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Unit 2 Bundle

by: Kara Nichols

Unit 2 Bundle CHM 111

Kara Nichols
GPA 3.7

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This bundle has everything you need to make an A on the unit 2 test! It includes all the notes, plus a hand made study guide!
General Chemistry 1
Mona Chaudhary
Chemistry, Stoicheometry
75 ?




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This 11 page Bundle was uploaded by Kara Nichols on Friday June 17, 2016. The Bundle belongs to CHM 111 at Calhoun Community College taught by Mona Chaudhary in Spring 2016. Since its upload, it has received 44 views.

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Date Created: 06/17/16
Lesson 11 Wednesday, June 15, 2016 5:29 PM • Atomic Weight / atomic mass ○ Decimal number on the periodic table ○ Weighted average of all the isotopes ○ We use it as grams ○ Weight and/or mass of one atom • Mole ○ Describes an amount of substance/compound ○ Mole is the SI unit used to represent 6.022 x 10^23 particles ○ 6.022 x 10^23= Avogadro's number ○ 1 mole of an atom - the weight (atomicmass) in grams ○ 1 mole of Cu = 63.54 grams  1 mole of Cu = 6.022 x 10^23atoms of Cu ○ 1 mole of O = 15.9994grams  1 mole of O = 6.022 x 10^23atoms of O ○ Molecular weight is used for molecules ○ Atomic weight is used for atoms • Formula / molecularweights ○ Formula weight (FW) = ionic compounds ○ Molecular Weight (MW) = covalent compounds ○ Calculation  The atomic weights of all the atomspresent in the compound • Calculations of grams, moles, molecules,and atoms ○ If you start with moles:  You will ALWAYS multiply  To get to grams: □ Multiply by the molecular weight of the compound or molecule  To get to the number of moleculesor atoms: □ Multiply by Avogadro's number (6.022 x 10^23) ○ If you start with grams:  To get to moles: □ DIVIDE by the molecular weight  To get to the number of moleculesor atoms:  To get to the number of moleculesor atoms: □ Divide by the molecular weight □ Then multiply by Avogadro's number ○ If you start with the number of moleculesor atoms  To get to moles: □ Divide by Avogadro's number  To get to grams: □ Divide by Avogadro's number □ Then multiply by the molecular weight ○ To find the number of atoms in a molecule  Multiply the number of moleculesby the subscript of the atom you are trying to find • Percent Composition ○ (Mass of the element/ mass of the compound ) x100 College Chemistry 1 – Mona Chaudhary 6/14/2016 Important Words or Definitions Important People Important Concepts Lesson 12  Formulas from element composition o Empirical formula  Simplest formula  Relative ration of atoms  Smallest whole number ratio of atoms o Molecular (true) Formula  True ratio of atoms  Multiple of empirical formula ratios o Molecular formula = (empirical formula x n)  Element Relative mass Relative # of atoms Divide by the Whole # of atoms smallest # ratio Symbol Change % Divide the relative mass by Divide by the Whole sign to Grams the atomic weight of the smallest relative number element value ratio  Example: o A compound contains the following: 40% carbon, 6.7% hydrogen, 53.3% oxygen. What is the empirical formula? o What is the molecular formula if the compound has a molecular mass of 60.0 g/mol? o Element Relative mass of Relative # of Divide by the Whole # atoms atoms smallest # ratio C 40 g 40/12.01 3.33/3.33 1 H 6.7g 6.7/1.01 6.63/3.33 2 O 53.3g 53.3/16 3.33/3.33 1 Empirical formula = CH2O Molecular formula: Molecular weight of empirical formula = 12.01+2.02+16 = 30.03 60/30.03 = 1.998 (almost 2) Multiply all the subscripts by 2 C 2 4 2 Lesson 13 ● Balancing chemical equations ○Reactants -> Products ○Conservation of matter: matter cannot be created nor destroyed ■ there must be the same number of molecules on each side ■ H + OH -2​H ● 2 hydrogen on the left, 2 hydrogen on the right ● 1 oxygen on the left, 1 oxygen on the right ○To balance chemical equations: ■ balance all molecules except Oxygen and Hydrogen ● Balance Hydrogen next ● then, balance Oxygen ■ Do not change subscripts (the small numbers) ■ only change the coefficients (the numbers in front) ● Chemical equations are ratios ○moles:moles ○molecules:molecules ○mass:mass ○Before you can do any calculations, the chemical equation must be balanced Lesson 14 ● Limiting Reactants ○the limiting reactant stops the reaction when it is used up ■ if you have 3 slices of bread, and 2 slices of cheese, how many sandwiches can you make? ■ 1, because you only have enough bread to make 1 sandwich ○The amount of product that is created depends on the limiting reactant ○Start wit​alanc​reaction, then use the fencing method to solve (shown below) ● Solve a problem: ○Balance the equation ○treat each reaction as if it were the limiting reactant ○the reactant that has the smallest number of product will be the limiting reactant ■the reactant with the large amount of product will be the excess reactant ○ Given the unbalanced equation: ■N2​ 4​ N2​ 4​ N2​+ H2​ ■what mass (g)2​ould be produced by the reaction of 850g of N2​ 4​d 550g2​ 4​ ■What is the limiting reactant? ■What is the excess reactant? ■How much of the excess reactant is left over? ■What is the theoretical yield? ■Of the reaction actually produc2​what is the percent yield for the reaction? Unit 2 Study Guide 1. How many moles are in 37.6 g of HCl? 22 2. What is the weight, in grams, of 3.7 x 10 molecules H O?2 3. How many molecules are in 89 moles of Fe O ? 2 3 4. Calculate the percent by mass of carbon in C H O : 6 12 6 5. Find the empirical and molecular formula of a compound that is 28.6% magnesium, 14.3% carbon, and 57.1% oxygen, and has a molecular weight of 337.28g. 6. What is the empirical formula of glucose? (C H O ) 6 12 6 7. Balance the chemical equations: a. P 4 O  2 O 4 10 b. Na + H O 2 NaOH + H 2 c. NH 4O  2 + H O2 2 8. Given the unbalanced reaction: L3 N + H 2  NH 3 LiOH a.How many grams of LiOH can (theoretically) be produced when .38 grams of lithium nitride reacts with water? b.What is the limiting reactant?______________________________ c. How many grams of the excess reactant will be used?________________________ d.If the reaction actually produces .67 grams of lithium hydroxide, what is the percent yield?


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