CH 3 - Chapter 3 Stoichiometry of Formulas and Equations...

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Stoichiometry of Formulas and Equations Chapter 3
Mole - Mass Relationships in Chemical Systems 3.5 Fundamentals of Solution Stoichiometry 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts of Reactant and Product

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mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. This amount is 6.022x10 23 . The number is called Avogadro’s number and is abbreviated as N. One mole (1 mol) contains 6.022x10 23 entities (to four significant figures)
12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 10 23 atoms Fe 32.07g S = 6.022 x 10 23 atoms S Figure 3.1 Counting objects of fixed relative mass.

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Water 18.02 g CaCO 3 100.09 g Oxygen 32.00 g Copper 63.55 g One mole of common substances. Figure 3.2
Table 3.1 Summary of Mass Terminology Term Definition Unit Isotopic mass Mass of an isotope of an element amu Atomic mass Molecular (or formula) mass (also called molecular weight) Molar mass ( M ) (also called atomic weight) (also called gram-molecular weight) amu amu g/mol Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units)

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Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 ( M = 180.16 g/mol) Oxygen (O) Mass/mole of compound 6 atoms 96.00 g Table 3.2 Carbon (C) Hydrogen (H) Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/molecule of compound 6 atoms 12 atoms 6 moles of atoms 12 moles of atoms 6 moles of atoms 6(6.022 x 10 23 ) atoms 12(6.022 x 10 23 ) atoms 6(6.022 x 10 23 ) atoms 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 6(16.00 amu) =96.00 amu 72.06 g 12.10 g
Interconverting Moles, Mass, and Number of Chemical Entities Mass (g) = no. of moles x no. of grams 1 mol No. of moles = mass (g) x no. of grams 1 mol No. of entities = no. of moles x 6.022x10 23 entities 1 mol No. of moles = no. of entities x 6.022x10 23 entities 1 mol g M

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MASS(g) of element AMOUNT(mol) of element ATOMS of element Summary of the mass-mole- number relationships for elements. M (g/mol) Avogadro’s number (atoms/mol) Figure 3.3
Sample Problem 3.1 Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element PROBLEM: PLAN: SOLUTION: amount(mol) of Ag mass(g) of Ag (a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag? (b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe? (a)

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CH 3 - Chapter 3 Stoichiometry of Formulas and Equations...

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