1 / 26

Stoichiometry

Stoichiometry. 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

lana
Télécharger la présentation

Stoichiometry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.)

More Related