chapter 5 quantities in chemistry malone and dolter - basic concepts of chemistry 9e2 setting the...

Chapter 5Chapter 5

Quantities in ChemistryQuantities in Chemistry

Malone and Dolter - Basic Concepts of Chemistry 9e

2

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


3

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


4

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


5

Objective for Section 5-1 Calculate the masses of equivalent

numbers of atoms of different elements


6

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


7

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


8

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


9

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


10

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


11

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


12

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


13

Objective for Section 5-2

Define the mole and relate this unit to numbers of atoms and to the atomic masses of the elements


14

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


15

Avogadro

Amedeo Avogadro, a pioneer in the investigation ofquantitative aspects of chemistry


16

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)


17

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


18

Moles of Elements

Here is 1 mole each of copper, iron, mercury and sulphur


19


Perform calculations involving masses, moles, and numbers of molecules or formula units for compounds


20

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)


21

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


22

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


23

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


24

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


25

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


26

Moles of Compounds

One mole of copper sulfate pentahydrate, sodium chloride, sodium chromate and water


27

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


28

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


29

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


30


Given the formula of a compound, determine the mole, mass, and percent composition of its elements


31

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.


32

Table 5-2


33

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


34

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


35

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


36

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%


37

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


38

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


39

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


40

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


41

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


42

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


43

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


44

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


45

Summary Chart for Part B (1)

Composition of Compounds

Mass ofcompound

Moles ofelements

Moles ofcompound

Mass ofelements

%Composition


46

Summary Chart for Part B (2)

Empirical and Molecular Formulas

%Composition Mass ofelements

Moles ofelements

Mole ratioof elements

Whole-numbermole ratio

Empiricalformula

Molecularformula

Molarmass


47

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


48

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.


49

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

chapter 5 quantities in chemistry malone and dolter - basic concepts of chemistry 9e2 setting the...

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