The MoleThe Mole

Chapter 6Chapter 6

The MoleThe Mole► Remember it is a Remember it is a

measurement for anythingmeasurement for anything► 1 mole means 6.022 x 101 mole means 6.022 x 102323

of anythingof anything► This is called Avogadro’s This is called Avogadro’s

NumberNumber► A sample of an element A sample of an element

with a mass equal to that with a mass equal to that element’s average atomic element’s average atomic mass expressed in grams mass expressed in grams contains 1 mol of atomscontains 1 mol of atoms

►26.98 g of Aluminum has 6.022 26.98 g of Aluminum has 6.022 x 10x 102323 atoms = 1mol atoms = 1mol

How can this help us?How can this help us?► If you know how many grams make 1 If you know how many grams make 1

mole, you can figure out how many mole, you can figure out how many moles you have in a given massmoles you have in a given mass If you have 0.00568 g of Silicon how many If you have 0.00568 g of Silicon how many

atoms do you have?atoms do you have? Know that 1 mol Si is 28.09 g and 1 mole Know that 1 mol Si is 28.09 g and 1 mole

is 6.022 x 10is 6.022 x 102323 atoms, these are your atoms, these are your conversion factorsconversion factors

So 0.00568 g x 1 mol/28.09 g = 0.000202 So 0.00568 g x 1 mol/28.09 g = 0.000202 mol Simol Si

0.000202 mol Si x 6.022 x 100.000202 mol Si x 6.022 x 102323 atoms/ 1 atoms/ 1 mol = 1.22 x 10mol = 1.22 x 102020 Si atoms Si atoms

Molar MassMolar Mass

►What is the mass of 1 mole of a compound? What is the mass of 1 mole of a compound? Simply add the mass of the elements moles Simply add the mass of the elements moles to get the compounds molar massto get the compounds molar mass

► CHCH44 is made up of 1 mol of C and 4 mol of H is made up of 1 mol of C and 4 mol of H 1 mol C = 1 x 12.01 g = 12.01 g1 mol C = 1 x 12.01 g = 12.01 g 4 mol H = 4 x 1.008 g = 4.032 g4 mol H = 4 x 1.008 g = 4.032 g So 1 mol of CHSo 1 mol of CH44 = 12.01 g + 4.032 g = 16.04 g = 12.01 g + 4.032 g = 16.04 g

► Try to find molar mass of CaCOTry to find molar mass of CaCO33

► Answer =Answer =

►Now figure out how many grams of CaCONow figure out how many grams of CaCO33 are in 4.86 mol are in 4.86 mol

► Answer = Answer =

Practice ProblemPractice Problem►How many moles and how many How many moles and how many

molecules of Cmolecules of C77HH1414OO22 is in a bee sting if is in a bee sting if 1 gram is released?1 gram is released? First find Molar MassFirst find Molar Mass AnswerAnswer Second find how many moles in 1 gram Second find how many moles in 1 gram AnswerAnswer Finally change moles to moleculesFinally change moles to molecules AnswerAnswer

Enough for 1 day, Watch the Enough for 1 day, Watch the Video “The Mole”Video “The Mole”

Percent composition of Percent composition of CompoundsCompounds

►Take mass of 1 element and divide it by Take mass of 1 element and divide it by the mass of entire compound and the mass of entire compound and multiply by 100multiply by 100

►Example: CExample: C22HH55OH (ethanol) What is the OH (ethanol) What is the mass percent of Carbon in this mass percent of Carbon in this compound?compound? What is mass of 1 mole of ethanol?What is mass of 1 mole of ethanol? AnswerAnswer What is the mass percent of carbon in What is the mass percent of carbon in

ethanol?ethanol? AnswerAnswer Now do Hydrogen and OxygenNow do Hydrogen and Oxygen AnswersAnswers

6.6 Formulas of Compounds6.6 Formulas of Compounds

►Empirical Formula – simplest formula and Empirical Formula – simplest formula and expresses the smallest whole-number expresses the smallest whole-number ratio of the atoms presentratio of the atoms present CC66HH1212OO66 and C and C44HH88OO44 both have the same both have the same

empirical formula CHempirical formula CH22OO►Molecular Formula – the actual formula of Molecular Formula – the actual formula of

a compound that tells you the a compound that tells you the composition of the molecules that are composition of the molecules that are presentpresent CC66HH1212OO66 is the molecular formula for glucose is the molecular formula for glucose

6.7 Calculation of Empirical 6.7 Calculation of Empirical FormulasFormulas

► Important to learn the chemical formula Important to learn the chemical formula of a new compoundof a new compound First figure out relative massesFirst figure out relative masses Second convert masses to number of molesSecond convert masses to number of moles Divide by the smallest number of moles foundDivide by the smallest number of moles found Multiply the numbers from 3Multiply the numbers from 3rdrd step by smallest step by smallest

number that will make them all whole number that will make them all whole numbers, this is the empirical formulanumbers, this is the empirical formula

Example ProblemExample Problem► An oxide of aluminum is formed by the An oxide of aluminum is formed by the

reaction of 4.151 g of aluminum with 3.692 g reaction of 4.151 g of aluminum with 3.692 g of oxygen. What is the empirical formula of of oxygen. What is the empirical formula of the compound formed?the compound formed? Relative Masses must be converted to molesRelative Masses must be converted to moles

►4.151 g Al x 1 mol Al/ 26.98 g Al = 0.1539 mol of Al atoms4.151 g Al x 1 mol Al/ 26.98 g Al = 0.1539 mol of Al atoms►3.692 g O x 1 mol O/ 16.oo g O= 0.2308 mol of O atoms3.692 g O x 1 mol O/ 16.oo g O= 0.2308 mol of O atoms

Divide by smallest number of molesDivide by smallest number of moles►0.1539 mol Al / 0.1539 = 1.000 mol Al0.1539 mol Al / 0.1539 = 1.000 mol Al►0.2308 mol O / 0.1539 = 1.500 mol O0.2308 mol O / 0.1539 = 1.500 mol O

Multiply by smallest number to make them all Multiply by smallest number to make them all wholewhole►1.500 O x 2 = 3 O atoms1.500 O x 2 = 3 O atoms►1.000 Al x 2 = 2 Al atoms1.000 Al x 2 = 2 Al atoms

So formula is AlSo formula is Al22OO33

Practice ProblemPractice Problem

► In a lab experiment it was observed that In a lab experiment it was observed that 0.6884 g of lead combines with 0.2356 g of 0.6884 g of lead combines with 0.2356 g of chlorine to form a binary compound. What is chlorine to form a binary compound. What is the empirical formula of this compound?the empirical formula of this compound? Step 1 Relative MassesStep 1 Relative Masses Step 2 Convert to molesStep 2 Convert to moles Step 3 Divide by smallest numberStep 3 Divide by smallest number Step 4 Multiply by smallest number that makes Step 4 Multiply by smallest number that makes

them all whole numbersthem all whole numbers► Rules apply no matter how many elements Rules apply no matter how many elements

are in the compoundare in the compound

Calculation of Molecular Calculation of Molecular FormulasFormulas

►Must know percent composition and Must know percent composition and the molar massthe molar mass

► It’s always a multiple of the empirical It’s always a multiple of the empirical formula (Empirical Formula)formula (Empirical Formula)nn = = molecular formula or Empirical molecular formula or Empirical Formula x n = molecular formulaFormula x n = molecular formula

►Compare the empirical formula to Compare the empirical formula to molar massmolar mass

ExampleExample► If Empirical formula is PIf Empirical formula is P22OO55 and molar and molar

mass is 283.88 g, what is compound’s mass is 283.88 g, what is compound’s molecular formula?molecular formula? Know we have 2 moles P and 5 moles OKnow we have 2 moles P and 5 moles O

►2 mol P = 2 x 30.97 g = 61.94 g2 mol P = 2 x 30.97 g = 61.94 g►5 mol O = 5 x 16.00 g = 80.00 g5 mol O = 5 x 16.00 g = 80.00 g►1 mol P1 mol P22OO55 = 141.94 g = 141.94 g

Know empirical formula x n = molecular Know empirical formula x n = molecular formula and that the molecular formula = formula and that the molecular formula = molar massmolar mass

This tells us n = molar mass/ empirical This tells us n = molar mass/ empirical formulaformula

So 283.88 g / 141.94 = n = 2So 283.88 g / 141.94 = n = 2 So (PSo (P22OO55))2 2 means Pmeans P44OO1010

Practice ProblemPractice Problem

►What are the empirical and molecular What are the empirical and molecular formulas of a compound with these formulas of a compound with these percent compositions 71.65% Cl, percent compositions 71.65% Cl, 24.27% C, 4.07% H and a molar mass 24.27% C, 4.07% H and a molar mass of 98.96 g?of 98.96 g?