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