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Summer 2007 Ch 3

# Summer 2007 Ch 3 - Chapter 3 Stoichiometry of Formulas and...

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Stoichiometry of Formulas and Equations Chapter 3

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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)
Water 18.02 g CaCO 3 100.09 g Oxygen 32.00 g Copper 63.55 g One mole of common substances.

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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 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

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Sample Problem 3.1 Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element PROBLEM: (a) How many moles of C are in 315 mg of carbon?
Sample Problem 3.2 Calculating the Moles and Number of Formula Units in a Given Mass of a Compound PROBLEM: What is the mass in grams of 4.65 x 10 22 molecules of tetraphosphorous decaoxide?

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Mass % of element X = (moles of X in formula) (molar mass of X (g/mol)) Mass (g) of 1 mol of compound x 100 Mass percent from the chemical formula
Sample Problem 3.3 Calculating the Mass Percents and Masses of Elements in a Sample of Compound PROBLEM: (a) Calculate the mass % of N in ammonium nitrate.

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Summer 2007 Ch 3 - Chapter 3 Stoichiometry of Formulas and...

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