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WVU CHEM 115 - Lecture 7

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1 Collection Terms A collection term states a specific number of items.  1 dozen donuts = 12 donuts  1 ream of paper = 500 sheets  1 case = 24 cans A mole is a collection that contains  the same number of particles as there are carbon atoms in 12.0 g of carbon 12C How many atoms is this? 12 g 12C x 1 amu x 1 atom 12C = 6.022x1023 atoms 1.6605402x10-24g 12 amu So…. 1 mol of anything = 6.022x1023 particles * particles can be atoms, molecules or formula units Avogadro’s Number (NA) = 6.022x1023 particles 1 mol 1 mole of Element Number of Atoms 1 mole of C = 6.022 x 1023 C atoms 1 mole of Na = 6.022 x 1023 Na atoms 1 mole of Au = 6.022 x 1023 Au atoms A Mole of Atoms2 A mole  of a covalent compound has Avogadro’s number of molecules 1 mole of CO2 = 6.022 x 1023 CO2 molecules 1 mole of H2O = 6.022 x 1023 H2O molecules  of an ionic compound contains Avogadro’s number of formula units 1 mole of NaCl = 6.022 x 1023 NaCl formula units 1 mole of K2SO4 = 6.022 x 1023 K2SO4 formula units A Mole of a Compound Using Avogadro’s Number, NA • Avogadro's number, NA, is measured as 6.0223×1023 • Microscale  measure chemicals in terms of atoms, molecules, and f.u. (atomic scale) • Macroscale  measure chemicals in terms of moles or mass (g, lb, kg…) (lab scale) • Use Avogadro’s number whenever convert macroscale  microscale. • Use molar mass whenever convert moles  mass. • Counting formula units by moles is no different than counting eggs by the dozen (12 eggs) or pens by the gross (144 pens) • Since the individual particle is very small, the mole is a more practical quantity • It is a group, in which 6.0223×1023 individuals comprise 1 mole3 Learning Check A. How many moles is 6.091x1032 molecules of water? B. How many molecules in 0.75 mol N2O5? C. How many atoms are present in 2.0 kg of silver? The molar mass is  the mass of one mole of a substance  Is numerically equal to the atomic mass (or formula/molecular mass).  the atomic mass of an element expressed in grams  The atomic masses (found on the periodic table) provide a moles per gram conversion factor. Ex: 1 atom Br = 79.904 amu 1 mol Br = 79.904 g So the conversion can be written as: 79.904 g Br OR 1 mol Br 1 mol Br 79.904 g Br Molar Mass4 Learning Check D. How many formula units are present in 10. mg of (NH4)2CO3? E. What is the mass of 5.2 mol beryllium? F. How many moles is 211 g molybdenum? Using The Chemical Formula • Subscripts within chemical formula can be used as conversion factors!! • Relate components of a compound to the compound quantity we look at the chemical formula • In Na2CO3 there are 3 relationships: – 2 mol Na: 1 mol Na2CO3 – 1 mol C: 1 mol Na2CO3 – 3 mol O: 1 mol Na2CO3 • We can also use these on the atomic scale ,e.g.: – 1 atom C:1 fu Na2CO35 Learning Check In 0.347 mol Cl2O7, how many atoms of oxygen are present? How many moles of calcium are combined with 7.67 mol N in Ca3N2? What is the mass of 10.57 mol sodium phosphate? Learning Check How many moles is 2.573 g tetraphosphorus decaoxide? What is the mass (in g) of 1 molecule of P4O10? (Give answer in 4 sf.) In a sample of Fe3O4 there are 11.7 g of oxygen. What mass of iron is combined with the oxygen?6 Percent Composition • Percent composition is a list of the mass percent of each element in a compound – Na2CO3 is 43.38% Na What is the sum of the 11.33% C percents composition of a 45.29% O compound? • How is it calculated: What is the % C in CO2? • Determine the molar mass of the compound – molst mass of CO2= 44.0 g/mol • Multiply the ratio of the mass of the element to the molar mass of the compound by 100 – (12.0107/44.0)×100= 27.3 %C – ((2×15.9994)/44.0) ×100= 72.7 %O Percent Composition (Theoretical Weight/Mass Percent) As an example…..In lab, a student attempted to prepare the compound iron(III) chloride. The student sent a small portion of the sample away for chemical analysis. The lab returned the following results of percentages of each element in the compound. Element Exp. % Fe 34.0% Cl 66.0% Did the student prepare iron(III) chloride?? To decide, calculate the theoretical mass percent of each element in the compound. Theoretical Mass Percent = mass of element in 1 mol x 100 mass of 1 mol of cmpd7 Examples: Calculate the theoretical mass percent of carbon in LSD (C20H25N3O). In 7.59 g CO2, what mass of carbon is present? What mass of oxygen is


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