ch. 3 the mole: relating the microscopic world of atoms to ... · 3.1 the mole conveniently links...

Ch. 3 The Mole: Relating the Microscopic World of Atoms to

Laboratory Measurements

Brady & Senese, 5th Ed.

2

Index

3.1 The mole conveniently links mass to number of atoms or molecules

3.2 Chemical formulas relate amounts of substances in a compound

3.3 Chemical formulas can be determined from experimental mass measurements

3.4 Chemical equations link amounts of substances in a reaction

3.5 The reactant in shortest supply limits the amount of product that can form

3.6 The predicted amount of product is not always obtained experimentally

3.1 The mole conveniently links mass to number of atoms or molecules 3

Particles Have Characteristics Masses

• The same mass may not represent the same number of molecules

• Suppose one rabbit has a mass of 250 g. What mass in kg would a case of 24 rabbits have?

1000gkg

rabbitg 250

rabbits 24

6.0 kg


Counting Atoms By Their Mass

• The mass of an atom is called its atomic mass

• Atomic mass provides a means to count atoms by measuring the mass of a sample

• The periodic table gives atomic masses of the elements in u per atom

• to reduce rounding errors, use the most precise values possible


Learning Check

• How many atoms of C are there in 3.5 × 108 u?

• What is the mass (in u) of 2.33 × 1016 atoms of H?

×u 12.0107

C atom 1u103.5

8

2.35 ×1016 u

atomic masses: C=12.0107 u; H=1.00794 u

×H atom 1

u .00794 1 atoms102.33

16

2.9 ×107


Your Turn!

Given that the atomic mass of Ba is 137.327u, what is the mass of 23 atoms of Ba?

A. 3.2×103 u

B. 3.2×10-4 u

C. 1.37×102 u

D. none of these


Your Turn!

A new element is discovered that has a mass of

3.2 ×102 u for15 atoms. What is the atomic mass?

A. 3.2 ×102

B. 0.047

C. 21

D. not enough information

E. None of these answers


Relationships

• 1.66×10-27 kg = 1 u (from the inside back cover of the book) may also be written as:

• 6.0223×1023 u = 1 g ( a form you will often use)

• We can use this as a conversion factor to convert between mass quantities in u, and those in g

grams (g)atomic mass units (u)

× u 106.0223g 1

23

×g 1

u 106.0223 23


Relationships

• Atomic Mass (AM) u = 1 particle

• We can use this as a conversion factor to convert between these quantities.

uAM particle 1

particleuAM

mass (u)particles


Learning Check

• How many u of Na are there in 55.2 kg Na?

• How many g Na are there in 3.2 x 1015 u of Na?

×

gu106.0223

kgg10

55.2kg233

××

u10 6.0223

1gu 103.2

23

15 5.3×10-9 g

3.32×1028 u


Your Turn!

Which of the following are not equivalent to a sample of 10.5×107 u of Cu?

A. 1.74×10-16 g

B. 1.65×106 atoms

C. 63.54 u

D. None of these


What Is The Formula Mass Of…?

• Ba3(PO4)2 :

• (NH4)2CO3:

atomic masses: Ba: 137.327(7)u; P:30.973761(2)u; O: 15.9994(3)u; H:1.00794u; N:14.00672u; C 12.0107(8)u

96.08603 u/fu

601.9261 u/fu


Relationships

• Formula mass (FM) u = 1 particle

• We can use this as a conversion factor to convert between these quantities.

u FM

particle 1

particleuFM

mass (u)particles


Counting Molecules By Their Masses

• The molecular mass allows counting of molecules by mass

• The molecular mass is the sum of atomic masses of the atoms in the compound’s formula

• Strictly speaking, ionic compounds do not have a molecular mass, we describe an analogous quantity- the formula mass- to cover all possibilities


Learning Check:

• How many molecules of CO2 are there in 3.5 ×108 u?

• What is the mass (in u) of 2.33 × 1016 molecules of H2 ?

×u 44.0095

CO molecule 1u103.5

6

28

4.70 ×1016 u

atomic masses: C=12.0107 u; H=1.00794 u; O=15.99943 u

×2

16

Hmolecule 1u .01588 2

atoms102.33

8.0 ×106 u


What Is a Mole?

• One mole of any substance contains the same number of units, called Avogadro’s number, N

• 1 mole formula units = 6.0223 x 1023 formula units

• It is a large quantity of particles because the particles described are so small.


Why is Molar Mass the Same as Formula Mass?

• suppose we start with 12.0107g of C. How many atoms of C are there?

• given that the atomic mass of C is 12.0107 u

• 12.0107 g C = 6.0223 x 1023 atoms

• thus for any substance, the formula mass (in g) corresponds to the same number of atoms, N

×u 12.0107

atom 1g

u106.022312.0107gC

23


Molar Mass

• One molecontains the same number of particles as the number of atoms in exactly 12 g of carbon-12

• The molar mass of a substance has the same numeric value as the formula mass

• The value is different because the units are different� Thus if the formula mass of Ba3(PO4)2 is 610.332 u/fu,

the molar mass of Ba3(PO4)2 is 610.332 g/mol


Relationships

• MM g = 1 mole

• Use this as a conversion factor to convert between these quantities

moleMass (g)

g MM

mole 1

mole 1

g MM


Learning Check: Converting Between Mass And MolesGiven that the molar mass of CO2 is 44.0098 g/mol

• What mass of CO2 is found in 1.55 moles?

• How many moles of CO2 are there in 10 g?

molg 44.0098

mol 1.55

g 44.0098

molg 10 0.2 mol

68.2 g


Your Turn!

What is the molar mass of Ca3(PO4)2 in g/mol?

Ca: 40.078 ; P: 30.973761 ; O:15.9994

A. 279.203

B. 215.205

C. 310.177

D. none of these


Your Turn!

What mass in g, of Ca3(PO4)2 (MM=310.1767) would a 3.2 mole sample have?

A. 1.0×10-3 g

B. 9.9×102 g

C. 6.0×1026 g

D. 1.6×10-21 g

E. None of these


Using Avogadro’s Number, N

• 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 mole

• The quantity, N, is Avogadro's number and is measured as 6.0223×1023


Relationships

• N particles = 1 mole

• We can use this as a conversion factor to convert between these quantities

particles

mole 1

N

mole

particles N

Molesparticle


Learning Check: Mole Conversions

• Calculate the formula units of Na2CO3 in 1.29 moles of Na2CO3

• How many moles of Na2CO3 are there in 1.15 x 105 formula units of Na2CO3?

×mol

fu 106.0223mol 1.29

23

××

fu106.0223mol

fu 101.15

235 1.91×10-19

mol

7.77×1023

fu


Relationships Between Quantities

moles

mass (u)

# particles

mass (g)

N

N

FM MM


Your Turn!

Which of the following is not a relationship, but is a sample size?

A. molar mass

B. Avogadro’s number

C. formula mass

D. Mass in u

E. None of these


Your Turn!

Given that you have a sample of 5.5 g Na2CO3 how many formula units are present?

A. 6.0×1023

B. 5.2×10-2

C. 3.2×10-23

D. 3.3×1024

E. None of these

moles

mass (u)

# particles

mass (g)

N

N

FM MM

Na: 22.989770 ; C: 12.011; O:15.9994

3.1×1022

3.2 Chemical formulas relate amounts of substances in a compound 29

Using The Chemical Formula

• To 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 Na2CO3


Learning Check:

• Calculate the number of moles of sodium in 2.53 moles of sodium carbonate

• Calculate the number of atoms of sodium in 2.53 moles of sodium carbonate

32CONa mol 1Na mol 2

mol 2.53 5.06 mol Na

×

Na mol 1

Na atoms 106.0223

CONa 1mol

Na mol 2mol 2.53

23

3 2

3.05×1024 atoms Na


Your Turn!

How many atoms of iron are in a 15.0 g sample of iron(III) oxide (MM 159.68859 g/mol)?

A. 1.13×1023

B. 9.39×10-2

C. 5.66×1022

D. 1.88×10-1

E. None of these

3.3 Chemical formulas can be determined from experimental mass measurements 32

Percent Composition

• Percent composition is a list of the mass percent of each element in a compound� Na2CO3 is

� 43.38% Na

� 11.33% C

� 45.29% O

• What is the sum of the percent composition of a compound?


Percent Composition: How Is It Calculated?

• What is the % C in CO2?

• Determine the molar mass of the compound� MM=44.00956 g/mol

• Multiply the ratio of the mass of the element to the molar mass of the compound by 100� (12.0107/44.009656)×100= 27.2911 %C

MM g/mol C:12.0107; O:15.99943


Learning Check

A sample was analyzed and found to contain 0.1417 g nitrogen and 0.4045 g oxygen. What is the percentage composition of this compound?

100 totalg 0.4045 0.1417g

N g 0.1417 ×

+

74.06% O25.94% N


Your Turn!

A 35.5 g sample is analyzed and found to contain 23.5% Si. What mass of Si is present in the sample?

A. 6.62×10-1 g

B. 8.88×101 g

C. 1.51×102 g

D. 8.34 g

E. None of these


Empirical vs. Molecular Formulas

• The empirical formula is the lowest whole number ratio of atoms in a compound

• Note that the molecular formula is a whole number multiple of the empirical formula.

C6H12O6

glucose

CH2O

C1x6H2x6 O1x6


Strategy

• Convert starting quantities to moles

• Divide all quantities by the smallest number of moles to get the smallest ratio of moles

• Convert any non-integers into integers� If any number ends in a common decimal equivalent of

a fraction, multiply by the least common denominator

� Otherwise, round the numbers to the nearest integers


Common Ratios And Their Decimal Equivalents

decimal Fraction equivalent

multiplier

.25 or .75

¼ or ¾

4

.3333 or

.6667

1/3 or 2/3 3

.50 ½ 2

For example:

5 4x 25.1 =


Learning Check:

• A 2.012 g sample of a compound contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical formula

1.4900.522mass(g)

ON

2 5integer ratio

14.00674 15.99943MM

0.037268 0.0931283mol

1 2.50lowest ratio


Determining The Multiplier, n

• Ratio of the molecular mass to the mass predicted by the empirical formula and round to an integer

• The actual molecule is larger by this amount� If the empirical formula is AxBy , the molecular formula

will be An×xBn×y

massformula empiricalmassformula molecular

n =

glucose 6 OCH g 30.0262OHC g 180.1572

n2

6126 ========


Example:

• The empirical formula of hydrazine is NH2, and its molecular mass is 32.0. What is its molecular formula?

• A substance is known to be 35.00% N, 5.05% H and 59.96% O. What is its EF? Determine the Molecular Formula if the MM of the compound is 80.06 g/mol

N2H4n=(32.0/16.02)=2

n=(80.06/80.043)=1

N2H4O3

EF: N2H4O3

MM: N:14.00674; H:1.00794; O:15.99943


Your Turn!

Given the composition analysis of lindane (a controversial pesticide ) what is its empirical formula?

A. C24H2Cl73

B. C2H2Cl2C. C142 HCl126

D. CHCl

E. None of these

73.14129%2.07943%24.77928%

ClHC


Your Turn!

We found that the empirical formula was CHCl. Given that the MM is 290.8316 g/mol, what is the molecular formula?

A. C6H6Cl6B. C8H17Cl5

C. C3H5Cl7

D. C5H18Cl7

E. none of these


Combustion Analysis:

• Empirical formulas may also be calculated indirectly

• When a compound made only from carbon, hydrogen, and oxygen burns completely in pure oxygen, only carbon dioxide and water are produced


Combustion Analysis:

Empirical formulas may be calculated from the analysis of combustion information� grams of C can be derived from amount of CO2

� grams of H can be derived from amount of H2O

� the mass of oxygen is obtained by difference:

� g O = g sample – ( g C + g H )


Learning Check:

The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO2 and 4.512 g of H2O. Calculate the empirical formula of the compound.

22

CO g 44.00956C g 12.0107

CO g .4067

H: 1.00794; C:12.0107; O: 15.99943

OH g 18.01531

H g 2.01588OH g 512.4

22 .504884 g H

2.02118 g C

5.217g- 2.02118 g C-.504884 g H= 2.69094 g O


Learning Check (con.):

Calculate the empirical formula of the compound.

H: 1.00794; C:12.0107; O: 15.99943

2.690940.5048842.02118mass

mol

12.0107

C

integer ratio

low ratio

15.999431.00794MM

OH

3 11

.500907 .16819.16828

2.97 11

CH3O


Your Turn!

Combustion analysis of 3.88 g of a compound containing C, H, and S reveals the following data. What is the empirical formula of the compound?

A. C6H5S

B. C9H2S

C. C5H5S

D. C3H9S2

E. None of these

1.59 g9.377 g

H2OCO2

3.4 Chemical equations link amounts of substances in a reaction 49

What Does The Balanced Equation Mean?

• 2CO(g) + O2(g)→2CO2(g)

• For every 2 CO reacted, 1 O2 is also reacted and 2 CO2 are also reacted


Using The Balanced Equation:

• The balanced equation gives the relationship between amounts of reactants used and amounts of products likely to be formed

• The numeric coefficient tells:� how many individual particles are needed in the

reaction on the microscopic level

� how many moles are necessary on the macroscopic level

� The stoichiometric coefficient


Stoichiometric Ratios

• Consider the reaction N2 + 3H2→ 2NH3

• What is the ratio between

• N2 and H2 ?� 1 mole N2: 3 mole H2

• N2 and NH3?� 1mole N2: 2 mole NH3

• H2 and NH3?� 3 mole H2 : 2 mole NH3


Learning Check:

• For the reaction N2 + 3 H2→ 2NH3, How many moles of N2 are used when 2.3 moles of NH3 are produced?

• If 0.575 mole of CO2 is produced by the combustion of propane, C3H8, how many moles of oxygen are consumed? The balanced equation is C3H8 + 5 O2→ 3 CO2 + 4 H2O

3

23

NH mol 2N mol 1

NH mol 2.3

2

22

CO mol 3

O mol 5CO mol 0.575 0.958 mol O2

1.2 mol N2


Learning Check

How many grams of Al2O3 are produced when 41.5 g Al react? 2Al(s) + Fe2O3(s)→ Al2O3(s) + 2 Fe(l)

MM (g/mol): Al: 26.9815; Al2O3:101.9613

32

3232

OmolAl 1

OgAl 101.9613

Al mol 2

OAl mol 1

Al g 26.9815

Al mol 1Al g 41.5

78.4 g Al2O3


Your Turn!

Given the reaction:

H2SO4 + 2KOH→2H2O + K2SO4,

How many moles of KOH are required to make 3.0 moles of K2SO4?

A. 3.0 moles

B. 6.0 moles

C. 1.5 moles

D. None of these


Your Turn!

Given the reaction:

H2SO4 + 2KOH→2H2O + K2SO4,

How many g of H2O (18.0153) would result from the complete reaction of 1.2 g H2SO4 (98.08)?

A. 2.4 g

B. 1.2 g

C. 0.60 g

D. 0.44 g

E. none of these


Balancing By Inspection

• Balance the most complex substance in the equation first

• Balance elements, H and O last• Use coefficients to adjust quantities, not subscripts • Some equations may be balanced using fractions,

but the most common approach allows only for integer coefficients

• If polyatomic ions remain intact in a reaction balance them as a group

• If you have an even/odd problem dilemma, multiply all previously balanced moieties by 2


Learning Check: Balance The Following:

____Ba(OH)2(aq)+____ Na2SO4(aq)→ ___BaSO4(s) + ____NaOH(aq)1 1 2

2___KClO3(S)→ ___KCl(s) +___ O2(g)2 3

1

___H3PO4(aq)+___ Ba(OH)2(aq)→ ___Ba3(PO4)2(s) + ___H2O(l)3 1 62

Your Turn!

Given the following reaction:

KCl + Hg2(NO3)2→KNO3 + Hg2Cl2 , when it is balanced, what is the coefficient for KCl?

A. 1

B. 2

C. 3

D. 4

E. none of these

3.5 The reactant in shortest supply limits the amount of product that can form 59

Limiting Reagent

• Consider the reaction of N2 with H2 to form NH3:

• N2(g) + 3H2(g)→ 2NH3(g)

• The stoichiometry suggests that for every mole of N2 we will need 3 moles of H2 to form 2 moles of NH3.

• So what happens if these proportions are not met? The reaction proceeds, to use up one of the reactants (the limiting reagent) and will not use all of the other reactant (it is in excess)


Limiting Reagents

• Note that in this reaction, some of the O2 is not consumed. This is because there is not enough CO to continue consuming the O2.

• Thus, CO is the limiting reagent.


Determining The Limiting Reagent (LR)

• There are several approaches to this. One method is to compare the quantities available to the quantities required.

• Any substance present in excess of the requirement cannot be limiting.


Learning Check:

• Ca(OH)2(aq)+ 2HCl(aq)→2 H2O(l) + CaCl2(s)when 1.00 g of each reactant is combined:

• What is the theoretical yield of H2O?

• The limiting reagent?

TY H2O (mol)

mol

MM (g/mol)18.0152836.4609474.09468

mass (g)1.001.00

H2O

0.013496 0.027427

O Hmol 0274.0 HClmol 2

O Hmol 2

1

HClmol 0.027427

O Hmol 0269.0Ca(OH) mol 1

O Hmol 2

1

Ca(OH) mol 0.0134

2272

2922

2296

=×

=×

0.026992

0.486277

Ca(OH)2 HCl

Ca(OH)2: 74.09468; HCl: 36.46094; H2O: 18.01528


Learning Check:

How many grams of NO can form when 30.0 g NH3and 40.0 g O2 react according to:4 NH3 + 5 O2→ 4 NO + 6 H2O

TY NO (mol)

mol

MM (g/mol)30.006131.998817.03052

mass (g)40.030.0

NOO2NH3

1.7615 1.2500 1.0000

30.0

NO mol1.00O mol 5

NO mol 4

1

O mol1.25

NO mol1.76NH mol 4

NO mol 4

1

NH mol 1.76

002

2 00

153

315

=×

=×

NH3: 17.03052; O2=31.9988; NO: 30.0061

Your Turn!

Given 1.0 g each of KCl and Hg2(NO3)2, what is the expected mass of Hg2Cl2 ?

A. 1.0 g

B. 2.0 g

C. 0.90 g

D. 3.2 g

E. none of these

MM (g/mol)472.086525.189974.5513

Hg2Cl2Hg2(NO3)2KCl

3.6 The predicted amount of product is not always obtained experimentally 65

Actual Yield

• Often we do not obtain the quantity expected

• This may be due to errors, mistakes, side reactions, contamination or a host of other events

• Thus we describe the actual yield, the amount obtained experimentally

3.6 The predicted amount of product is not always obtained experimentally 66

Percent Yield

• The amount of product, predicted by the limiting reagent is termed the theoretical yield

• Percent yieldrelates the actual yield to the theoretical yield

• It is calculated as:

• If a cookie recipe predicts a yield of 36 cookies and yet only 24 are obtained, what is the % yield?

100yield ltheoretica

yield actual% x

=

10036

24% x

=

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