1. 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. 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?
6.0 kg
3.1 The mole conveniently links mass to number of atoms or molecules

4. 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
Chem FAQs: How do I count atoms in a sample of element using atomic masses?
3.1 The mole conveniently links mass to number of atoms or molecules

5. 3.1 The mole conveniently links mass to number of atoms or molecules
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?
2.9 ×107 u
2.35 ×1016 u
Chem FAQs: How do I count atoms in a sample of element using atomic masses?
atomic masses: C= u; H= u
3.1 The mole conveniently links mass to number of atoms or molecules

6. 3.1 The mole conveniently links mass to number of atoms or molecules
Your Turn!
Given that the atomic mass of Ba is u, what is the mass of 23 atoms of Ba?
3.2×103 u
3.2×10-4 u
1.37×102 u
none of these
3.1 The mole conveniently links mass to number of atoms or molecules

7. 3.1 The mole conveniently links mass to number of atoms or molecules
Your Turn!
A new element is discovered that has a mass of
3.2 ×102 u for15 atoms. What is the atomic mass?
3.2 ×102
0.047 21
not enough information
None of these answers
3.1 The mole conveniently links mass to number of atoms or molecules

8. 3.1 The mole conveniently links mass to number of atoms or molecules
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)
3.1 The mole conveniently links mass to number of atoms or molecules

9. 3.1 The mole conveniently links mass to number of atoms or molecules
Relationships
Atomic Mass (AM) u = 1 particle
We can use this as a conversion factor to convert between these quantities.
mass (u)
particles
3.1 The mole conveniently links mass to number of atoms or molecules

10. 3.1 The mole conveniently links mass to number of atoms or molecules
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?
3.32×1028 u
5.3×10-9 g
3.1 The mole conveniently links mass to number of atoms or molecules

11. 3.1 The mole conveniently links mass to number of atoms or molecules
Your Turn!
Which of the following are not equivalent to a sample of 10.5×107 u of Cu?
1.74×10-16 g
1.65×106 atoms
63.54 u
None of these
3.1 The mole conveniently links mass to number of atoms or molecules

12. What Is The Formula Mass Of…?
Ba3(PO4)2 :
(NH4)2CO3:
u/fu
u/fu
atomic masses: Ba: (7)u; P: (2)u;
O: (3)u; H: u; N: u; C (8)u
3.1 The mole conveniently links mass to number of atoms or molecules

13. 3.1 The mole conveniently links mass to number of atoms or molecules
Relationships
Formula mass (FM) u = 1 particle
We can use this as a conversion factor to convert between these quantities.
mass (u)
particles
3.1 The mole conveniently links mass to number of atoms or molecules

14. 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
Note that the sig figs come from the decimal places of the individual atomic masses.
3.1 The mole conveniently links mass to number of atoms or molecules

15. 3.1 The mole conveniently links mass to number of atoms or molecules
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 ?
8.0 ×106 u
Chem FAQ: How do I count molecules in a sample of compound using molecular masses?
4.70 ×1016 u
atomic masses: C= u; H= u; O= u
3.1 The mole conveniently links mass to number of atoms or molecules

16. 3.1 The mole conveniently links mass to number of atoms or molecules
What Is a Mole?
One mole of any substance contains the same number of units, called Avogadro’s number, N
1 mole formula units = x 1023 formula units
It is a large quantity of particles because the particles described are so small.
Chem FAQs: What is a mole?
What is Avogadro’s number
How do I convert moles of an element into atoms?
How do I convert moles of compound into molecules?
3.1 The mole conveniently links mass to number of atoms or molecules

17. Why is Molar Mass the Same as Formula Mass?
suppose we start with g of C. How many atoms of C are there?
given that the atomic mass of C is u
g C = x 1023 atoms
thus for any substance, the formula mass (in g) corresponds to the same number of atoms, N
Since the mole is larger than the individual in the same proportion that the g is larger than the amu, the numbers are identical.
Units: MM (g/mol); FM (amu/particle)
Chem FAQs: What is molar mass?
3.1 The mole conveniently links mass to number of atoms or molecules

18. 3.1 The mole conveniently links mass to number of atoms or molecules
Molar Mass
One mole contains 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 u/fu, the molar mass of Ba3(PO4)2 is g/mol
3.1 The mole conveniently links mass to number of atoms or molecules

19. 3.1 The mole conveniently links mass to number of atoms or molecules
Relationships
MM g = 1 mole
Use this as a conversion factor to convert between these quantities
Mass (g)
mole
3.1 The mole conveniently links mass to number of atoms or molecules

20. Learning Check: Converting Between Mass And Moles
Given that the molar mass of CO2 is g/mol
What mass of CO2 is found in 1.55 moles?
How many moles of CO2 are there in 10 g?
68.2 g
Chem FAQs: How do I convert masses into moles?
How do I convert moles into masses?
0.2 mol
3.1 The mole conveniently links mass to number of atoms or molecules

21. 3.1 The mole conveniently links mass to number of atoms or molecules
Your Turn!
What is the molar mass of Ca3(PO4)2 in g/mol?
Ca: ; P: ; O:
none of these
3.1 The mole conveniently links mass to number of atoms or molecules

22. 3.1 The mole conveniently links mass to number of atoms or molecules
Your Turn!
What mass in g, of Ca3(PO4)2 (MM= ) would a 3.2 mole sample have?
1.0×10-3 g
9.9×102 g
6.0×1026 g
1.6×10-21 g
None of these
3.1 The mole conveniently links mass to number of atoms or molecules

23. 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 ×1023 individuals comprise 1 mole
The quantity, N, is Avogadro's number and is measured as ×1023
3.1 The mole conveniently links mass to number of atoms or molecules

24. 3.1 The mole conveniently links mass to number of atoms or molecules
Relationships
N particles = 1 mole
We can use this as a conversion factor to convert between these quantities
particle
Moles
3.1 The mole conveniently links mass to number of atoms or molecules

25. 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?
7.77×1023 fu
1.91× mol
3.1 The mole conveniently links mass to number of atoms or molecules

26. Relationships Between QuantitiesMM FM
1mole = N counting units (particles)
1mole  Molar Mass (g)
1g = N amu
1 counting unit  Formula Mass (amu)
How do I convert grams of substance to particles of substance? N
3.1 The mole conveniently links mass to number of atoms or molecules

27. 3.1 The mole conveniently links mass to number of atoms or molecules
Your Turn!
Which of the following is not a relationship, but is a sample size?
molar mass
Avogadro’s number
formula mass
Mass in u
None of these
3.1 The mole conveniently links mass to number of atoms or molecules

28. 3.1 The mole conveniently links mass to number of atoms or molecules
Your Turn!
Na: ; C: ; O:
Given that you have a sample of 5.5 g Na2CO3 how many formula units are present?
6.0×1023
5.2×10-2
3.2×10-23
3.3×1024
None of these N FM MM
3.1×1022
3.1 The mole conveniently links mass to number of atoms or molecules

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
Chem FAQ: How can I relate moles of elements within a compound, given the compound's formula?
3.2 Chemical formulas relate amounts of substances in a compound

30. 3.2 Chemical formulas relate amounts of substances in a compound
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
5.06 mol Na
Note the number of sodium atoms in each formula unit.
3.05×1024 atoms Na
3.2 Chemical formulas relate amounts of substances in a compound

31. 3.2 Chemical formulas relate amounts of substances in a compound
Your Turn!
How many atoms of iron are in a 15.0 g sample of iron(III) oxide (MM g/mol)?
1.13×1023
9.39×10-2
5.66×1022
1.88×10-1
None of these
3.2 Chemical formulas relate amounts of substances in a compound

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?
Chem FAQ: How can I relate masses of elements within a compound, given the compound's formula?
How can I compute the mass of an element within a compound, given the compounds formula?
3.3 Chemical formulas can be determined from experimental mass measurements

33. Percent Composition: How Is It Calculated?
What is the % C in CO2?
Determine the molar mass of the compound
MM= g/mol
Multiply the ratio of the mass of the element to the molar mass of the compound by 100
( / )×100= %C
Chem FAQs: How can I compute the percentage composition of a compound from element masses?
How can I compute the percentage composition of a compound from its formula?
MM g/mol C: ; O:
3.3 Chemical formulas can be determined from experimental mass measurements

34. Learning Check
A sample was analyzed and found to contain g nitrogen and g oxygen. What is the percentage composition of this compound?
25.94% N
74.06% O
3.3 Chemical formulas can be determined from experimental mass measurements

35. 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?
6.62×10-1 g
8.88×101 g
1.51×102 g
8.34 g
None of these
3.3 Chemical formulas can be determined from experimental mass measurements

36. Empirical vs. Molecular Formulas
glucose
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
CH2O
Ionic compounds are always written in empirical form, but molecular compounds are not.
C1x6H2x6 O1x6
3.3 Chemical formulas can be determined from experimental mass measurements

37. 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
3.3 Chemical formulas can be determined from experimental mass measurements

38. Common Ratios And Their Decimal Equivalents
For example:
3.3 Chemical formulas can be determined from experimental mass measurements

39. Learning Check:
A g sample of a compound contains g of nitrogen and g of oxygen. Calculate its empirical formula N O
mass(g)
0.522
1.490 MM
mol
lowest ratio 1
2.50
Subscripts indicate insignificant digits not yet rounded
integer ratio 2 5
3.3 Chemical formulas can be determined from experimental mass measurements

40. 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
glucose
3.3 Chemical formulas can be determined from experimental mass measurements

41. Example: EF: N2H4O3 n=(80.06/80.043)=1 N2H4O3 N2H4 n=(32.0/16.02)=2
The empirical formula of hydrazine is NH2, and its molecular mass is 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 g/mol
n=(32.0/16.02)=2
N2H4
EF: N2H4O3
n=(80.06/80.043)=1
N2H4O3
MM: N: ; H: ; O:
3.3 Chemical formulas can be determined from experimental mass measurements

42. Your Turn!
Given the composition analysis of lindane (a controversial pesticide ) what is its empirical formula?
C24H2Cl73
C2H2Cl2
C142 HCl126
CHCl
None of these C H Cl % % %
3.3 Chemical formulas can be determined from experimental mass measurements

43. Your Turn!
We found that the empirical formula was CHCl. Given that the MM is g/mol, what is the molecular formula?
C6H6Cl6
C8H17Cl5
C3H5Cl7
C5H18Cl7
none of these
3.3 Chemical formulas can be determined from experimental mass measurements

44. 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
3.3 Chemical formulas can be determined from experimental mass measurements

45. 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 )
3.3 Chemical formulas can be determined from experimental mass measurements

46. Learning Check:
The combustion of a g sample of a compound of C, H, and O gave g CO2 and g of H2O. Calculate the empirical formula of the compound.
g C
g H
Chem FAQ: How can I calculate an empirical formula from a combustion analysis?
5.217g g C g H= g O
H: ; C: ; O:
3.3 Chemical formulas can be determined from experimental mass measurements

47. Learning Check (con.): CH3O
Calculate the empirical formula of the compound.
H: ; C: ; O: C H O
mass MM
mol
low ratio
integer ratio
.16828
.16819
CH3O 1
2.97 1 1 3 1
3.3 Chemical formulas can be determined from experimental mass measurements

48. 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?
C6H5S
C9H2S
C5H5S
C3H9S2
None of these
CO2
H2O
9.377 g
1.59 g
3.3 Chemical formulas can be determined from experimental mass measurements

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
3.4 Chemical equations link amounts of substances in a reaction

50. 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
Chem FAQs
How can I obtain mole-to-mole conversion factors form a chemical equation?
How can I relate moles of substances involved in a chemical reaction?
How can I relate grams of substances involved in chemical reactions?
3.4 Chemical equations link amounts of substances in a reaction

51. 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
3.4 Chemical equations link amounts of substances in a reaction

52. 3.4 Chemical equations link amounts of substances in a reaction
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 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
1.2 mol N2
Use the stoichiometric ratio as a conversion factor, making certain that the units cancel, top & bottom.
0.958 mol O2
3.4 Chemical equations link amounts of substances in a reaction

53. 3.4 Chemical equations link amounts of substances in a reaction
Learning Check
How many grams of Al2O3 are produced when 41.5 g Al react? 2Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(l)
78.4 g Al2O3
MM (g/mol): Al: ; Al2O3:
3.4 Chemical equations link amounts of substances in a reaction

54. 3.4 Chemical equations link amounts of substances in a reaction
Your Turn!
Given the reaction:
H2SO4 + 2KOH→2H2O + K2SO4,
How many moles of KOH are required to make 3.0 moles of K2SO4?
3.0 moles
6.0 moles
1.5 moles
None of these
3.4 Chemical equations link amounts of substances in a reaction

55. 3.4 Chemical equations link amounts of substances in a reaction
Your Turn!
Given the reaction:
H2SO4 + 2KOH→2H2O + K2SO4,
How many g of H2O ( ) would result from the complete reaction of 1.2 g H2SO4 (98.08)?
2.4 g
1.2 g
0.60 g
0.44 g
none of these
3.4 Chemical equations link amounts of substances in a reaction

56. 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
Chem FAQ: How do I balance a chemical equation?
3.4 Chemical equations link amounts of substances in a reaction

57. Learning Check: Balance The Following:1
____Ba(OH)2(aq) +____ Na2SO4(aq) → ___BaSO4(s) + ____NaOH(aq) 1 2 1 2
___KClO3(S) → ___KCl(s) +___ O2(g) 2 3
___H3PO4(aq) +___ Ba(OH)2(aq) → ___Ba3(PO4)2(s) + ___H2O(l) 2 3 1 6
3.4 Chemical equations link amounts of substances in a reaction

58. Your Turn! Given the following reaction:
KCl + Hg2(NO3)2→KNO3 + Hg2Cl2 , when it is balanced, what is the coefficient for KCl? 1 2 3 4
none of these

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)
Imagine a simple bolt assembly : 1 nut + 1 bolt = 1 bolt assembly.
If we don’t have the same number of nuts as bolts, there will be leftover parts and not entirely product.
3.5 The reactant in shortest supply limits the amount of product that can form

60. 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.
3.5 The reactant in shortest supply limits the amount of product that can form

61. 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.
3.5 The reactant in shortest supply limits the amount of product that can form

62. Learning Check: Ca(OH)2 HCl 0.486277 0.013496 0.027427 0.026992
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?
H2O
1.00
mass (g)
MM (g/mol)
mol
TY H2O (mol)
Ca(OH)2
HCl
Chem FAQs: How do I determine which reactant is limiting?
How do I compute grams of product, when I have grams of more than one reactant?
Ca(OH)2: ; HCl: ; H2O:
3.5 The reactant in shortest supply limits the amount of product that can form

63. Learning Check:
How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to: 4 NH3 + 5 O2 → 4 NO + 6 H2O
NH3 O2 NO
30.0
40.0
mass (g)
MM (g/mol)
mol
TY NO (mol)
30.0
1.7615
1.2500
1.0000
NH3: ; O2= ; NO:
3.5 The reactant in shortest supply limits the amount of product that can form

64. Your Turn!
Given 1.0 g each of KCl and Hg2(NO3)2, what is the expected mass of Hg2Cl2 ?
1.0 g
2.0 g
0.90 g
3.2 g
none of these
KCl
Hg2(NO3)2
Hg2Cl2
MM (g/mol)

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 yield relates 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?
Chem FAQs: How do I compute the percent yield of a reaction?
3.6 The predicted amount of product is not always obtained experimentally