Chapter 3 Molecules, Compounds, and Chemical...

Chapter 3

Molecules, Compounds, and

Chemical Composition

Elements and Compounds

Elements combine together to make an almost limitless number of compounds.

The properties of the compound are totally different

from the constituent elements.

Chemical Bonds

Compounds are made of atoms held together by bonds.

Chemical bonds are forces of attraction between atoms.

The bonding attraction comes from attractions between protons and electrons.

Bond TypesTwo general types of bonding between atoms found in

compounds, ionic and covalent.

Ionic bonds result when electrons have been transferred between atoms, resulting in oppositely charged ions that attract each other.

Generally found when metal atoms bond to nonmetal atoms

Covalent bonds result when two atoms share some of their electrons.

Generally found when nonmetal atoms bond together

Formation of an Ionic Compound

}Na0

Cl0

⤸Neutral Atoms Undergo

Electron Transfer

Na+Cl-

Charged Ions

An Orderly Aggregate Called an Ionic Crystal

Formation of an Covalent Compound

Formation of an Covalent Compound

Representative Covalent Compounds

Chemical Formulas Describe Compounds

A compound is a distinct substance that is composed of atoms of two or more elements.

We describe the compound by describing the number and type of each atom in the simplest unit of the compound.

Each element is represented by its letter symbol.

The number of atoms of each element is written to the right of the element as a subscript.

Polyatomic ions are placed in parentheses.(if more than one is present)

Types of Formula: Empirical FormulaAn empirical formula gives the relative number of atoms

of each element in a compound.

It does not describe how many atoms, the order of attachment, or the shape.

For example:

1) The empirical formula for the ionic compound fluorspar is CaCl2. This means that there is 1 Ca2+ ion for every 2 Cl− ions in the compound.

2) The empirical formula for the molecular compound oxalic acid is CHO2. This means that there is 1 C atom and 1 H atom for every 2 O atoms in the molecule.

Types of Formula: Molecular Formula

A molecular formula gives the actual number of atoms of each element in a molecule of a compound.

It does not describe the order of attachment, or the shape.

The empirical formula for the molecular compound oxalic acid is CHO2.

The actual molecular formula is C2H2O4

Types of Formula: Structural Formula

A structural formula uses lines to represent covalent bonds and shows how atoms in a molecule are connected or bonded to each other.

Structural Formulas of Oxalic Acid

O C

O

C

O

O HH

Molecular Models

Practice — Find the empirical formula for each of the following

The ionic compound that has two aluminum ions for every three oxide ions

arabinose, C5H10O5

pyrimidine

ethylene glycol

C

C

N

C

N

C

H

HH

H

Al2O3

CH2O

C2H2N

CH3O

Classifying Elements & Compounds

Atomic elementselements whose particles are single atoms

Molecular elementselements whose particles are multi-atom molecules

Molecular compoundscompounds whose particles are molecules

Ionic compoundscompounds whose particles are cations and anions

Molecular View of Elements and Compounds

ElementsMOST ELEMENTS

Single atoms are the constituent particles.The atoms may be physically attracted to each other, but

are not chemically bonded together.

A FEW ELEMENTS Molecules are the constituent particles.

The molecules are made of two or more atoms chemically bonded together by covalent bonds.

Molecular Elements

H2

Cl2

Br2

I2

7

7A

N2 O

2 F

2

20

Compounds

SOME COMPOUNDS Composed of ions arranged in a 3-dimensional pattern

These are called ionic compounds.

OTHER COMPOUNDS Composed of individual molecule units

Each molecule contains atoms of different elements chemically attached by covalent bonds

These are called molecular compounds.

Ionic vs. Molecular Compounds

Propane – contains individual C3H8

moleculesTable salt – containsan array of Na+ ions

and Cl- ions

Classify Each of the Following as Either an Atomic Element, Molecular Element, Molecular Compound,

or Ionic Compound

Aluminum, AlAluminum chloride, AlCl3

Chlorine, Cl2

Acetone, C3H6OCarbon monoxide, COCobalt, Co

atomic elementionic compound

molecular elementmolecular compound

molecular compoundatomic element

Formula Units vs. Molecules

Compound must have no total charge, therefore we must balance the numbers of cations and anions in a compound to get 0 charge.

Practice — What are the formulas for compounds made from the following ions?

Potassium ion with a nitride ion

Calcium ion with a bromide ion

Aluminum ion with a sulfide ion

K3N

CaBr2

Al2S3

K+ with N3−

Ca2+ with Br−

Al3+ with S2−

calcium bromide

aluminum sulfide

potassium nitride

Formula MassThe mass of an individual molecule or formula unit

Also known as molecular mass or molecular weight

Sum of the masses of the atoms in a single molecule or formula unit

mass of 1 molecule of H2O

= 2(1.01 amu H) + 16.00 amu O = 18.02 amu

mass of 1 formula unit of MgCl2 = 2(35.45 amu Cl) + 24.30 amu Mg = 95.20 amu

Molar Mass

The relative masses of molecules can be calculated from atomic masses.

Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu

1 mole of H2O contains 2 moles of H and 1 mole of O.

molar mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g

so the Molar Mass of H2O is 18.02 g/mole

Molar Mass of Compounds

Practice — How many moles are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00

g PbO2 mol PbO2

Pb = 1 x 207.2 = 207.2O = 2 x 16.00 = 32.00PbO2 = 239.2 g/mol

Example: Find the number of CO2 molecules in 10.8 g of dry ice

g CO2 mol CO2 molec CO2

1 mol CO2 = 44.01 g, 1 mol = 6.022 x 1023

Practice — How many formula units are in 50.0 g of PbO2? (PbO2 = 239.2)

g PbO2 mol PbO2 units PbO2

1 mol PbO2 = 239.2 g,1 mol = 6.022 x 1023

Practice — What is the mass of 4.78 x 1024 NO2 molecules?

molecules mol NO2 g NO2

1 mol NO2 = 46.01 g, 1 mol = 6.022 x 1023

Percent Composition

Percent Composition

Percentage of each element in a compound by mass

Can be determined from

1. the formula of the compound2. the experimental mass analysis of the compound

Find the mass percent of Cl in C2Cl4F2

Practice — Determine the mass percent composition of the following CaCl2

Mass Percent as a Conversion Factor

If NaCl is 39% sodium, find the mass of table salt containing 2.4 g of Na.

g Na g NaCl

Find the mass of sodium in 6.2 g of NaCl

g NaCl mol NaCl mol Na g Na

58.44 g NaCl

Empirical Formulas

Empirical Formula

Simplest, whole-number ratio of the atoms of elements in a compound

Can be determined from elemental analysis

Finding an Empirical Formula from % Composition

1. Convert the percentages to grams

2. Convert grams to moles

3. Write a pseudoformula using moles as subscripts

4. Divide all by smallest number of moles

5. Multiply all mole ratios by number to make all whole numbers

Example:Find the empirical formula of aspirin with the given mass percent composition

Given:" C = 60.00%" " " H = 4.48% " " " O = 35.53%

Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O"

g C g H g O

mol C mol H mol O

pseudoformulaCxHyOz

empirical formulaCxHyOz

Manipulate subscripts to obtain whole-number ratio

Calculate the moles of each element

Write a pseudoformula

C4.996H4.44O2.220

Find the mole ratio

C2.25H2.00O1.00Multiply subscripts by factor to give whole number

C9H8O4

(x 4)

÷ 2.220

C4.996H4.44O2.220

Practice — Determine the empirical formula of stannous fluoride, which

contains 75.7% Sn (118.70 g/mol) and the rest fluorine (19.00 g/mol)

g Sn mol Sn

g F mol F

pseudo-

formula

empirical

formula

mole

ratio

wholenumber

ratio

Given: 75.7% Sn, (100 – 75.3) = 24.3% F

ElementRatio in Grams

Molar MassRatio in Moles

Ratio in Moles

Sn 75.7g1mol

118.7g0.6377 1.000

F 24.3g1mol

19.00g1.279 2.005

Practice — Determine the empirical formula of stannous fluoride, which contains 75.7% Sn

(118.70 g/mol) and the rest fluorine (19.00 g/mol)

SnF2

X

X

Practice — Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85)

and the rest oxygen (16.00)

Given: 72.4% Fe, (100 – 72.4) = 27.6% O

g Fe mol Fe

g O mol O

pseudo-

formula

empirical

formula

mole

ratio

whole

number

ratio

Practice — Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)

ElementRatio in Grams

Molar MassRatio in Moles

Ratio in Moles

Ratio in Moles

Fe 72.4g1mol

55.85g1.296 1.000 3

O 27.6g1mol

16.00g1.725 1.33 4

Fe3O4

X

X

Molecular Formulas

Molecular Formulas

The molecular formula is a multiple of the empirical formula.

To determine the molecular formula you need to know theempirical formula and the molar mass of the compound.

Find the molecular formula of butanedione if its empirical formula is C2H3O and its molar mass (MM) is 86.03 g/mol.

Factor of 2

Practice – Benzopyrene has a molar mass of 252 g and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01)

Molecular formula = {C5H3} x 4 = C20H12

C5 =" 5(12.01 g) = 60.05 gH3 =" 3(1.01 g) = 3.03 gC5H3 "" = 63.08 g

252

?

Combustion Analysis

Combustion Analysis

A known mass of compound is burned in oxygen and the masses of the products formed (CO2 and H2O) are determined.

By knowing the masses of the products and composition of constituent elements in the product, the original amount of constituent elements can be determined.

It is assumed that all of the carbon in the original sample is converted to carbon dioxide and all of the hydrogen in the sample is converted to water.

(Generally used for organic compounds containing C, H, O)

Combustion Analysis

Example of Combustion AnalysisCombustion of a 0.8233 g sample of a compound containing only

carbon, hydrogen, and oxygen produced the following:

CO2 = 2.445 g H2O = 0.6003 g

! Determine the empirical formula of the compound.

g

CO2, H2O

mol

ratio

empirical

formula

mol

CO2, H2O

mol

C, H

g

C, H

g

O

mol

O

mol

C, H, O

pseudo

formula

This came from C.This came from H.

1 mole CO2 = 44.01 g CO2 1 mole H2O = 18.02 g H2O

1 mole C = 12.01 g C1 mole H = 1.008 g H 1 mole O = 16.00 g O

1 mole CO2 = 1 mole C 1 mole H2O = 2 mole H

In the original sample

In the original sample

In the original sample

C0.05556H0.06662O0.00556

Pseudo formula

÷ 0.00556

Empiricalformula

Combustion of 0.844 g of caproic acid produced 0.784 g of H2O and 1.92 g of CO2.

If the molar mass of caproic acid is 116.2 g/mol,what is the molecular formula of caproic acid?

moles

g

0.0145 0.0870 0.0436

0.232 0.0877 0.524

O H C

Molecular formula = {C3H6O} x 2 = C6H12O2