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CHEMISTRY The Central Science 9th Edition

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Presentation on theme: "CHEMISTRY The Central Science 9th Edition"— Presentation transcript:

1 CHEMISTRY The Central Science 9th Edition
Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations David P. White Prentice Hall © 2003 Chapter 3

2 Chemical Equations Lavoisier: mass is conserved in a chemical reaction. Chemical equations: descriptions of chemical reactions. Two parts to an equation: reactants and products: 2H2 + O2  2H2O Prentice Hall © 2003 Chapter 3

3 Chemical Equations The chemical equation for the formation of water can be visualized as two hydrogen molecules reacting with one oxygen molecule to form two water molecules: 2H2 + O2  2H2O Prentice Hall © 2003 Chapter 3

4 Chemical Equations 2Na + 2H2O  2NaOH + H2 2K + 2H2O  2KOH + H2
Prentice Hall © 2003 Chapter 3

5 Chemical Equations Stoichiometric coefficients: numbers in front of the chemical formulas; give ratio of reactants and products. Prentice Hall © 2003 Chapter 3

6 Chemical Equations Prentice Hall © 2003 Chapter 3

7 Chemical Equations Law of conservation of mass: matter cannot be lost in any chemical reactions. Prentice Hall © 2003 Chapter 3

8 Some Simple Patterns of Chemical Reactivity
Combination and Decomposition Reactions Combination reactions have fewer products than reactants: 2Mg(s) + O2(g)  2MgO(s) The Mg has combined with O2 to form MgO. Decomposition reactions have fewer reactants than products: 2NaN3(s)  2Na(s) + 3N2(g) (the reaction that occurs in an air bag) The NaN3 has decomposed into Na and N2 gas. Prentice Hall © 2003 Chapter 3

9 Some Simple Patterns of Chemical Reactivity
Combination and Decomposition Reactions Prentice Hall © 2003 Chapter 3

10 Some Simple Patterns of Chemical Reactivity
Combination and Decomposition Reactions Prentice Hall © 2003 Chapter 3

11 Some Simple Patterns of Chemical Reactivity
Combustion in Air Combustion is the burning of a substance in oxygen from air: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Prentice Hall © 2003 Chapter 3

12 Formula Weights Formula and Molecular Weights
Formula weights (FW): sum of AW for atoms in formula. FW (H2SO4) = 2AW(H) + AW(S) + 4AW(O) = 2(1.0 amu) + (32.0 amu) + 4(16.0) = amu Molecular weight (MW) is the weight of the molecular formula. MW(C6H12O6) = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) Prentice Hall © 2003 Chapter 3

13 Formula Weights Percentage Composition from Formulas
Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100: Prentice Hall © 2003 Chapter 3

14 The Mole Mole: convenient measure chemical quantities.
1 mole of something =  1023 of that thing. Experimentally, 1 mole of 12C has a mass of 12 g. Molar Mass Molar mass: mass in grams of 1 mole of substance (units g/mol, g.mol-1). Mass of 1 mole of 12C = 12 g. Prentice Hall © 2003 Chapter 3

15 The Mole

16 The Mole Prentice Hall © 2003 Chapter 3

17 The Mole This photograph shows one mole of solid (NaCl), liquid (H2O), and gas (N2). Prentice Hall © 2003 Chapter 3

18 molar mass of N2 = 2  (molar mass of N).
The Mole Interconverting Masses, Moles, and Number of Particles Molar mass: sum of the molar masses of the atoms: molar mass of N2 = 2  (molar mass of N). Molar masses for elements are found on the periodic table. Formula weights are numerically equal to the molar mass. Prentice Hall © 2003 Chapter 3

19 Empirical Formulas from Analyses
Start with mass % of elements (i.e. empirical data) and calculate a formula, or Start with the formula and calculate the mass % elements. Prentice Hall © 2003 Chapter 3

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21 Empirical Formulas from Analyses
Molecular Formula from Empirical Formula Once we know the empirical formula, we need the MW to find the molecular formula. Subscripts in the molecular formula are always whole-number multiples of subscripts in the empirical formula Prentice Hall © 2003 Chapter 3

22 Empirical Formulas from Analyses
Combustion Analysis Empirical formulas are determined by combustion analysis: Prentice Hall © 2003 Chapter 3

23 Quantitative Information from Balanced Equations
Balanced chemical equation gives number of molecules that react to form products. Interpretation: ratio of number of moles of reactant required to give the ratio of number of moles of product. These ratios are called stoichiometric ratios. NB: Stoichiometric ratios are ideal proportions Real ratios of reactants and products in the laboratory need to be measured (in grams and converted to moles). Prentice Hall © 2003 Chapter 3

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25 Limiting Reactants If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess). Limiting Reactant: one reactant that is consumed Prentice Hall © 2003 Chapter 3

26 Limiting Reactants Prentice Hall © 2003 Chapter 3

27 Limiting Reactants Theoretical Yields
The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield. The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield: Prentice Hall © 2003 Chapter 3

28 End of Chapter 3: Stoichiometry: Calculations with Chemical Formulas and Equations
Prentice Hall © 2003 Chapter 3


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