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© 2012 Pearson Education, Inc. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Ashley Warren Kings High School Lecture Presentation.

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Presentation on theme: "© 2012 Pearson Education, Inc. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Ashley Warren Kings High School Lecture Presentation."— Presentation transcript:

1 © 2012 Pearson Education, Inc. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Ashley Warren Kings High School Lecture Presentation

2 © 2012 Pearson Education, Inc. Stoichiometry The area of study that examines the quantities of substances consumed and produced in chemical reactions. Stoicheion means element and metron means measure.

3 Stoichiometry © 2012 Pearson Education, Inc. Law of Conservation of Mass We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends. --Antoine Lavoisier, 1789

4 Stoichiometry © 2012 Pearson Education, Inc. Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Reactants appear on the left side of the equation. Products appear on the right side. The states of the reactants and products are written in parentheses to the right of each compound. Coefficients are inserted to balance the equation.

5 Stoichiometry © 2012 Pearson Education, Inc. Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule. Coefficients tell the number of molecules.

6 Stoichiometry Interpreting and Balancing Chemical Equations The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms. –Write the chemical formulas for the reactants and products. –Write a balanced chemical equation for the reaction. –Is the diagram consistent with the Law of conservation of mass? © 2012 Pearson Education, Inc.

7 Stoichiometry Interpreting and Balancing Chemical Equations In the diagram above, the white spheres represent hydrogen atoms and the blue spheres represent nitrogen atoms. To be consistent with the law of conservation of mass, how many NH 3 molecules should be shown in the right (product) box? © 2012 Pearson Education, Inc.

8 Stoichiometry © 2012 Pearson Education, Inc. Reaction Types

9 Stoichiometry © 2012 Pearson Education, Inc. Combination Reactions Examples: –2Mg(s) + O 2 (g) 2MgO(s) –N 2 (g) + 3H 2 (g) 2NH 3 (g) –C 3 H 6 (g) + Br 2 (l) C 3 H 6 Br 2 (l) In combination reactions two or more substances react to form one product.

10 Stoichiometry © 2012 Pearson Education, Inc. In a decomposition reaction one substance breaks down into two or more substances. Decomposition Reactions Examples: –CaCO 3 (s) CaO(s) + CO 2 (g) –2KClO 3 (s) 2KCl(s) + O 2 (g) –2NaN 3 (s) 2Na(s) + 3N 2 (g)

11 Stoichiometry © 2012 Pearson Education, Inc. Combustion Reactions Examples: –CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) –C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) Combustion reactions are generally rapid reactions that produce a flame. Combustion reactions most often involve hydrocarbons reacting with oxygen in the air.

12 Stoichiometry More Reactions © 2012 Pearson Education, Inc.

13 Stoichiometry More Reactions © 2012 Pearson Education, Inc.

14 Stoichiometry Final Reactions

15 Stoichiometry © 2012 Pearson Education, Inc. Formula Weights

16 Stoichiometry © 2012 Pearson Education, Inc. Formula Weight (FW) A formula weight is the sum of the atomic weights for the atoms in a chemical formula. So, the formula weight of calcium chloride, CaCl 2, would be Ca: 1(40.08 amu) + Cl: 2( amu) amu Formula weights are generally reported for ionic compounds. What is the formula weight for sulfuric acid?

17 Stoichiometry © 2012 Pearson Education, Inc. Molecular Weight (MW) A molecular weight is the sum of the atomic weights of the atoms in a molecule. For the molecule ethane, C 2 H 6, the molecular weight would be C: 2( amu) amu + H: 6( amu)

18 Stoichiometry © 2012 Pearson Education, Inc. Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % Element = (number of atoms)(atomic weight) (FW of the compound) x 100

19 Stoichiometry © 2012 Pearson Education, Inc. Percent Composition So the percentage of carbon in ethane (C 2 H 6 ) is %C = (2)( amu) ( amu) amu amu = x 100 = %

20 Stoichiometry © 2012 Pearson Education, Inc. Moles

21 Stoichiometry © 2012 Pearson Education, Inc. Avogadros Number 1 mole = 6.02 x 10 23, symbolized N A. 1 mole of 12 C has a mass of g. Abbreviation = mol

22 Stoichiometry Avogadros Number See exercise 3.7 and 3.8 © 2012 Pearson Education, Inc.

23 Stoichiometry © 2012 Pearson Education, Inc. Molar Mass By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol). –The molar mass of an element is the mass number for the element that we find on the periodic table. –The formula weight (in amus) will be the same number as the molar mass (in g/mol). *The atomic weight of an element in atomic mass units is numerically equal to the mass in grams of 1 mol of that element.

24 Stoichiometry © 2012 Pearson Education, Inc. Mole Relationships One mole of atoms, ions, or molecules contains Avogadros number of those particles. One mole of molecules or formula units contains Avogadros number times the number of atoms or ions of each element in the compound.

25 Stoichiometry Molar Mass Which has more mass, a mole of water or a mole of glucose? Which contains more molecules, a mole of water or a mole of glucose? Calculate the molar mass of glucose. Calculate the molar mass of calcium nitrate. © 2012 Pearson Education, Inc.

26 Stoichiometry © 2012 Pearson Education, Inc. Using Moles Moles provide a bridge from the molecular scale to the real-world scale. Lets think about units in all of these cases! If you know the units you will ALWAYS be able to figure out what to do!

27 Stoichiometry Masses and Moles Calculate the number of moles of water in grams of water. Calculate the mass, in grams, of mol of potassium chlorite. How many molecules are in grams of C 3 H 8 ? © 2012 Pearson Education, Inc.

28 Stoichiometry © 2012 Pearson Education, Inc. Finding Empirical Formulas

29 Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition.

30 Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

31 Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas Assuming g of para-aminobenzoic acid, C:61.31 g x = mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= mol N O:23.33 g x = mol O 1 mol g 1 mol g 1 mol 1.01 g 1 mol g

32 Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= H:= N:= O:= mol mol 5.09 mol mol mol mol

33 Stoichiometry © 2012 Pearson Education, Inc. Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

34 Stoichiometry Molecular Formulas from Empirical Mesitylene, a hydrocarbon found in crude oil, has an empirical formula of C 3 H 4, and an experimentally determined molecular weight of 121 amu. What is its molecular formula? The formula weight of C 3 H 4 = 40.0 amu Whole number multiple = molecular weight empirical formula weight = 121 = This means the molecular formula is triple the empirical. Therefore the molecular formula is C 9 H 12. © 2012 Pearson Education, Inc.

35 Stoichiometry © 2012 Pearson Education, Inc. Combustion Analysis Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like the one shown in Figure –C is determined from the mass of CO 2 produced. –H is determined from the mass of H 2 O produced. –O is determined by difference after the C and H have been determined. ***See Exercise 3.15

36 Stoichiometry © 2012 Pearson Education, Inc. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products as well as the relative numbers of molecules.

37 Stoichiometry © 2012 Pearson Education, Inc. Stoichiometric Calculations Starting with the mass of Substance A, you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if its a product) or used (if its a reactant).

38 Stoichiometry © 2012 Pearson Education, Inc. Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 …(using molar mass) use the coefficients to find the moles of H 2 O…(using mole to mole ratio) and then turn the moles of water to grams…(using molar mass again) C 6 H 12 O O 2 6 CO H 2 O

39 Stoichiometry Stoichiometric Calculations If given the following reaction, how many grams of O 2 can be prepared from 4.50 grams of KClO 3 ? 2KClO 3 (s) 2KCl (s) + 3 O 2 (g) © 2012 Pearson Education, Inc.

40 Stoichiometry © 2012 Pearson Education, Inc. Limiting Reactants

41 Stoichiometry © 2012 Pearson Education, Inc. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount. –In other words, its the reactant youll run out of first (in this case, the H 2 ). –In the example below, the O 2 would be the excess reagent.

42 Stoichiometry © 2012 Pearson Education, Inc. Theoretical Yield The theoretical yield is the maximum amount of product that can be made. –In other words, its the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount one actually produces and measures.

43 Stoichiometry © 2012 Pearson Education, Inc. Percent Yield One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield): Percent yield =x 100 actual yield theoretical yield

44 Stoichiometry Limiting Reactant/Theoretical Yield Calculations See Exercise 3.18, 3.19, and 3.20 © 2012 Pearson Education, Inc.

45 Stoichiometry Chapter 3 Homework Problems 1, 3, 5, 6, 8, 10, 12, 13, 15, 17, 19, 22, 23, 25, 27, 30, 31, 33, 35, 37, 40, 41, 43, 46, 47, 49, 52, 53, 55, 61, 63, 65, 67, 69, 71, 76, 77, 80, 81, 83, 87, 89a,b, 97, 101 © 2012 Pearson Education, Inc.


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