in compound. Cr2 = 104g • 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.

  20. Chemical Equations • Definitions ex. 2H2(g) + O2(g) 2H2O(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 H2O)

  21.  symbol can be interpreted as makes, produces, yields, decomposes materials next to arrow: MnO2 ex. H2O2 H2O + O2 chemicals = catalysts e- = electricity Δ = heat

  22. Types of Chemical Reactions • Single Displacement Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq) • Double Displacement CoCl2(aq) + 2NaOH(aq) Co(OH)2(s) + 2NaCl(aq) • Decomposition 2H2O2 2H2O + O2 (NH4)2Cr2O7(s) N2(g) + Cr2O3(s) + 4H2O(g) • Redox (Oxidation-reduction) Fe2O3 + 3CO  2Fe + 3CO2 • Synthesis 2H2(g) + O2(g) 2H2O(g)

  23. 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.

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