Transcript 1.2 The Mole Concept

1.2 The Mole Concept
Mrs. Page
IB SL Chemistry
2015-2016
Understandings
• The mole is a fixed number of particles and refers to the amount, n, of
substance.
• Masses of atoms are compared on a scale relative to 12C and are expressed as
relative atomic mass (A r) and relative formula/molecular mass (M r).
• Molar mass (M) has the units g mol-1.
• The empirical formula and molecular formula of a compound give the
simplest ratio and the actual number of atoms present in a molecule
respectively.
Application & Skills
• Calculation of the molar masses of atoms, ions, molecules and formula units.
• Solution of problems involving the relationships between the number of particles,
the amount of substance in moles and the mass in grams.
• Interconversion of the percentage composition by mass and the empirical formula.
• Determination of the molecular formula of a compound from its empirical formula
and molar mass.
• Obtaining and using experimental data for deriving empirical formulas from
reactions involving mass changes.
SI Units
• Systeme International de Unites
• Worldwide Standard set of Units
• 7 Base Units
• The International Bureau of
Weights and Measures (BIPM)
monitors the correct use of SI
Units in all applications of science
Property
Base Unit
Symbol
kilogram
kg
Temperature
kelvin
K
Time
second
s
Amount
mole
mol
Electric
current
ampere
A
Luminosity
candela
cd
meter
m
Mass
Length
See table 3 p. 13 for prefixes &
scales
The Mole
• Avogadro’s constant (L or NA) = 6.02 x 1023 mol-1
• Allows us to make comparisons between different
chemical species (atoms, molecules, formula units,
ions, electrons)
TOK
• The magnitude of Avogadro’s constant is
beyond the scale of our everyday
experience. How does our everyday
experience limit our intuition?
Relative Atomic & Molecular Mass
• Isotopes: atoms of the same element with different number of neutrons
(& therefore different atomic masses)
• Relative Abundance: the percent of each isotope present in a sample of
an element
• Relative Atomic Mass (Ar): the weighted average of the relative
abundance of an isotope and its atomic mass
• Relative Molecular Mass (Mr): the total mass of all of the relative
atomic masses of the elements in a compound
• REMEMBER: The masses of atoms are compared with each other based
on 12 units of C-12, therefore Ar and Mr are ratios and have no units!
Calculate the Relative Atomic Mass
• Magnesium has 3 isotopes. Use the following data to calculate the
relative atomic mass.
ISOTOPE
Relative
Mass
Relative
Abundance
24Mg
23.99
24.89
25.98
78.7%
10.13%
11.17%
25Mg
26Mg
23.99 ∙ 0.787 + 24.89 ∙ 0.1013 + 25.98 ∙ 0.1117 =
24.3
Calculate the Relative Molar Mass
• C2H5OH
2 12.01 + 6 1.01 + 16.00 = 46.08
• (NH4)3PO4
3 14.01 + 12 1.01 + 30.97 + 4(16.00) = 149.12
63.55 +32.07 + 4 16.00 + 10 1.01 + 5(16.00) = 249.72
Molar Mass
• Molar Mass: the mass of one mole of a substance (units: g mol-1)
• Note: because this is the mass of 1 mole this means that this is the mass
of 6.02 x 1023 atoms of a element or 6.02 x 1023 molecules/formula units
of a compound!
• Calculate the molar mass of copper (II) sulfate pentahydrate
CuSO4●5H2O
63.55 +32.07 + 4 16.00 + 10 1.01 + 5 16.00 = 249.72𝑔 𝑚𝑜𝑙 −1
Mole Calculations
• Calculate the number of moles in 4.00 g of sodium hydroxide, NaOH?
4.00 𝑔𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙
∙
40.00𝑔
= 0.100 𝑚𝑜𝑙
Mole Calculations
• Calculate the amount of water that contains 1.80 x 1024 molecules.
1.80 × 1024 𝐻2 𝑂 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
1 𝑚𝑜𝑙
∙
6.02 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
= 2.99 𝑚𝑜𝑙
Mole Calculations
• Calculate how many hydrogen atoms are present in 3.0 moles of ethanol,
C2H5OH
3.0 𝑚𝑜𝑙 𝐶2 𝐻5 𝑂𝐻 6.02 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
6 𝐻 𝑎𝑡𝑜𝑚𝑠
∙
∙
1 𝑚𝑜𝑙
1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝐶2 𝐻5 𝑂𝐻
= 1.08 × 1025 H atoms
Mole Calculations
• Calculate the mass in g of 0.250 mol of magnesium chloride (MgCl2)
0.250 𝑚𝑜𝑙 𝑀𝑔𝐶𝑙2 95.21 𝑔
∙
1 𝑚𝑜𝑙
= 23.8 𝑔 𝑀𝑔𝐶𝑙2
Empirical & Molecular Formulas
• “Empirical” = information from experimental data
• Chemists frequently attempt to determine the composition of a substance through
chemical analysis
• Qualitative analysis – what elements are in a compound
• Quantitative analysis – calculate relative masses to determine exact composition
• Empirical Formula: simplest whole-number ratio of atoms in a compound
• Molecular Formula: actual number of atoms in a compound
• For example for glucose the molecular formula is C6H12O6 but the empirical formula
is CH2O
• Sometimes the empirical and molecular formulas are the same. For ionic compounds
they are always the same since the formula is the simplest ratio of ions.
Percentage Composition
• The percentage by mass of each element in a compound
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑝𝑟𝑒𝑠𝑒𝑛𝑡
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
• % 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 =
× 100%
• Find the percent composition of butanoic acid, C4H8O2
4(12.01)
%𝐶 =
× 100% = 54.52%
88.12
8(1.01)
%𝐻 =
× 100% = 9.17%
88.12
2(16.00)
%𝑂 =
× 100% = 36.31%
88.12
Finding Empirical Formula from
Experimental Data
• Find mass (or %) of each element.
• Find moles of each element.
• Divide moles by the smallest # to find the smallest
ratio
• If ratio does not come out to whole numbers,
multiply both numbers by 2, 3, or 4 to get whole a #’s.
Empirical & Molecular Formulas From Data
• A 2.765 g sample of lead oxide was heated in a stream of hydrogen
gas and completely converted into elemental lead with a mass of
2.401 g. What is the empirical formula of the oxide?
PbO2
Mass loss is due to loss of oxygen.
Pb
Mass/ g
Moles
Simplest Ratio
2.401
O
2.765-2.401 =0.364
1𝑚𝑜𝑙
2.401 𝑔 ∙
= 0.01159
207.19𝑔
1𝑚𝑜𝑙
0.364 𝑔 ∙
= 0.0228
16.00𝑔
0.01159
=1
0.01159
0.0228
= 1.97 ≈ 2
0.01159
Empirical & Molecular Formulas From Data
A compound is 43.4%Na, 11.3%C, and 45.3% O. What is the
empirical formula for this compound?
Assume 100 grams
Mass/ g
Moles
Simplest Ratio
Na
C
O
43.4
11.3
45.3
1𝑚𝑜𝑙
43.3 𝑔 ∙
= 1.88
22.99𝑔
1𝑚𝑜𝑙
11.3 𝑔 ∙
= 0.942
12.00𝑔
1𝑚𝑜𝑙
45.3 𝑔 ∙
= 2.83
16.00𝑔
1.88
= 1.99 ≈ 2
0.942
0.942
=1
0.942
Na2CO3
2.83
=3
0.942
Finding Molecular Formula from
Empirical Formulas
• Find the empirical formula.
• Determine the empirical formula mass.
• Divide the molecular mass by the empirical formula mass to
determine the multiple.
• Multiply the empirical formula by the multiple to find the molecular
formula.
MF mass = n
EF mass
(EF)n = molecular formula
The empirical formula for ethylene is CH2. Find the molecular
formula if the molecular mass is 28.1g/mol.
C = 1 x 12.01 = 12.01
H = 2 x 1.01 = +2.01
14.02 g/mol = empirical formula mass
28.1 g/mol
14.02 g/mol
= 2
(CH2)2  C2H4