ch 3 6e - Chapter 3 Stoichiometry of Formulas and Equations...

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3-1Stoichiometry of Formulas and Equations Chapter 3
3-2Mole - Mass Relationships in Chemical Systems3.5 Fundamentals of Solution Stoichiometry3.1 The Mole3.2 Determining the Formula of an Unknown Compound3.3 Writing and Balancing Chemical Equations3.4 Calculating Quantities of Reactant and Product
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3-4One mole (6.022x1023entities) of some familiar substances.Figure 3.1
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3-7Table 3.1Information Contained in the Chemical Formula of Glucose C6H12O6( M= 180.16 g/mol)Carbon (C)Hydrogen (H)Oxygen (O)Atoms/molecule of compound6 atoms12 atoms6 atomsMoles of atoms/mole of compound6 mol of atoms12 mol of atoms6 mol of atomsAtoms/mole of compound6(6.022x1023) atoms12(6.022x1023) atoms6(6.022x1023) atomsMass/molecule of compound6(12.01 amu) = 72.06 amu12(1.008 amu) = 12.10 amu6(16.00 amu) = 96.00 amuMass/mole of compound72.06 g12.10 g96.00 g
3-8Interconverting Moles, Mass, and Number of Chemical EntitiesMass (g) = no. of moles x no. of grams1 molNo. of moles = mass (g) xno. of grams1 molNo. of entities = no. of moles x6.022x1023entities1 molNo. of moles = no. of entities x 6.022x1023entities1 molgM
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Figure 3.2
3-10Sample Problem 3.1Calculating the Mass of a Given Amount of an ElementPROBLEM:Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342 mol of Ag?SOLUTION:amount (mol) of Agmass (g) of AgPLAN:To convert mol of Ag to mass of Ag in g we need the molar mass of Ag.multiply by Mof Ag (107.9 g/mol)0.0342 mol Ag x1 mol Ag107.9 g Ag= 3.69 g Ag
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