1. Stoichiometry and the Mole (Part 2) Converting—Particles and Grams

2.  Using atomic masses in grams/mole, chemical formulas, and the mole concept, one may calculate the molar mass of both ionic and covalent compounds.  The general procedure is as before: (A) Count (B) Multiply (C) Add  The only difference between the formula mass and the molar mass is the units. Review: Calculating Molar Masses

3.  Use the Periodic Table to find Atomic Mass.  Round to the nearest tenth of a unit.  Count:1 P @ 31.0 g/mol 3 H @ 1.0 g/mol  Multiply:(1 x 31.0) = 31.0 g/mol (3 x 1.0)= 3.0 g/mol  Add:31.0 + 3.0 = 34.0 g/mol Calculate the Molar Mass of PH 3

4.  By multiplying the atomic masses or formula masses (in amu’s) by Avogadro’s number, one gets quantities that one can measure in grams in the laboratory.  The concept is defined with respect to pure carbon-12 (the isotope of carbon with 6 protons and 6 neutrons).  Thus, one mole of carbon-12 has a mass of 12.0000000000000000000000 grams. Review: 6.02 x 10 23 is called Avogadro’s Number

5.  1 mole contains 6.02 x 10 23 particles.  1 mole of carbon contains 6.02 x 10 23 atoms.  1 mole of lithium contains 6.02 x 10 23 atoms.  1 mole of H 2 contains 6.02 x 10 23 molecules.  1 mole of CO 2 contains 6.02 x 10 23 molecules.  1 mole of NaCl contains 6.02 x 10 23 formula units.  1 mole of (NH 4 ) 2 CO 3 contains 6.02 x 10 23 formula units. Examples of what one mole means

6.  Moles ↔ Grams (Both ways) This type of problem uses a Molar Mass as a conversion factor.  Moles ↔ Particles (Both ways) This type of problem uses Avogadro’s number as a conversion factor.  Both types of problems used dimensional analysis (DA). Two Types of Conversion Problems Using the Mole Concept

7.  Ex.: Convert 2.5 moles of CuCl 2, which is copper (II) chloride, to grams of CuCl 2 ?  Find Molar mass first: (1 x 63.55 g/mol for Cu) + (2 x 35.45 g /mol for Cl) = 134.45 g /mol for CuCl 2  (2.5 mol CuCl 2 ) (134.45 g CuCl 2 ) = 336.1 g (1 mol of CuCl 2 ) of CuCl 2 Converting Moles to Grams

8.  Ex.: Convert 25.7 grams of AlBr 3, which is aluminum bromide, to moles of AlBr 3 ?  Find Molar mass first: (1 x 26.98 g/mol for Al) + (3 x 79.90 g /mol for Br) = 266.7 g /mol for AlBr 3  (25.7 g AlBr 3 ) (1 mol of AlBr 3 ) = 0.0964 mol (266.7 g of AlBr 3 ) of AlBr 3 Converting Grams to Moles

9.  Ex.: Convert 4.24 moles of CO 2, which is carbon dioxide, to molecules of CO 2 ?  Use Avogadro’s number in this type of conversion problem  (4.24 mol CO 2 ) (6.02 x 10 23 molecules of CO 2 ) (1 mol of CO 2 ) = 2.55 x 10 24 molecules of CO 2 Converting Moles to Particles

10.  Ex.: Convert 3.51 x 10 21 atoms of B, which is boron, to moles of B?  Use Avogadro’s number in this type of conversion problem  (3.51 x 10 21 atoms of B) ( 1 mol of B ) (6.02 x 10 23 atoms of B) = 0.00583 mol of B Converting Particles to Moles

11.  Refer to your notes (that you have just taken) for your first try at doing these conversions.  Use the logic of DA (UP-DOWN cancellation) to set up and solve the problems.  Make sure that one uses a correctly calculated molar mass on Moles ↔ Grams problems. One Needs to use DA to solve these Problems