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Chemical Reactions, Chemical Equations, and Stoichiometry

Chemical Reactions, Chemical Equations, and Stoichiometry. Brown, LeMay Ch 3 AP Chemistry. 3.2: Types of reactions. http://web.fuhsd.org/kavita_gupta/July.html. 3.3: Atomic, Molecular & Formula Weights. Atomic mass units: 1 amu = 1.66054 x 10 -24 g or 1 g = 6.02214 x 10 23 amu

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Chemical Reactions, Chemical Equations, and Stoichiometry

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  1. Chemical Reactions, Chemical Equations, and Stoichiometry Brown, LeMay Ch 3 AP Chemistry

  2. 3.2: Types of reactions http://web.fuhsd.org/kavita_gupta/July.html

  3. 3.3: Atomic, Molecular & Formula Weights Atomic mass units: • 1 amu = 1.66054 x 10-24 g or 1 g = 6.02214 x 1023 amu • 1 atom of 12C isotope defined as weighing exactly 12 amu Average atomic mass(or atomic weight): • 12C: 98.892% x 12 amu = 13C: 1.108% x 13.00335 amu = 11.867 + 0.1441 AW = 12.011 amu….... Therefore 12.011 g C = 1 mol C

  4. Molecular mass or weight (MM or MW): sum of atomic masses of each atom in the chemical formula. Ex: H2SO4 MW = (2)(1.0079) + 32.06 + (4)(16.00) = 98.08 amu • Formula weight (FW):same as MW, except for ionic substances in which no “molecule” exists. Then, FW is the simplest integer ratio of moles of each element (ions) present. Ex: NaCl is a 3D array of ions FW = 22.99 + 35.453 = 58.44 amu l

  5. 3.4 – 3.7: Mass Relationships • Percent Composition: Ex: What is the percentage of oxygen in phosphoric acid? Ex: What is the percent of O in CuSO4.5H2O? 57.7%

  6. The Mole and Molar Mass (MM) • From Latin moles: “a mass” • Avogadro’s Number = 6.022 x 1023 atoms in exactly 12.000 g of 12C = 1 mol • Moles important because they can be used for ratios. Mass can not be used for ratios. Ex. CO2 etc. • Converting: grams → moles → molecules Ex: How many molecules of H2O in 100.0 g H2O? =3.343 x 1024 molecules H2O

  7. Ex: What is the empirical formula of a compound that is composed of 80.% Carbon and 20.% Hydrogen? If the molar mass is found to be 30 g/mol, what is the molecular formula? Strategy: Assume 100. g of unknown C:H mole ratio is 6.7:20. ≈1:3. So, empirical formula is CH3, and the molecular formula is C2H6. Percent Composition Tutorial

  8. Practice Problem 1: One of the most commonly used white pigments in the paint is a compound of Titanium and Oxygen that contains 59.9% Ti by mass. Determine the empirical formula of this compound. Ans; TiO2

  9. Practice problem2:Benzene contains only C and H and is 7.74% H by mass; the molar mass of benzene is 184 g/mol. What are the empirical and molecular formula of this compound?Ans: CH, C14H14

  10. EF from Combustion Data: To find out the EF of a compound, the compound is burned in the air. The equation for combustion of these compounds is all follows.For Hydrocarbons: CxHy + O2 CO2 + H2OFor Compounds Containing CHN: CxHyNz+ O2 CO2 + H2O+ NO2For Compounds containing C, H and OCxHyOz + O2 CO2 + H2O In such cases to figure out mass of O,Mass of O = total mass of compound – (mass of C + mass of H) Animation: Must See!!

  11. Important Fact to remember for Combustion: • Complete combustion produces CO2 as opposed to incomplete combustion which produces CO.

  12. Practice Problem on EF from Combustion Data: Many homes in rural America are heated by propane gas, a compound that contains only carbon and hydrogen. Complete combustion of a sample of propane produced 2.641 g of carbon dioxide and 1.442 g of water as the only products. Find the empirical formula of propane. Answer: C3H8

  13. Practice Problem Level 2: EF from Combustion Data Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of 47.6 mg cumene produces some CO2 and 42.8 mg water. The molar mass of cumene produces between 115 and 125 g/mol. Determine the empirical and molecular formulas. Answer: C3H4, C9H12

  14. Practice Problem 3: A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this compound? Ans: CH2O

  15. Limiting Reagent (or reactant): Ex: If 6.0 g hydrogen gas reacts with 40.0 g oxygen gas, what mass of water will be produced? 2 H2 (g) + O2 (g) → 2 H2O (g) O2 is limiting reagent, H2 is excess. Limiting Reactant Movie

  16. Percent yield & Percent error: Ex: If in the previous example, only 40.0 g water were formed, what is the percent yield and percent error?

  17. Determining Formula of a Hydrate: To determine the formula of a hydrate, a certain mass of hydrate is heated to drive off the water. Then the mass of water driven-off is calculated. Now the mole ratio of water to the compound is calculated. For more information visit the following website: Movie on Determining formula of a Hydrate • Questions to ponder over: • What might be a good procedure to calculate percent of water in a hydrate? • How do you identify hydrates? Do you remember how to name hydrates?

  18. Practice Problem: A student is given three unknowns labeled A, B, and C. The student is told they are Na2CO3.H2O; Na2CO3.7H2O; and Na2CO3.10H2O, not necessarily in that order.a. When 2.000 gram of A was heated, 0.914 gram of anhydrous residue remained.Formula for sample A. Ans: Na2CO3.7H2O   Sodium carbonate heptahydrate

  19. Practice Problem 2 b. When 4.000 grams of B was heated, 1.48 gram of anhydrous residue remained.Formula for sample B.Chemical name for sample B.Ans: Na2CO3.10H2O   Sodium carbonate decahydrate c. When 1.000 gram of C was heated, 0.855 gram of anhydrous residue remained.Formula for sample C.Chemical name for sample C. Ans: Na2CO3.H2O   Sodium carbonate monohydrate

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