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Chapter 5Chapter 5
Quantities in ChemistryQuantities in Chemistry
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Setting the Stage - Bioavailability of Nitrogen
Plants need nitrogen as either ammonia or nitrate in order to use it for biosynthesis
While air is 80% nitrogen, it is in the form of the relatively inert gas N2, hence plants cannot use it directly
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Type and Efficacy of Nitrogen Sources
Main types of fertilizer are ammonia gas (NH3) and ammonium nitrate (NH4NO3)
100 kg of NH3 delivers 82 kg of N
100 kg of NH4NO3 only delivers 35 kg of N The mass of N delivered is related to the
formula of the compound and the relative masses of each element in the formula
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Setting a Goal – Part AThe Measurement of Masses of Elements and Compounds
You will become proficient at working with the units of moles, mass, and numbers of atoms and molecules, and at converting between each of these
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Objective for Section 5-1 Calculate the masses of equivalent
numbers of atoms of different elements
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5-1 - Relative Masses of Elements
When the masses of samples of any two elements are in the same ratio as that of their atomic masses, the samples have the same number of atoms
We can therefore use the atomic masses and the masses of samples of chemical substances to “count” the number of atoms or molecules
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The Mass of an Atom
Recall that an atom has an unbelievably small mass
12C is used as the standard and is assigned a mass of exactly 12 amu.
Other isotopes are present in natural samples (i. e. C has an overall atomic mass of 12.01) so the periodic table lists masses that are the weighed average mass of the natural sample
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Relative Masses of the Elements
The amu has no practical value in a laboratory situation:
12.01 amu = 1.994 10-23 g The best balance made can detect no less
than 10-5 g, so we have to scale up the masses to something we can measure
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Counting by Weighing
Hardware stores often count by weighing If we want 175 bolts and assume that the
mass of an average bolt is 10.5 g, then the mass of bolts will be:
175 bolts
bolts
10.5 g 1 kg
103 g
1.85 kg=x x
Bolts Nuts
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Counting by Weighing
Nuts, being smaller, will have a lower average mass (2.25 g)
What weight of nuts will provide a nut for each bolt?
175 nuts
nuts
2.25 g 1 kg
103 g
0.394 kg=x x
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Counting by weighing By using ratios of the average masses
of bolts and nuts, or the masses of a fixed number of items, we can get equivalent numbers of bolts and nuts:
bolts:nuts =10.5 g
2.25 g=
1.84 kg
0.394 kg= 4.67
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From Hardware to Atoms
We need to know the relative numbers of atoms of different elements present and the relative masses of the individual atoms
A 4He atom has a mass of 4.00 amu and a 12C atom has a mass of 12.00 amu.
If 4He and 12C are present in a 4.00:12.00 mass ratio, regardless of the units of mass, the number of atoms is the same
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Objective for Section 5-2
Define the mole and relate this unit to numbers of atoms and to the atomic masses of the elements
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5-2 The Mole and the Molar Mass of Elements
A mole is: The number of atoms in exactly 12 g
(12.00 recurring) of 12C Avogadro’s number of atoms
(6.022 1023) The number of atoms in one atomic
mass of an element expressed in grams
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Avogadro
Amedeo Avogadro, a pioneer in the investigation ofquantitative aspects of chemistry
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The Mole
One mole of an element implies The atomic mass expressed in grams, it
is different for each element It contains Avogadro’s number (6.022 x
1023) of atoms, which is the same for all elements
A conversion factor between mass and numbers of things (allows us to count atoms by weighing)
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Mass and Number of Things
For our purposes, assume that oranges are identical in mass; 12 have a mass of 2.71 kg
We can do the same with atomic mass
8.13 kg x12 oranges
2.71 kg= 36 oranges
3.00 mol Na x22.99 g Na
1 mol Na= 69.0 g Na
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Moles of Elements
Here is 1 mole each of copper, iron, mercury and sulphur
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Objective for Section 5-3
Perform calculations involving masses, moles, and numbers of molecules or formula units for compounds
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5-3 The Molar Mass of Compounds
Molecular compounds chemical formula represents a discrete
molecular unit (e. g. CO2)
Ionic compounds chemical formula represents a formula
unit (the whole number ratio of cations to anions; e. g. K2SO4)
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Calculation of Formula Weight
The sum of the atomic masses of all the atoms in a molecule
This is often referred to as the molecular weight
Consider CO2
1 C (12.01 amu) + 2 O (2 16.00 amu)= 44.01 amu
One molar mass of a compound contains Avogadro’s number of molecules
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Calculation of Formula Weight
Second example, using a salt, Fe2(SO4)3
Atom Number of atoms in molecule
Atomic mass Total mass of atoms in a molecule
Fe 2 55.85 amu = 111.70 amu S 3 32.07 amu = 96.21 amu
O 12 16.00 amu = 192.00 amu
Formula weight of Fe2SO4
399.91 amu
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Hydrates
Some ionic compounds can have water molecules attached within the structure
These compounds are termed hydrates and have properties distinct from the unhydrated form
The formula weight of a hydrate includes the mass of the water molecules
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Hydrates
Examples CuSO4 - [copper(II) sulfate] - a pale
green solid CuSO4•5H2O - [copper(II) sulfate
pentahydrate - a dark blue solid Often, the waters of hydration can be
removed by heating
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The Molar Mass of a Compound
The mass of one mole (6.022 × 1023 molecules or formula units) is referred to as the molar mass of the compound
It is the formula weight expressed in grams
For example, 44.0 g of CO2 is the molar mass of CO2 and is the mass of 6.022 × 1023 molecules of CO2
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Moles of Compounds
One mole of copper sulfate pentahydrate, sodium chloride, sodium chromate and water
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Summary Chart for Part A (1)
Relationships of One mole
1.000 mol
Atoms Molecules Ionic Formula Units
6.022 x 1023
atoms:atomic massin grams
6.022 x 1023
molecules:molecular weightin grams
6.022 x 1023 formulaunits:formula weightin grams
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Summary Chart for Part A (2)
Conversions involving the Mole
mass
mole (mol) mole (mol)
number
x g/molx mol/g
x 1 mol
6.022 x 1023
x 6.022 x 1023
g
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Setting a Goal – Part BThe Component Elements of Compounds
You will learn about the relationship between the formula of a compound and its elemental composition
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Objective for Section 5-4
Given the formula of a compound, determine the mole, mass, and percent composition of its elements
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5-4 The Composition of Compounds
Table 5-2 relates one mole of a compound
(H2SO4) to all its component parts. All of
these relationships can be used to construct
conversion factors between the compound
and its elements.
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Table 5-2
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Mole composition is the number of moles of each of the elements that make up 1 mole of the compound
CO2 – one mole of C and two moles of O
H2SO4 – one mole of S, two moles of H, and four moles of O
The Mole Composition of a Compound
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Mass composition is the mass of each element in the compound
CO2 – 12.01 g of C and 32.00 g of O
H2SO4 – 2.016 g of H, 32.07 g of S, and 64.00 g of O
The Mass Composition of a Compound
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Percent Composition of a Compound
mass of each element per 100 mass units of compound in 100 g of NH3, there are 82.2 g of N therefore, the mass percentage of N is
82.2% N
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CO2
Calculation of % composition of carbon dioxide requires determining the number of grams of each element (C and O) in one mole
% composition =Total mass of component element
Total mass of molecule
%C =12.0 g
x 100
44.0 gx 100 = 27.3% %O =
32.0 g
44.0 gx 100 = 72.7%
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Objectives for Section 5-5
Use percent or mass composition to determine the empirical formula of a compound
Given the molar mass of a compound and its empirical formula, determine its molecular formula
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5-5 Empirical and Molecular Formulas
Empirical formula - simplest whole number ratio of atoms in the compound
Procedure to find empirical formula from % composition data Convert percent composition to an actual
mass Convert mass to moles of each element Find the whole number ratio of the moles of
different elements
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Empirical Formula of Laughing Gas
Contains 63.6% N and 36.4% O Assume 100 g of substance, so you have 63.6 g
of N and 36.4 g of O Calculation gives an empirical formula of N2O
63.6 g N x1 mol N
14.0 g N= 4.54 mol
36.4 g O x1 mol O
16.0 g O= 2.28 mol
N:4.54 mol
2.28 mol= ~2
O: 2.28 mol
2.28 mol= 1
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Molecular Formula
The actual number of each atom in a formula unit
Consider acetylene and benzene both have the empirical formula CH, but
different molecular formulas: acetylene is actually C2H2
benzene is actually C6H6
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Molecular Formula Determination
Needs the molecular mass, which must be determined from an independent measurement (e.g. via mass spectrometry)
First determine the mass of the empirical formula
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Molecular Formula Determination…contd.
Divide the empirical formula mass into the molecular mass
The resulting number (which should be a small whole number or close to it) is the number of times the empirical formula unit appears in the molecular formula
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Acetylene and Benzene
The empirical formula mass for both substances is 12.0 g + 1.0 g = 13.0 g
The actual molar mass of acetylene is 26.0 g, so the empirical formula mass divides into the actual mass two times - C2H2
Benzene’s actual molar mass is 78.0 g, so the empirical formula mass divides into the actual mass six times - C6H6
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Acetylene and Benzene: Same Empirical Formula, but DifferentMolecular Formulas
C
C
C
C
C
C
H
H
H
HH
H
Benzene C6H6
C C HH
Acetylene C2H2
These are structuralformulas
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Summary Chart for Part B (1)
Composition of Compounds
Mass ofcompound
Moles ofelements
Moles ofcompound
Mass ofelements
%Composition
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Summary Chart for Part B (2)
Empirical and Molecular Formulas
%Composition Mass ofelements
Moles ofelements
Mole ratioof elements
Whole-numbermole ratio
Empiricalformula
Molecularformula
Molarmass
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Summary of Types of FormulaUnknown pure substance
% Elementalcomposition +atomic masses
EMPIRICALFORMULA
Gives whole numberratio of elements
Molar massGives number of atomsof each element inmolecule
MOLECULARFORMULA
STRUCTURALFORMULA
Physical andchemical data(especiallyspectroscopicdata)
Gives connectivitybetween atoms: tellsus how atoms arebonded in molecule
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Worked Example
Nicotine is a compound containing C, H and N only. Its molar mass is 162 g. A 1.50 g sample of nicotineis found to contain 1.11 g of C. Analysis of anothersample indicates that nicotine has 8.70% by mass of H.Determine the molecular formula of nicotine.
Solution. We should convert the data to masses or %, and find the mass or % of N by difference.
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Worked Example Continued
Data
As given:
C H N
1.11 g/1.50 g 8.70%
As masses 1.11 g 0.13 g 0.26 g
As % 74.0 8.70 17.3
Take the % data, convert to g and divide by the relevantmolar mass (C = 12.01; H = 1.01; N = 14.00), gives
6.16 mol C; 8.61 mol H; 1.23 mol N.Divide throughout by 1.23, gives mole ratio 5C: 7H: 1N, orthe empirical formula of nicotine is C5H7N (= 81 g/mol).Hence the molecular formula is C10H14N2 (= 162 g/mol)
HC
HCN
CH
C
HC HC
N
CH2
CH2H2C
CH3