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Chapter 3 Notes

by: Corey Stewart

Chapter 3 Notes Chem 1110 - 02

Corey Stewart

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These are notes on stoichiometry. Since this chapter is a lot of calculations, there aren't many notes. I'm going to go over a lot of this on the study guide. If you need examples before then email...
General Chemistry
Dr. Greg Love
Class Notes
General Chemistry, Stoicheometry
25 ?




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This 2 page Class Notes was uploaded by Corey Stewart on Sunday September 25, 2016. The Class Notes belongs to Chem 1110 - 02 at East Tennessee State University taught by Dr. Greg Love in Fall 2016. Since its upload, it has received 17 views. For similar materials see General Chemistry in Chemistry at East Tennessee State University.


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Date Created: 09/25/16
CH. 3 The Mole and Stoichiometry  What is a mole? - The SI Unit for amount - A way to deal with atoms, molecules, and/or compounds in larger quantities so as to not to have to deal with them on the microscopic level - The number of atoms in 12 grams of a sample C – 12 (Carbon – 12) - The same for all substances 23 - Avogadro’s number (6.022 x 10 ) - Molar mass is the same as atomic mass, but in grams - Two moles of two different atoms have the same amount of atoms (a 1 mol of H = 6.022 x 10 atoms of H;1 mol of He = 6.022 x 10 of He); however, the masses are different (1 mol of H = 1.01g; 1 mol He 4g) Molar Molecules  To find the molar mass of a molecule or compound add the mass of the atoms in the compound/molecule (Molar mass of water (H O) = 18g2 2H (2g) + 1O (16g) = water (18g)) Mole as A Conversion Factor  To convert between singular units (atoms of an element or number of molecules/formula units 23 (compounds)) to moles divide by 6.022 x 10 ; to convert from moles to singular units multiply by 6.022 x 10 23  To convert to molar mass of a substance, convert the quantity of a substance you have to moles and multiply by the molar mass; to convert from molar mass to moles divide by the molar mass (and if you need to go to singular units once you get the moles multiply by 6.022 x 10 ) 23 Stoichiometric Equivalencies  Ratios of atoms in a chemical equation is = to the ratio of moles in a molar chemical equation; 1 molecule of H O2= 2 hydrogen atoms and 1 oxygen atom……… 1 mole of H O = 2 moles 2f hydrogen atoms and 1 mole of oxygen atoms Mass-to-Mass Calculations  To find what mass of reactant A is needed to completely react with reactant B, take the mass of reactant A and convert it to moles by dividing by the molar mass of A; use the molar ratio between A and B to convert from moles of A to moles of B; then multiply by the molar mass of B  Carbon Monoxide is CO. How many grams of oxygen is needed to completely react with 0.56g of oxygen?  .56g of oxygen = .56/32 (use 32 because oxygen exists as O in natur2)  .56g O = 02018 mol O 2  There is a 1:1 ratio between Carbon and Oxygen  0.018 mol O = 0.218 mol C Limiting Reagent (Reactant)  The limiting reagent is the reagent that there is less of; This means the reaction will take place until there is none of this reagent left  Excess Reactant - The excess reactant is the reactant that there is too much of - There will be some of this reactant left Limiting Reagent Calculations - Dr. Love had a good explanation on how to do this. - Say you need to know how much water is produced when you have 10.0g H and 15.0g of 2 O 2 - 2H 2 O →22H O 2 - 10g H2(1 mol )/(H202g ) * (H2mol H2O)/(1 mol )H2 4.95 mol H2O - 15g * (1 mol )/(32.00g ) * (2 mol )/(1 mol ) = .938 mol O2 O2 O2 H2O O2 H2O - So O 2s your limiting reagent


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