To understand the concept of average mass Objectives To understand the concept of average mass To learn how counting can be done by massing To understand atomic mass and learn how it is determined To understand the mole concept and Avogadro’s number To learn to convert among moles and mass.

Objects do not need to have identical masses to be counted by massing. A. Counting by Massing Objects do not need to have identical masses to be counted by massing. All we need to know is the average mass of the objects. To count the atoms in a sample of a given element by massing we must know the mass of the sample and the average mass for that element. How many books do you have if you have 100 kg of books and each book has an average mass of 10 kg?

A. Counting by Massing (Remember the Bean Lab) Averaging the Mass of Similar Objects Example: What is the mass of 1000 beans? Not all beans have the same mass. Suppose we mass 10 beans and find: Now find the average mass of one bean.

A. Counting by Massing (Remember the Bean Lab) Now find the average mass of one bean. Finally we can multiply by the average mass to find the mass of 1000 beans! 5.0 g/bean X 1000 beans = 5000 g Why don’t all beans (atoms) have the same mass?

C. The Mole !!!! One mole of anything contains 6.022 x 1023 units of that substance. (Eggs, Shoes, Cars, etc.) (12 g of C-12) Avogadro’s number is 6.022 x 1023 (Al Foil) The Atomic Mass (molar mass) of an element indicates the mass in grams of 1 mole (6.022 x 1023) of those atoms. (grams/mole)

C. The Mole !!!!!

C. The Mole !!!!!! Let’s use the Periodic Table to determine the Atomic Mass of the following: (Rounding?) H Co Rn Fr W Ne How many atoms do each of the above atomic mass values contain? What are the units?

C. How many is a mole? 6.022 x 1023 Casio EXP below 3, TI30 XA EE above 7, XIIS 2ndEE above 7, XZ multiview x10n 3 above #8 Al Foil 1 mole of marbles would cover the earth To a depth of 50 miles Would you accept $1 million ($1 x 106) to count to a mole? If you counted 1 per second, it would take you 2 x 1016 years to finish Your hourly wage would be $5 x 10-15 per hour It would take hundreds of millions of years to earn $0.01

C. How to convert using unit analysis. First, let’s learn to convert from moles to grams. What is the atomic mass (molar mass) of Fe? What are the units of atomic mass (molar mass)? How many grams are contained in 1 mol of Fe? How many grams are contained in 2 mol of Fe? How many grams are contained in 0.5 mol of Fe? How many grams are contained in 2.4 mol of Fe? 2.4 mol Fe 55.85 g Fe = 1 mol Fe Know to Go approach, Unit Analysis!!

C. How to convert using unit analysis. Use Unit Analysis to convert from moles to grams: 0.250 mol Al 1.28 mol Ca 0.371 mol P 10.24 mol Au To convert from moles to grams, multiply by the molar mass.

C. How to convert using unit analysis. Second, let’s learn to convert from grams to moles What is the atomic mass (molar mass) of Al? What are the units of atomic mass (molar mass)? How many moles of atoms do 27 g of Al contain? How many moles of atoms do 54 g of Al contain? How many moles of atoms do 108 g of Al contain? How many moles of atoms do 56 g of Al contain? 56 g Al 1 mole Al = 27 g Al Know to Go approach, Unit Analysis!!

C. How to convert using unit analysis. Use Unit Analysis to convert from grams to moles: 220 g Al 568 g Si 3.61 mg He 26.7 g Li To convert from grams to moles, divide by the molar mass.

1 mole = 6.02 X 1023 = Avogadro’s Number Grams  moles X g or 1 mol Conversions Help Page! To convert from moles to grams, multiply by the molar mass. To convert from grams to moles, divide by the molar mass. 1 mole = 6.02 X 1023 = Avogadro’s Number Grams  moles X g or 1 mol 1 mol X g

To understand the concept of average mass Objectives To understand the concept of average mass To learn how counting can be done by massing To understand atomic mass and learn how it is determined To understand the mole concept and Avogadro’s number To learn to convert among moles and mass. Page 210 #15, 14 skip f *watch out for mg #13

Objectives Page 210 15a. 112 g Fe 14a. 0.133 mol Au b. 30.6 g Ni b. 1.04 mol Ca c. 0.240 g Pt c. 2.44 X 10-3 mol Ba d. 1.5 X 104 g Pb d. 1.33 X 1021 mol Pd e. 0.0248 g Mg e. 5.20 X 1020 mol Ni f. 1.31 X 105 g Al f. 8.12 mol Fe g. 1468 g Li g. 1.00 mol C h. 3.95 X 10-5 g Na 13. 0.50 mol O

To learn to convert among moles, mass, and # of atoms. Objectives To learn to convert among moles, mass, and # of atoms. Unit Analysis! Multi Step!

1 mole = 6.02 X 1023 = Avogadro’s Number Grams  moles  # atoms Conversions Help Page! To convert from moles to grams, multiply by the molar mass. To convert from grams to moles, divide by the molar mass. 1 mole = 6.02 X 1023 = Avogadro’s Number Grams  moles  # atoms X g or 1 mol 1 mol X g 6.02 X 1023 atoms or 1 mol . 1 mol 6.02 X 1023 atoms

Know to Go approach, Unit Analysis!! Let it Guide you! Grams  moles  # atoms How many atoms in 79 g of Se? (Know to go?) 79 g Se 1 mol Se 6.02 X 1023 atoms Se = 78.96 g Se 1 mol Se How many atoms in 2.35 mol of #%^$? 2.35 mol #%^$ 6.02 X 1023 atoms #%^$ = 1 mol #%^$ Remember that a mole measures quantity!!

Know to Go approach, Unit Analysis!! Let it Guide you! Grams  moles  # atoms How many atoms are contained in 4.17 mol of Al? How many atoms are contained in 4.17 mol of Si? How many atoms are contained in 1.67 mol of Al? How many atoms are contained in 2 X 10 -4 mol of Al? How many moles are present in 2.13 X 1024 C atoms? What is the mass in g of the above moles of C? What is the mass in g of 9.4 X 1025 atoms of Mg? How many atoms are in 365 g of V? Work Session: Pg 211 #6, 16, 11, 10 (amu = grams)

Pg 211#6, 16, 11, 10 6a. 10.81 amu = 10.81 g = 1 mol = 6.02 X 1023 B atoms b. 10 mol S = 6.02 X 1024 S atoms c. 100 mol Au = 6.02 X 1025 Au atoms d. 150 mol Xe = 9.03 X 1025 Xe atoms e. 133 mol Al = 8.0 X 1025 Al atoms 16a. 1.05 X 1019 Co atoms b. 6.20 X 1020 Co atoms c. 0.0467 mol Co d. 0.00995 mol Co e. 249 g Co f. 2.55 X 1024 Co atoms g. 4.32 X 1022 Co atoms 11. 177 g Co 10. 20.05 g Ca: 40.08 g Ca

To understand the definition of molar mass for atoms and compounds Objectives To understand the definition of molar mass for atoms and compounds To learn to convert between moles and mass for compounds To learn to calculate the mass percent of an element in a compound

1. Molar Mass A compound is a collection of atoms bound together. The molar mass of a compound is obtained by summing the masses of the component atoms.

1. Molar Mass For compounds containing ions the molar mass is obtained by summing the masses of the component ions.

Conversions Help Page! 1 mole = 6.02 X 1023 = Avogadro’s Number Grams  moles  # atoms or molecules For molecules… 6.02 X 1023 molecules or 1 mol . 1 mol 6.02 X 1023 molecules

2. Calculations Using Molar Mass (~Atomic Mass) What is the Molar Mass of methane gas CH4? Use Unit Analysis to convert: How many molecules are contained within 1 mole of methane gas CH4? How many C atoms are contained within 1 mole of methane gas CH4? How many H atoms are contained within 1 mole of methane gas CH4? How many g of C are contained within 1 mole of methane gas? How many g of H are contained within 1 mole of methane gas?

2. Calculations Using Molar Mass (~Atomic Mass) Find the Molar Mass of SO2 Find the Molar Mass of NaCl Find the Molar Mass of CaCO3 What would the mass of 4.86 mol CaCO3 be? Find the Molar Mass of Na2SO4 How many moles in 300 g of Na2SO4? (know-go?) SO2: 64 g/mol NaCl: 58.5 g/mol CaCO3: 100 g/mol, 486 g Na2SO4: 142 g/mol, 2.11 mol

2. Calculations Using Molar Mass (~Atomic Mass) How many moles in 1.56 g of Juglone C10H6O3? (Dye and herbicide made from black walnuts) 8.96 X 10 -3 mol C10H6O3

A. Calculations Using Molar Mass (~Atomic Mass) Bees release 1 X 10-6 g of isopentyl acetate (C7H14O2) during a sting. (Also a banana scent!) How many moles of isopentyl acetate are released? 7.69 X 10-9 mol C7H14O2 How many molecules of isopentyl acetate are released? 5 X 1015 molecules C7H14O2

2. Calculations Using Molar Mass (~Atomic Mass) How many molecules are contained in 135 g of Teflon (C2F4)? 8.127 X 1023 molecules of C2F4

Work Session: pg 195 2-5 skip #4 2) a. 17.03 g/mol b. 18.02 g/mol 5) Moles Mass g #molecules c. 98.09 g/mol 4.7 80 2.8 X 10 24 d. 164.1 g/mol 0.0999 1.80 6.02 X 10 22 3) a. 50 g NH3 0.0865 1.39 5.21 X 10 22 b. 4.5 g H2O c. 490 g H2SO4 d. 120 g Ca(NO3)2

3. Percent Composition of Compounds Percent composition consists of the mass percent of each element in a compound: % composition boys in class? (part over total) % composition girls in class? (part over total) Mass percent =

3. Percent Composition of Compounds Rules to determine the percent composition by mass of each element: 1. Determine the molar mass of the compound 2. Divide the mass of each element type by the total mass of the compound (#1) 3. Multiply by 100 for percent Mass percent =

3. Percent Composition of Compounds Determine the percent composition by mass of each element in methane gas CH4 Molar Mass Each Element X 100 75% C, 25% H

3. Percent Composition of Compounds Determine the percent composition by mass of each element in SO2 Molar Mass Each Element X 100 50/50

3. Percent Composition of Compounds Determine the percent composition by mass of each element in NaCl Molar Mass Each Element X 100 39% Na, 61% Cl

3. Percent Composition of Compounds Determine the percent composition by mass of each element in CaCO3 Molar Mass Each Element X 100 40% Ca, 12% C, 48% O

3. Percent Composition of Compounds Determine the percent composition by mass of each element in: Na2SO4 C10H6O3 C7H14O2 C2F4 C2H5OH C10H14O Work Session pg 195 (% Comp for #2)

3. Percent Composition of Compounds Answers: Na2SO4 32% Na, 23% S, 45% O C10H6O3 69% C, 3% H, 28% O C7H14O2 65% C, 11% H, 24% O C2F4 24% C, 76% F C2H5OH 52% C, 13% H, 35% o C10H14O 80% C, 9% H, 11% O NH3 82% N, 18% H H2O 11% H, 89% O H2SO4 2% H, 33% S, 65% O Ca(NO3)2 24% Ca, 17% N, 59% O

To understand the definition of molar mass for atoms and compounds Objectives Review To understand the definition of molar mass for atoms and compounds To learn to convert between moles and mass for compounds To learn to calculate the mass percent of an element in a compound

To understand the meaning of empirical formula Objectives To understand the meaning of empirical formula To learn to calculate empirical formulas To learn to calculate the molecular formula of a compound

1. Empirical Formulas The empirical formula of a compound is the simplest whole number ratio of the atoms present in the compound. The empirical formula can be found from the percent composition of the compound.

2. Calculation of Empirical Formulas

2. Calculation of Empirical Formulas Patch’s Interpretation: Calculate moles if you don’t already have them. Divide all by the smallest # of moles Get to Whole Number Multiples.

2. Calculation of Empirical Formulas Determine the empirical formula if an analysis of a C, H, and O compound resulted in the following : 0.0806 g C 0.01353 g H 0.1074 g O 0.00671 mol C, 0.01353 mol H, 0.00671 mol O

Analyzing for Carbon and Hydrogen Copyright © Cengage Learning. All rights reserved Analyzing for Carbon and Hydrogen Device used to determine the mass percent of each element in a compound.

2. Calculation of Empirical Formulas Now, let’s find the empirical formula: 0.00671 mol C 0.01353 mol H 0.00671 mol O CH2O (simplest) C2H4O2 ; C4H8O4 ;C5H10O5

2. Calculation of Empirical Formulas Determine the empirical formula if an analysis of a Ni and O compound resulted in the following : 0.2636 g Ni 0.0718 g O NiO

2. Calculation of Empirical Formulas AlxOy 4.151 g Al 3.692 g O Al2O3

2. Calculation of Empirical Formulas It is observed that 0.664 g of lead combined with 0.2356 g of chlorine. Determine the empirical formula. PbCl2

2. Calculation of Empirical Formulas Determine the empirical formula of cisplatin, a cancer tumor treatment drug that has the following % composition by mass: 65.02% Pt 9.34% N 2.02% H 23.63% Cl Assume some mass (100g) PtN2H6Cl2

2. Calculation of Empirical Formulas The most common form of nylon (Nylon-6) is 63.68% C, 12.38% N, 9.80% H, and 14.4% O by mass. Calculate the empirical formula for Nylon-6. C6NH11O now find the molar mass

3. Calculation of Molecular Formulas The molecular formula is the exact formula of the molecules present in a substance. The molecular formula is a whole number multiple (n) of the empirical formula. n = molar mass molecular formula molar mass empirical formula Find n (empirical formula) n = molecular formula **You have to be given the molar mass of the molecular formula!

3. Calculation of Molecular Formulas Calculate the molecular formula for a compound if the empirical formula is P2O5 and the molecular molar mass is 283.88 g/mol. P4O10 Check the mm=(4(31)+10(16))=284

Calculate the molar mass: MgSO4 * 7H2O Mg + S + 4(O) + 7(H2O) Hydrate Lab Calculate the molar mass: MgSO4 * 7H2O Mg + S + 4(O) + 7(H2O) Calculate the % mass of waters… 7(H2O) . X 100 MM = 246 g/mol : % H2O = 51.22% Silica Gel Packaging…

To understand the meaning of empirical formula Objectives Review To understand the meaning of empirical formula To learn to calculate empirical formulas To learn to calculate the molecular formula of a compound Work Session: pg 207 practice # 6-15 pg 208 4, 5 pg 213 46, 47

Formula Summary For the sugar glucose