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Chapter 3: Stoichiometry

Chapter 3: Stoichiometry. Atomic Masses. Naturally occurring chlorine has the following properties 75.78% Cl-35 (34.969 amu) 24.22% Cl-37 (36.966 amu) Calculate the average atomic mass. Avogadro’s Number. The molecular formula for aspartame is C 14 H 18 N 2 O 5. (molar mass = 294.30 g/mol)

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Chapter 3: Stoichiometry

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  1. Chapter 3: Stoichiometry

  2. Atomic Masses • Naturally occurring chlorine has the following properties • 75.78% Cl-35 (34.969 amu) • 24.22% Cl-37 (36.966 amu) • Calculate the average atomic mass

  3. Avogadro’s Number • The molecular formula for aspartame is C14H18N2O5. (molar mass = 294.30 g/mol) • How many moles are present in 10.0 grams of aspartame? • Calculate the mass in grams of 1.56 moles of aspartame. • How many atoms of nitrogen are in 1.2 grams of aspartame? • What is the mass in grams of 1.0 x 109 molecules of aspartame?

  4. Molar Mass and Percent Composition • What is the molar mass of C2H6O? • What is the mass percent of each atom in this molecule?

  5. Empirical and Molecular Formulas • Adipic acid is an organic compound composed of 49.31% C, 43.79% O and the rest hydrogen. • Find the empirical formula • If the molar mass is 146.1 g/mol, what is the molecular formula?

  6. Combustion Analysis • A combustion of 3.000 grams of acetylsalicyclic acid (made of carbon, hydrogen and oxygen) yields the following  1.200 grams of H2O and 6.60 grams of CO2. What is the empirical formula of this compound?

  7. Balancing Chemical Equations • Balance the following__ NH3 + __ O2  __ NO + __ H2O __ Mg3N2 + __ H2SO4  __ MgSO4 + __ (NH4) 2SO4

  8. Types of Equations • Combination: A + B  CCaO(s) + H2O(l)  Ca(OH)2(s) • Decomposition: C  A + B2 KClO3(s)  2 KCl(s) + 3 O2 (g) • Combustion: CxHyOz + O2(g) CO2(g) + 4H2O(g)C3H8 + 5 O2(g) 3 CO2(g) + 4H2O(g)

  9. Reaction Stoichiometry • K2PtCl4(aq) + 2 NH3(aq) Pt(NH3)2Cl2(s) + 2 KCl (aq)If you have 100. grams of K2PtCl4 with excess NH3 how much Pt(NH3)2Cl2 and KCl can we make?

  10. Limiting Reactants • 6 Li(s) + N2(g) 2 Li3N(s)If 5.00 grams of each reactant undergoes a reaction with an 88.5% yield, how many grams of Li3N are obtained from the reaction?

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