Chapter 6Chapter 6

Chemical QuantitiesChemical Quantities

HomeworkHomework

RecommendedRecommended ““Questions” 1-15 (odd)Questions” 1-15 (odd)

RequiredRequired ““Problem” 17-97 (odd)Problem” 17-97 (odd) ““Cumulative Problems” 99-123 (odd)Cumulative Problems” 99-123 (odd)

RecommendedRecommended ““Highlight Problems” 125, 127Highlight Problems” 125, 127

Counting Particles By WeighingCounting Particles By Weighing If a person requests 500 quarter inch If a person requests 500 quarter inch

hexagonal nuts for purchasehexagonal nuts for purchase How would you count 500 hex nuts?How would you count 500 hex nuts?

a)a) You could count the hex nuts individually You could count the hex nuts individually

b)b) Or, count the number of hex nuts by weightOr, count the number of hex nuts by weight First, find the average weight by weighing out 10 First, find the average weight by weighing out 10

hex nuts and obtaining the total weight (10 hex hex nuts and obtaining the total weight (10 hex nuts weigh 105 g) nuts weigh 105 g)

nuthexgnutshex

g/5.10

10

105 Average

Weight

Counting Particles By WeighingCounting Particles By Weighing

What size (wt.) will What size (wt.) will contain 500 hex contain 500 hex nuts?nuts?

Calculate what Calculate what weight will contain weight will contain 500 hex nuts500 hex nuts

So, weigh out 5.25 So, weigh out 5.25 kg of the hex nuts kg of the hex nuts

gnuthex

gnutshex 5250

5.10500

Counting Atoms by the GramCounting Atoms by the Gram Many chemical calculations require counting Many chemical calculations require counting

atoms and moleculesatoms and molecules There are two common methods for specifying the There are two common methods for specifying the

quantity of material in a sample of a substance: in quantity of material in a sample of a substance: in terms of mass and in terms of amountterms of mass and in terms of amount

It is difficult to do chemical calculations in terms It is difficult to do chemical calculations in terms of atoms or formula unitsof atoms or formula units

Since atoms are so small, extremely large Since atoms are so small, extremely large numbers are needed in calculationsnumbers are needed in calculations

Need to use a special counting unit just as used Need to use a special counting unit just as used for other itemsfor other items A A reamream of paper of paper One One dozendozen donuts donuts A A pairpair of shoes of shoes

Counting Atoms by the GramCounting Atoms by the Gram It is more convenient to use a special counting It is more convenient to use a special counting

unit for such large quantities of particles unit for such large quantities of particles MoleMole: A unit that contains : A unit that contains 6.022 6.022 хх 10 1023 23 objectsobjects It is used due to the extremely small size of It is used due to the extremely small size of

atoms, molecules, and ionsatoms, molecules, and ions 6.022x106.022x102323 particles in 1 mole particles in 1 mole Called Called Avogadro’s NumberAvogadro’s Number

Periodic TablePeriodic Table The The average atomic massaverage atomic mass inin amuamu (one atom)(one atom) The The weight of 1 moleweight of 1 mole of the element in of the element in gramsgrams Avogadro’s number provides the connecting Avogadro’s number provides the connecting

relationship between molar masses and relationship between molar masses and atomic massesatomic masses

Converting Between Moles and Converting Between Moles and Number of AtomsNumber of Atoms

Avogadro’s number relates the Avogadro’s number relates the number of particles present to moles number of particles present to moles presentpresent

Use when solving chemical Use when solving chemical calculations involving atoms, calculations involving atoms, molecules, or ions present in a given molecules, or ions present in a given amount of a substance.amount of a substance.

1 mole

Number of Atoms(Particles)

Avogadro’sNumber

Moles of substance

6.022 × 10232 moles 1.204 × 102410 moles 6.022 × 1024

Converting Between Moles and Converting Between Moles and Number of MoleculesNumber of Molecules

How many molecules of bromine are present in How many molecules of bromine are present in 0.045 mole of bromine gas? 0.045 mole of bromine gas?

223

2

2

223

10022.6

10022.6

Brmolecules

Brmoland

Brmol

Brmolecules

223

2 Br molecules 10022.6Br mol 1

222 Br molecules 107.2

Given: 0.045 mol Br2 Find: molecules of Br2

Avogadro’s number

2

223

2 10022.6045.0

Brmol

BrmoleculesBrmol

Conversion factors:

Equality:

Solution:

Converting Between Grams and Moles of an Element:Converting Between Grams and Moles of an Element:The Atomic Mass UnitThe Atomic Mass Unit

Atoms are too small to conveniently count Atoms are too small to conveniently count individuallyindividually

The usual standards of mass (g or lbs.) are not The usual standards of mass (g or lbs.) are not convenient for use with atomsconvenient for use with atoms

To avoid such small numbers scientists use a To avoid such small numbers scientists use a mass value for atoms relative to a standard mass value for atoms relative to a standard instead of an actual mass valueinstead of an actual mass value

The standard is the atomic mass unit (amu) with The standard is the atomic mass unit (amu) with all masses measured relative to carbon-12all masses measured relative to carbon-12

The mass of this isotope is set at 12.00 amuThe mass of this isotope is set at 12.00 amu

The mass of one atom of uranium-238 = 3.95 × 10-22 g

Converting Between Grams and Moles of an Element:Converting Between Grams and Moles of an Element: The Atomic Mass UnitThe Atomic Mass Unit

One amu is defined as One amu is defined as the mass of one-the mass of one-twelfth the mass of a twelfth the mass of a carbon-12 atom carbon-12 atom (nearly equal to one (nearly equal to one proton)proton)

The masses of all other The masses of all other atoms are determined atoms are determined relative to carbon-12relative to carbon-12

The masses of these The masses of these particles in amu brings particles in amu brings the mass of the proton the mass of the proton and neutron close to 1.and neutron close to 1.

g 101.66amu 1 24

Converting Between Grams and Moles of an Element:Converting Between Grams and Moles of an Element: The Atomic Mass UnitThe Atomic Mass Unit

Atomic masses are Atomic masses are determined on a determined on a relative scalerelative scale

The standard scale The standard scale references the references the carbon-12 isotope carbon-12 isotope = 12.00 amu= 12.00 amu

All other atomic All other atomic masses are masses are determined determined relativerelative to carbon-12to carbon-12

Converting Between Grams and Moles of an Element: Converting Between Grams and Moles of an Element:

The Atomic Mass UnitThe Atomic Mass Unit Atoms of the same Atoms of the same

element do not element do not necessarily have necessarily have exactly the same massexactly the same mass

Each element exists as Each element exists as a mixture of isotopes a mixture of isotopes so use an so use an average average massmass

To calculate the To calculate the average mass of a average mass of a sample of atomssample of atoms Use a “weighted Use a “weighted

average” for the average” for the atomic massatomic mass

Number on the bottom Number on the bottom of each square in the of each square in the periodic table is the periodic table is the average weightaverage weight of the of the element (element (in amuin amu))

Converting Between Grams and Moles of an Element: Converting Between Grams and Moles of an Element:

The Atomic Mass UnitThe Atomic Mass Unit

Using Atomic Mass to Count AtomsUsing Atomic Mass to Count Atoms Calculating the number of atoms in a specific Calculating the number of atoms in a specific

massmass If you have a sample of an element, can If you have a sample of an element, can

calculate the number of atoms in that samplecalculate the number of atoms in that sample From the atomic mass per one atom a From the atomic mass per one atom a

conversion factor can be madeconversion factor can be made For example: One nitrogen atom has an atomic For example: One nitrogen atom has an atomic

mass of 14.01 amumass of 14.01 amu

amu14.01

atomN1and

atomN1

amu14.01

amu14.01atomN1

Calculating Atomic Mass in amuCalculating Atomic Mass in amu(Sample Calculation)(Sample Calculation)

Calculate the mass (Calculate the mass (in amuin amu) of 1.0 ) of 1.0 хх 101044 carbon atoms carbon atoms

amu12.01

atomC1

atom C1

amu12.01and

C atoms 101.0 4

amu 12.01= C atom 1

amu101.2 5

1) Given:

2) Plan: Convert from atoms to amu

3) CF

atomC1

amu12.01atomC101.0 4

4) Set Up Problem

Find: total mass of C atoms (amu)

Converting Between Grams and Converting Between Grams and Moles of an ElementMoles of an Element

The molar mass of an element (monatomic) The molar mass of an element (monatomic) is the mass of one mole of atoms of the is the mass of one mole of atoms of the elementelement

The molar mass of The molar mass of one moleone mole of carbon-12 of carbon-12 atoms 12.01 gatoms 12.01 g

The molar mass (in grams) of any element The molar mass (in grams) of any element has the same numerical value as the atomic has the same numerical value as the atomic mass (in amu) of the same elementmass (in amu) of the same element

This mass contains 6.022 This mass contains 6.022 хх 10 1023 23 particles of particles of that elementthat element

Use the periodic table to obtain the molar Use the periodic table to obtain the molar mass of any elementmass of any element

Converting Between Grams and Converting Between Grams and Moles of an ElementMoles of an Element

When the number of grams (weighed out) of When the number of grams (weighed out) of a substance equals the formula mass of that a substance equals the formula mass of that substance, substance, Avogadro’s numberAvogadro’s number of atoms or of atoms or molecules of that substance are presentmolecules of that substance are present

Formula mass of water molecule

= 18.02 amu

Formula mass ofI2 molecule

= 253.8 amu

Molar mass of water = 18.02 g

Molar mass ofI2 = 253.8 g

Converting Between Grams and Converting Between Grams and Moles of an ElementMoles of an Element

The three quantities most often The three quantities most often calculated in chemical problems calculated in chemical problems Number of particlesNumber of particles Number of moles Number of moles Number of gramsNumber of grams

Using molar mass as a conversion Using molar mass as a conversion factor is one of the most useful in factor is one of the most useful in chemistrychemistry Can be used for gram to mole and mole Can be used for gram to mole and mole

to gram conversions to gram conversions

Converting Between Grams and Converting Between Grams and Moles of an ElementMoles of an Element

The The molar massmolar mass provides a relationship provides a relationship between the number of grams of an between the number of grams of an element and the number of moles of the element and the number of moles of the same elementsame element

The molar mass becomes part of the The molar mass becomes part of the conversion factor used to convert from conversion factor used to convert from grams to moles or moles to gramsgrams to moles or moles to gramsSolution map

Moles of Element

Molar Mass

Grams of Element

Converting Between Grams of an Converting Between Grams of an Element and Number of ParticlesElement and Number of Particles

The grams, moles, and number of particles of a substance The grams, moles, and number of particles of a substance are interrelated through the conversion factors of molar are interrelated through the conversion factors of molar mass and Avogadro’s number mass and Avogadro’s number

These two relationships can be combined into a single These two relationships can be combined into a single diagram which is useful to convert grams of a substance diagram which is useful to convert grams of a substance to the number of particles (atoms) of a substanceto the number of particles (atoms) of a substance

Particles of substance

Moles of substance

Grams of substance

Molar Mass

Moles of substance

Avogadro’sNumber

Converting Between Grams of an Element Converting Between Grams of an Element and Number of Atoms (Sample Calculation)and Number of Atoms (Sample Calculation)

How many lead atoms are there in a lead How many lead atoms are there in a lead weight with a mass of 1.2 g? weight with a mass of 1.2 g?

Pbmol1

atomsPb10 6.022

Pb g 207.19

Pb mol 1Pb g 1.2 23

Given: 1.2 g Pb Find: Pb atoms

Conversion Factors

1.2 g Pb number of Pb atomsmol PbAvog. NumberMolar mass

1 mol Pb= 207.19 g 1 mol Pb= 6.022 × 1023 Pb atoms

Solution Map

3.5 × 1021 Pb atoms

Solution

Converting Between Grams and Moles Converting Between Grams and Moles of a Compoundof a Compound

The molar mass provides a relationship The molar mass provides a relationship between the number of grams of a between the number of grams of a substance and the number of moles of the substance and the number of moles of the same substancesame substance

The molar mass of any substance is The molar mass of any substance is always numerically equal to its formula always numerically equal to its formula mass in amumass in amuSolution map

Grams of substance

Molar Mass

Moles of substance

Converting Between Grams and Moles Converting Between Grams and Moles of a Compoundof a Compound

Calculate the mass (in grams) of 0.555 Calculate the mass (in grams) of 0.555 mol of Almol of Al22OO33

Given: 0.555 mol Al2O3 Find: g Al2O3

Conversion Factors

mol Al2O3 g Al2O3

Molar mass

1 mol Al2O3 = 101.96 g

Solution Map

56.6 g Al2O3

Solution

OAl mol 1

g 101.96OAl mol 0.555

32

32

Converting Between Grams of an Converting Between Grams of an Compound and Number of MoleculesCompound and Number of Molecules

The molar mass provides a relationship between the The molar mass provides a relationship between the number of grams of a substance and the number of number of grams of a substance and the number of moles of the same substancemoles of the same substance

Avogadro’s number provides a relationship between the Avogadro’s number provides a relationship between the number of particles of a substance and the number of number of particles of a substance and the number of moles of the same substancemoles of the same substance

These two relationships can be combined into a single These two relationships can be combined into a single diagram which is useful to convert grams of a substance diagram which is useful to convert grams of a substance to the number of particles (molecules) of a substanceto the number of particles (molecules) of a substance

Particles of substance

Moles of substance

Avogadro’sNumber

Moles of substance

Grams of substance

Molar Mass

Converting Between Grams of an Element Converting Between Grams of an Element and Number of Molecules (Sample and Number of Molecules (Sample

Calculation)Calculation) How many molecules of COHow many molecules of CO22 are in a 5.61 g are in a 5.61 g

sample of COsample of CO22? ?

2

223

2

22

OCmol1

moleculesCO10 6.022

CO g 44.01

CO mol 1CO g 5.61

Given: 5.61 g CO2 Find: CO2 molecules

Conversion Factors

5.61 g CO2 number of CO2 moleculesmol CO2

Avog’s No.Molar mass

1 mol CO2= 44.01 g1 mol CO2 = 6.022 × 1023 CO2 molec

Solution Map

7.68 × 1022 CO2 molecules

Solution

Converting Between Grams of an Element Converting Between Grams of an Element and Number of Molecules (Sample and Number of Molecules (Sample

Calculation)Calculation) Ethylene glycol has the chemical formula:Ethylene glycol has the chemical formula: C C22HH66OO22..

Find the mass of 1.24 × 10Find the mass of 1.24 × 102424 molecules of ethylene molecules of ethylene glycol (antifreeze)glycol (antifreeze)

262

262

26223

26226224

OHCmol1

OHC g 62.05

OHC molecules 10022 6.

OHC mol 1OHC molecules 10 1.24

Given: 1.24 × 1024 molecules C2H6O2 Find: g of C2H6O2

Conversion Factors

mol C2H6O2

molar massAvog No.

1 mol C2H6O2= 62.05 g1 mol C2H6O2= 6.022 × 1023 C2H6O2 molec

Solution Map

128 g C2H6O2

Solution

molecules C2H6O2 g of C2H6O2

Chemical Formulas as Conversion FactorsChemical Formulas as Conversion Factors

The subscripts in a chemical formula indicate the The subscripts in a chemical formula indicate the number of atoms number of atoms of each element present in a single of each element present in a single molecule or formula unit of a compoundmolecule or formula unit of a compound

The subscripts in a chemical formula can also indicate The subscripts in a chemical formula can also indicate the the number of molesnumber of moles of atoms of each element present of atoms of each element present in in one moleone mole of a compound of a compound

e.g. In one e.g. In one moleculemolecule of glucose (C of glucose (C66HH1212OO66) there are ) there are 66 atoms of carbon, atoms of carbon, 1212 atoms of hydrogen, and atoms of hydrogen, and 66 atoms of atoms of oxygenoxygen

e.g. In one e.g. In one mole mole of glucose (Cof glucose (C66HH1212OO66) there are ) there are 66 moles moles of carbon, of carbon, 1212 moles of hydrogen, and moles of hydrogen, and 66 moles of oxygen moles of oxygen

Converting Between Moles of a Converting Between Moles of a Compound and Moles of a Constituent Compound and Moles of a Constituent

ElementElement Some problems require a calculation of the moles of a Some problems require a calculation of the moles of a

particular element in a chemical formula of the compound particular element in a chemical formula of the compound The subscript of the particular element in the compound The subscript of the particular element in the compound

becomes part of the conversion factor used to convert becomes part of the conversion factor used to convert moles of compound to moles of element within the moles of compound to moles of element within the compound compound

Solution map

Moles of compound

Formulasubscript

Moles of Element

(within compound)

Converting Between Moles of a Compound and Converting Between Moles of a Compound and Moles of a Constituent Element (Sample Moles of a Constituent Element (Sample

Calculation)Calculation) How many moles of carbon atoms are present in How many moles of carbon atoms are present in

1.85 moles of glucose?1.85 moles of glucose?

6126

6126

OHCmol1

C mol 6OHC mol 1.85

Given: 1.85 mol C6H12O6 Find: mol C

Conversion Factors

FormulaSubscript

1 mol C6H12O6= 6 mol C

Solution Map

11.1 mol C

Solution

mol C6H12O6 mol C

Converting Between Grams of a Compound and Converting Between Grams of a Compound and Moles of a Constituent Element (Sample Moles of a Constituent Element (Sample

Calculation)Calculation) International Foods Coffee contains 3.0 mg of International Foods Coffee contains 3.0 mg of

sodium chloride per cup of coffee. How many sodium chloride per cup of coffee. How many moles of sodium are in each cup of coffee?moles of sodium are in each cup of coffee?

NaCl mol 1

Na mol1

NaCl g 58.44

NaCl mol1

NaClmg1

NaCl g10NaCl mg 3.0 -3

Given: 3.0 mg NaCl Find: mol Na

Conversion Factors 1 mg NaCl = 10-3 g NaCl

Solution Map

5.13 × 10-5 mol Na

Solution

mg NaCl

1 mol NaCl= 58.44 g NaCl1 mol Na =1 mol NaCl

g NaCl mol NaCl mol NaNaClg 58.44

NaClmol 1NaClmg

NaClg 10-3

1 NaClmol1

Namol1

Converting Between Grams of a Compound and Converting Between Grams of a Compound and Grams of a Constituent ElementGrams of a Constituent Element

In chemical problems, the three quantities most often calculated In chemical problems, the three quantities most often calculated for a substance are number of particles (atoms, molecules, or for a substance are number of particles (atoms, molecules, or formula units), number of moles, number of gramsformula units), number of moles, number of grams

All of these quantities are interrelated and the conversion factors All of these quantities are interrelated and the conversion factors which deal with these relationships are Avogadro’s number, which deal with these relationships are Avogadro’s number, molar mass, and chemical formula subscriptsmolar mass, and chemical formula subscripts

The three concepts can be combined into a single diagram that The three concepts can be combined into a single diagram that is useful for problem solvingis useful for problem solving

0.003 g NaCl

1 mol Na = 1 mol NaCl

1 mol = 58.44 g NaCl

mol Na mol NaCl

Converting Between Grams of a Compound and Converting Between Grams of a Compound and Grams of a Constituent ElementGrams of a Constituent Element

Throughout Chapter 6, there are calculations for which Throughout Chapter 6, there are calculations for which the initial information is given: moles, grams, or particles.the initial information is given: moles, grams, or particles.

Additional information is then requested concerning a Additional information is then requested concerning a component of that same substance (find): moles, grams, component of that same substance (find): moles, grams, or particlesor particles

Now, use concepts illustrated in previous problems to Now, use concepts illustrated in previous problems to solve problems in which a given mass of a sample solve problems in which a given mass of a sample substance is given and we need to calculate the mass of substance is given and we need to calculate the mass of a constituent elementa constituent element

Grams of compound

Molarmass

Moles of compound

Moles of element

Grams of element

Formulasubscript

Molar Mass

CConverting Between Grams of a Compound onverting Between Grams of a Compound and Grams of a Constituent Elementand Grams of a Constituent Element

Caffeine is the stimulant found in coffee and tea. It has the Caffeine is the stimulant found in coffee and tea. It has the chemical formula Cchemical formula C88HH1010NN44OO22. How many grams of carbon are . How many grams of carbon are present in a 50.0 g sample? present in a 50.0 g sample?

Cmol 1

C g 12.01

ONHCmol 1

C mol 8

ONHCg 194

ONHC mol 1 ONHC g 50.0

2410824108

2410824108

Given: 50.0 g C8H10N4O2 Find: g of C

Conversion Factors

Molar Mass

1 mol C8H10N4O2= 194 g C8H10N4O2

Solution Map

24.8 g C

Solution

50.0 g C8H10N4O2 mol of C

8 mol C= 1 mol C8H10N4O2

1 mol C= 12.01 g C

mol C8H10N4O2 g of C

Mol-mol ratio

Molar Mass

Mass Percent Composition of Mass Percent Composition of CompoundsCompounds

The mass percent of an element may be The mass percent of an element may be obtained by:obtained by:

Decomposition of a sample of a known massDecomposition of a sample of a known mass Then, the masses of the constituent elements Then, the masses of the constituent elements

present in each sample are then obtainedpresent in each sample are then obtained Each element’s mass, and the original sample Each element’s mass, and the original sample

mass, is used to calculate the percent compositionmass, is used to calculate the percent composition Mass data obtained from the combination of Mass data obtained from the combination of

elements which combine to form a compoundelements which combine to form a compound Each element’s mass, and the product’s mass, is Each element’s mass, and the product’s mass, is

used to calculate the percent composition used to calculate the percent composition

Mass Percent Composition of Mass Percent Composition of CompoundsCompounds

The composition of compounds is determined by The composition of compounds is determined by their formulas (e.g. the formula of carbon dioxide their formulas (e.g. the formula of carbon dioxide tells us it’s composed of carbon and oxygen (2 tells us it’s composed of carbon and oxygen (2 oxygen atoms and 1 carbon atom)oxygen atoms and 1 carbon atom)

Another way to describe the composition of Another way to describe the composition of compounds is in terms of the mass percentcompounds is in terms of the mass percent of of each element in the compoundeach element in the compound

The The mass percentmass percent composition is the percent by composition is the percent by mass of each element in a compoundmass of each element in a compound

%100compoundtheofsampleofMass

compound theof sample ain of Masselement ofpercent Mass X

X

Mass Percent Composition from a Mass Percent Composition from a Chemical FormulaChemical Formula

The The mass percentmass percent composition is the percent by composition is the percent by mass of each element in a compoundmass of each element in a compound

To determine a compound’s mass percent To determine a compound’s mass percent composition: composition: its chemical formula is neededits chemical formula is needed the molar mass of each element that the molar mass of each element that

composes the compound composes the compound the subscript numbers in the formula are the subscript numbers in the formula are

equal to the number of moles of each element equal to the number of moles of each element in in oneone mole of the compound (its molar mass) mole of the compound (its molar mass)

%100compoundofmol1ofMass

compound ofmol1in elementof Masselement ofpercent Mass X

X

Mass Percent Composition of Mass Percent Composition of CompoundsCompounds

Percent numbers are always in terms of parts Percent numbers are always in terms of parts per hundredper hundred

For example, the percent composition of For example, the percent composition of water is 88.81% oxygen and 11.19% water is 88.81% oxygen and 11.19% hydrogenhydrogen The percent composition for each element is The percent composition for each element is

determined by dividing the mass contributed by determined by dividing the mass contributed by each element, by the molar mass and multiplied by each element, by the molar mass and multiplied by 100100

The sum of the percents of all of the elements in a The sum of the percents of all of the elements in a compound equals 100%compound equals 100%

If the mass percent of a particular element in a If the mass percent of a particular element in a compound is known, you can calculate the amount compound is known, you can calculate the amount of the element present in any sample size of that of the element present in any sample size of that compoundcompound

Mass Percent Composition from a Mass Percent Composition from a Chemical FormulaChemical Formula

Calculate the mass percent composition of each Calculate the mass percent composition of each element in NHelement in NH44OH?OH?

g05.35)g00.16(1g) 1.008 (5 g)14.01(1

%100g35.05

g14.01 :N %100

g35.05

g5.040 :H %100

g35.05

g16.00 :ON 39.97% H 14.38% O 45.65%

Determine the contribution of each element

Molar mass

100%compoundofmol1ofMass

compound of mol 1 in of Mass element of percent Mass X

X

Determine the mass of 1 mol of compound

16.00 g O14.01 g N 5.040 g H

Calculating Empirical Formulas for Calculating Empirical Formulas for CompoundsCompounds

Chemical formulas are determined by Chemical formulas are determined by calculation using experimentally obtained calculation using experimentally obtained informationinformation

Depending on the amount of experimental Depending on the amount of experimental data available, two types of formulas can data available, two types of formulas can be obtainedbe obtained Empirical formulas (simplest formula)Empirical formulas (simplest formula) Molecular formulas (true formula)Molecular formulas (true formula)

Calculating Empirical Formulas for Calculating Empirical Formulas for CompoundsCompounds

The The empirical formula empirical formula gives the simplest ratio gives the simplest ratio of elements in a compoundof elements in a compound

It uses the smallest possible whole number It uses the smallest possible whole number ratio of atoms present in a formula unit of a ratio of atoms present in a formula unit of a compoundcompound

The subscripts in the formula cannot be The subscripts in the formula cannot be reduced any further to a simpler ratio by reduced any further to a simpler ratio by division with a small integerdivision with a small integer

If the percent composition of a compound is If the percent composition of a compound is known, an empirical formula can be calculatedknown, an empirical formula can be calculated

Calculating an Empirical Formula from Calculating an Empirical Formula from Experimental Data Experimental Data

To Determine the empirical formula:To Determine the empirical formula:1)1) Write down or calculate the given number of grams Write down or calculate the given number of grams

of each element present in a sample of the of each element present in a sample of the compound compound

2)2) Convert the mass of each element to moles of Convert the mass of each element to moles of element using the appropriate molar masselement using the appropriate molar mass

3)3) Using the calculated moles of each of the elements: Using the calculated moles of each of the elements: Write down a Write down a pseudoformula pseudoformula using the moles of using the moles of

each element as a subscripteach element as a subscript

4)4) Find the lowest whole number mole ratioFind the lowest whole number mole ratio Divide all the subscripts in the formula by the Divide all the subscripts in the formula by the

smallest subscriptsmallest subscript

Obtaining an Empirical Formula from Experimental Data Obtaining an Empirical Formula from Experimental Data

Converting Decimal Numbers to Whole Converting Decimal Numbers to Whole NumbersNumbers

The subscripts in a “true” formula are The subscripts in a “true” formula are expressed as whole numbers, not as decimalsexpressed as whole numbers, not as decimals

The resulting decimal numbers from a The resulting decimal numbers from a calculation represent each element’s subscriptcalculation represent each element’s subscript

Arithmetic may need to be applied to convert Arithmetic may need to be applied to convert decimals to whole numbersdecimals to whole numbers

If the number(s) are If the number(s) are NOTNOT whole numbers, whole numbers, multiply each number by the same small multiply each number by the same small integer (2, 3, 4, 5, or 6) until a whole number is integer (2, 3, 4, 5, or 6) until a whole number is obtained for each subscript numberobtained for each subscript number

Obtaining an Empirical Formula Obtaining an Empirical Formula from Experimental Datafrom Experimental Data

Lactic acid has a molar mass of 90.08 Lactic acid has a molar mass of 90.08 g and has this percent composition:g and has this percent composition:

40.0% C, 6.71% H, 53.3% O40.0% C, 6.71% H, 53.3% O What is the What is the empiricalempirical and and molecularmolecular

formula of lactic acid?formula of lactic acid? Assume a 100.0 g sample sizeAssume a 100.0 g sample size

Convert percent numbers to gramsConvert percent numbers to grams

Obtaining an Empirical Formula from Obtaining an Empirical Formula from Experimental DataExperimental Data

CmolCg

CmolCg 33.3

0.120.40 Hmol

Hg

HmolHg 66.6

008.171.6 Omol

Og

OmolOg 33.3

00.163.53

Empirical formula is

CH2O

Step 3: Pseudoformula C3.33H6.66 O3.33

Step 2: Convert mass to moles

Step 4: Divide by smallest subscript

Step 1: Mass of each element in a 100 g

sample of the compound40.0 g C 6.71 g H 53.3 g O

00.13.33

3.33 :O 00.1

3.33

3.33 :C 00.2

3.33

6.66 :H

Calculating Molecular Formulas for Calculating Molecular Formulas for CompoundsCompounds

Molecular formulas Molecular formulas are the formulas of are the formulas of moleculesmolecules

They illustrate all of the atoms of each element They illustrate all of the atoms of each element in the moleculein the molecule

It does not necessarily have the smallest set of It does not necessarily have the smallest set of subscript numbers: It can be the same as or a subscript numbers: It can be the same as or a whole number multiplier of its empirical whole number multiplier of its empirical formulaformula

To determine a molecular formula, the To determine a molecular formula, the compound’s molecular mass is neededcompound’s molecular mass is needed

Relating Empirical and Molecular FormulasRelating Empirical and Molecular Formulas Empirical FormulaEmpirical Formula

Smallest possible set of subscript numbers Smallest possible set of subscript numbers Smallest whole number ratioSmallest whole number ratio All ionic compounds are given as empirical formulasAll ionic compounds are given as empirical formulas

Molecular FormulasMolecular Formulas The actual formulas of moleculesThe actual formulas of molecules When an empirical formula is known, the molar mass When an empirical formula is known, the molar mass

is needed to determine if the empirical formula and is needed to determine if the empirical formula and the molecular formula are the samethe molecular formula are the same

It shows all of the atoms present in a moleculeIt shows all of the atoms present in a molecule It may be the same as the EF or a whole-number It may be the same as the EF or a whole-number

multiple of its EFmultiple of its EF

Molecular formula = n х Empirical formula

Calculating Molecular Formulas for Calculating Molecular Formulas for CompoundsCompounds

nn represents a whole number multiplier represents a whole number multiplier from 1 to as large as necessaryfrom 1 to as large as necessary

Calculate the empirical formula and the Calculate the empirical formula and the mass of the empirical formulamass of the empirical formula

Divide the given molecular mass by the Divide the given molecular mass by the calculated empirical formula molar masscalculated empirical formula molar mass Answer is a Answer is a whole number multiplierwhole number multiplier

)/(

)/(

molgmassmolarformulaEmpirical

molgmassMolarn

Relating Empirical and Relating Empirical and Molecular FormulasMolecular Formulas

Multiply each Multiply each subscriptsubscript in the empirical in the empirical formula by the formula by the whole number multiplierwhole number multiplier to to get the molecular formulaget the molecular formula

For molecular compounds, the two types For molecular compounds, the two types of formulas may or may not be the sameof formulas may or may not be the same

Relating Empirical and Molecular Relating Empirical and Molecular FormulasFormulas

For ionic compounds For ionic compounds The empirical and molecular formula are the The empirical and molecular formula are the

samesame For molecular compounds For molecular compounds

The two types of formulas may be the The two types of formulas may be the same but not necessarilysame but not necessarily

Calculating Molecular Formulas for Calculating Molecular Formulas for CompoundsCompounds

What is the What is the molecularmolecular formula of lactic acid? formula of lactic acid? Obtain the value of Obtain the value of nn (whole number multiplier) (whole number multiplier) Multiply the empirical formula by the multiplierMultiply the empirical formula by the multiplier

)/(

)/(

molgmassmolarformulaempirical

molgmassmolarn 3

/03.30

/08.90

molg

molg

Molecular formula = n х empirical formula

Molecular formula = 3 (CH2O) C3H6O

3

Empirical formula is

Empirical formula molar mass = 12.01 g + 2 (1.008 g) + 16.00 g = 30.03 g/mol

CH2O

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