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

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

2 3.: Types of reactions

3 3.3: Atomic, Molecular & Formula Weights Atomic mass units: 1 amu = x 10-4 g or 1 g = x 10 3 amu 1 atom of 1 C isotope defined as weighing exactly 1 amu Average atomic mass (or atomic weight): 1 C: 98.89% x 1 amu = C: 1.108% x amu = AW = amu... Therefore g C = 1 mol C 3

4 Molecular mass or weight (MM or MW): sum of atomic masses of each atom in the chemical formula. Ex: H SO 4 MW = ()(1.0079) (4)(16.00) = 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 = = amu l 4

5 : Mass Relationships Percent Composition: Ex: What is the percentage of oxygen in phosphoric acid? 4( AW O) % O = 100 = MW H 3 PO 4 4(16.00) 3(1.0079) (16.00) 100 = 65.31% Ex: What is the percent of O in CuSO4.5HO? 57.7% 5

6 The Mole and Molar Mass (MM) From Latin moles: a mass Avogadro s Number = 6.0 x 10 3 atoms in exactly g of 1 C = 1 mol Moles important because they can be used for ratios. Mass can not be used for ratios. Ex. CO etc. Converting: grams moles molecules Ex: How many molecules of H O in g H O? g H 1 3 O 1mol H O molecules H 18.0 g HO 1mol HO molar mass Avogadro's number = = x 10 4 molecules H O O 6

7 Ex: What is the empirical formula of a compound that is composed of 80.% Carbon and 0.% Hydrogen? If the molar mass is found to be 30 g/mol, what is the molecular formula? Strategy: Assume 100. g of unknown 80.g C 1 0.g H 1 1mol C g C 1mol H g H 6.7 mol C:H mole ratio is 6.7:0. 1:3. So, empirical formula is CH 3, and the molecular formula is C H 6. Percent Composition Tutorial = C = 0. mol H 7

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; TiO 8

9 Practice problem: 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 9

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.

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 + Oà CO + HO For Compounds Containing CHN: CxHyNz+ Oà CO + HO+ NO For Compounds containing C, H and O CxHyOz + Oà CO + HO 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!! 10

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

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.641 g of carbon dioxide and 1.44 g of water as the only products. Find the empirical formula of propane. Answer: C3H8 1

13 Practice Problem Level : 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 CO and 4.8 mg water. The molar mass of cumene produces between 115 and 15 g/mol. Determine the empirical and molecular formulas. Answer: C3H4, C9H1 13

14 Practice Problem 3: A g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce g of CO and g of H O. What is the empirical formula of this compound? Ans: CH O 14

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? H (g) + O (g) H O (g) 6.0 g H 1mol H g O 1mol O = 1.5 mol g O O is limiting reagent, H is excess. 1.5 mol O 1 =.976 = 3.0 mol H.0158 g H O mol H O g H O 1mol O 1mol HO = Limiting Reactant Movie 45.0 g H O 15

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? Actual yield % Yield = 100 Theoretical yield 40.0 g H O 100 = 88.9 % 45.0 g H O Accepted value - Measured result % Error = 100 Accepted value 45.0 g H O g H 45.0 g H O O 100 = 5.0 g HO 100 = 11.% 45.0 g H O 16

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: 1. 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? 17

18 Practice Problem: A student is given three unknowns labeled A, B, and C. The student is told they are NaCO3.HO; NaCO3.7HO; and NaCO3.10HO, not necessarily in that order. a. When.000 gram of A was heated, gram of anhydrous residue remained. Formula for sample A. Ans: NaCO3.7HO Sodium carbonate heptahydrate 18

19 Practice Problem b. When grams of B was heated, 1.48 gram of anhydrous residue remained. Formula for sample B. Chemical name for sample B. Ans: NaCO3.10HO Sodium carbonate decahydrate c. When gram of C was heated, gram of anhydrous residue remained. Formula for sample C. Chemical name for sample C. Ans: NaCO3.HO Sodium carbonate monohydrate 19

Chapter 3. Mass Relationships in Chemical Reactions

Chapter 3. Mass Relationships in Chemical Reactions Chapter 3 Mass Relationships in Chemical Reactions In this chapter, Chemical structure and formulas in studying the mass relationships of atoms and molecules. To explain the composition of compounds and