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CHEM Chapter 8 Notes- Chemical Composition

by: Kayra Reyes

CHEM Chapter 8 Notes- Chemical Composition Chem 1301

Marketplace > University of Houston > Chemistry > Chem 1301 > CHEM Chapter 8 Notes Chemical Composition
Kayra Reyes

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These notes cover Chapter 8 from "Introductory Chemistry" by Zumdahl and Decoste, 7th/8th edition. There's examples and vocabulary, plus some helpful hints. This chapter seems a bit tricky but...
Foundations of Chemistry
Roman S. Czernuszewicz
Class Notes
Chemistry, Molecular, compounds, empirical, atoms, atomic, mass, Mole
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This 4 page Class Notes was uploaded by Kayra Reyes on Sunday October 9, 2016. The Class Notes belongs to Chem 1301 at University of Houston taught by Roman S. Czernuszewicz in Fall 2015. Since its upload, it has received 4 views. For similar materials see Foundations of Chemistry in Chemistry at University of Houston.


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Date Created: 10/09/16
th th Chapter 8 from “Introductory Chemistry” by Zumdahl and Decoste, 7 /8 edition Pg 167. COUNTING BY WEIGHING Average mass = total mass Number of item/variables Pg 170. ATOMIC MASSES: COUNTING ATOMS BY WEIGHING Want to know how many molecules are needed to make carbon dioxide? C (s) + O2(g)  CO 2g) 1 atom + 1 molecule  1 molecule *because atoms weigh so little and kilograms would be too large of a measurement, chemists use Atomic Mass Unit (amu) In terms of grams: 1 amu = 1.66 x 10 -24g Average Atomic Mass: (means what its named…duh) Ex: Mass of 1000 natural C atoms = (1000 atoms) (12.01 (amu/atom)) = 12,010 amu = 12.01 x 10 amu Conversion factor: 1 Carbon atom 12.01 amu 20 19 3.00 x 10 amu x 1 Carbon atom = 2.50 x 10 Carbon atoms 12.01 amu Ex #2: Mass of 75 Al atoms Average Mass for 1 Al atom: 26.98 amu Equivalence Statement: 1 Al atom = 26.98 amu 75 Al atoms x 26.98 amu = 2024 amu 1 Al atom Pg 172. THE MOLE *”samples in which the ratio of the masses is the same as the ratio of the masses of the individual atoms always contain the same number of ratios” Mole: the quantity of anything that has the same number of particles Avogadro’s Number: the number of “elementary” particles (molecules, atoms, compounds, etc.) per mole of a substance equal to 6.022×10 mol -1 Ex: One mole of water = 6.022 x 10 H O 2olecules *”any element that weighs a number of grams equal to the average atomic mass of that 23 element contains 6.022 x 10 atoms (1 mol) of that element” Pg 177. LEARNING TO SOLVE PROBLEMS Conceptual Problem Solving: solving problems in a creative way based on your past knowledge (of basic chemistry, in this case) to solve future, unfamiliar problems Principles of Problem Solving: 1. Analyze the parts of the problem and set an end goal 2. Answer the questions that would help lead you to your goal (ex: reactants? Products? Is the equation balanced?) 3. Check if your final answer is reasonable Pg 180. MOLAR MASS Molar Mass: the mass of 1 mole of a substance  found by summing up masses of separate atoms of a compound Ex: find the mass of 1 mole of SO2molecules Mass of 1 mol of S = 1 x 32.07 = 32.07 g Mass of 2 mol of O = 2 x 16.00 = 32.00 g Mass of 1 mol SO 2 = 64.07 g  molar mass 1 mole of SO 2s 64.07 g Pg 182. Formula Weight: alternative of molar mass for ionic compounds Pg 187. PERCENT OF COMPOSITION OF COMPOUNDS Mass fraction for a given element = mass of the element present in 1 mole of compound Mass of 1 mole of compound *becomes mass percent when multiplied by 100 Ex: using ethanol Mass of C = 2 mol x 12.01 (g/mol) = 24.02 g Mass of H = 6 mol x 1.008 (g/mol) = 6.048 g Mass of 1 mol of C H2O5 Mass of O = 1 mol x 16.00 (g/mol) = 16.00 g = 46.07 g =molar mass Mass percent of C = mass of C in 1 mol C2H5OH x 100% (do independently per element Mass of 1 mol C2H5OH C, H, and O) C: (24.02 g/46.07 g) x 100% = 52.14% H: (6.048 g/ 46.07 g) x 100% = 13.13%  mass percents O: (16.00 g/46.07 g) x 100% = 34.73% 52.14 + 13.13 + 34.73 = 100% FORMULAS OF COMPOUNDS Pg 190. *when we separate the elements in a compound and “count” the atoms, we are actually only counting the relative number of atoms Empirical Formula: simplest formula with smallest whole number ration Molecular Formula: actual formula that gives composition of molecules present CALCULATION OF EMPIRICAL FORMULAS Step 1: find the relative masses of the elements present, in grams Step 2: determine the number of moles of each type of atom present Step 3: divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. Step 4: multiply the number you found in step 3 by the smallest number that will get rid of any fractions Ex: what is the empirical formula of aluminum oxide, x yO Aluminum oxide is made up of 4.151g of Al and 3.692g of O Atomic Mass of Al is 26.98 g/mol 4.151g Al x (1mol/26.98g) = 0.1539 mol of Al atoms Atomic Mass of O is 16.00 g/mol 3.692g O x (1mol/16.00g) = 0.2308 mol of O atoms *note we need whole numbers so we must find the ratio of the atoms by dividing both by the smallest number 1) 0.1539 mol Al = 1.000 mol Al atoms 2) 0.1539  0.2308 mol O = 1.500 mol O 0.2308 mol O = 1.500 mol O atoms 0.1539 mol Al 1.000 mol Al 0.1539 3) 0.2308 O = 1.500 O = (3/2) O = 3 O empirical formula is A2 3 0.1539 Al = 1.000 Al = 1 Al = 2 Al Pg 198. CALCULATION OF MOLECULAR FORMULAS Ex: the empirical formula is2 5O Molar mass is 283.88 g/mol 2 mol of P: 2 x 30.97g = 61.94g Atomic Mass of Phosphorous is 30.97 g/mol 5 mol of O: 5 x 16.00g = 80.00g Atomic Mass of Oxygen is 16.00 g/mol = 141.94g Molecular Formula = (empirical formula) n = 238.88g =2 Molecular Formula = n x empirical formula 141.94g Molar mass = n x empirical formula mass 2 empirical formula units jjjjjjjjjjMolar mass so molecular formula is (P2 5 2 n= Empirical formula mass or P O 4 10


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