2. Molecules Ions Stoic

8/11/2019 2. Molecules Ions Stoic

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1.Molecules, Ions and

their Compounds

2. Chemical Equationsand Stoichiometry


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MOLECULAR FORMULAS

Formula for glycine is C2H5NO2

In one molecule there are

2 C atoms

5 H atoms

1 N atom

2 O atoms


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WRITING FORMULAS

Can also write glycine formula as

H2NCH2COOH

to show atom ordering

or in the form of a structuralformula

C

H

H C

H

H

O

O HN


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Empirical and Molecular Formulas

Empirical Formula -

Molecular Formula -

The simplest formula for a compound that agrees withthe elemental analysis and gives rise to the smallest set

of whole numbers of atoms.

The formula of the compound as it exists, it may be a

multiple of the empirical formula.


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MOLECULAR WEIGHT AND

MOLAR MASSMolecular weight= sum of the atomic

weights of all atoms in the molecule.

Molar mass= molecular weight in grams


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How many molesof alcohol are therein a standard can of beer if thereare 21.3 g of C2H6O?

(a) Molar mass of C2H6O= 46.08 g/mol

(b) Calc. moles of alcohol

21.3 g 1 mol

46.08 g = 0.462 mol


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How many moleculesof alcohol arethere in a standard can of beer ifthere are 21.3 g of C2H6O?

= 2.78 x 1023molecules

We know there are 0.462 mol of C2H6O.

0.462 mol 6.022 x 10

23molecules

1 mol

There are 2.78 x 1023molecules.

Each molecule contains 2 C atoms.

Therefore, the number of C atoms is

2.78 x 1023

molecules 2 C atoms

1 molecule

= 5.57 x 1023C atoms


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Table 3.1 Summary of Mass Terminology

Term Definition Unit

Isotopic mass Mass of an isotope of an element amu

Atomic mass

Molecular

(or formula) mass

(also calledmolecular weight)

Molar mass (M)

(also called

atomic weight)

(also called

gram-molecular weight)

amu

amu

g/mol

Averageof the masses of the naturallyoccurring isotopes of an element

weighted according to their abundance

Sum of the atomic masses of the atoms(or ions) in a molecule (or formula unit)

Mass of 1 mole of chemical entities(atoms, ions, molecules, formula units)


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Information Contained in the Chemical Formula of Glucose

C6H12O6( M = 180.16 g/mol)

Oxygen (O)

Mass/mole ofcompound

6 atoms

96.00 g

Table 3.2Carbon (C) Hydrogen (H)

Atoms/moleculeof compound

Moles of atoms/mole of compound

Atoms/mole ofcompound

Mass/moleculeof compound

6 atoms 12 atoms

6 moles

of atoms

12 moles

of atoms

6 moles

of atoms

6(6.022 x 1023)

atoms

12(6.022 x 1023)

atoms

6(6.022 x 1023)

atoms

6(12.01 amu)

=72.06 amu

12(1.008 amu)

=12.10 amu

6(16.00 amu)

=96.00 amu

72.06 g 12.10 g


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Interconverting Moles, Mass, and Number of Chemical Entities

Mass (g) = no. of moles xno. of grams

1 mol

No. of moles = mass (g) x

no. of grams

1 mol

No. of entities = no. of moles x6.022x1023entities

1 mol

No. of moles = no. of entities x6.022x1023entities

1 mol

g

M


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IONS AND IONIC COMPOUNDS

IONSare atoms or groups of atoms with a positive or negative charge.

Taking away an electron from an atom gives a CATIONwith a positive

charge

Adding an electron to an atom gives an ANIONwith a negative charge.

Mg --> Mg2++ 2 e- F + e- --> F-


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PREDICTING ION CHARGES

In general

metals(Mg) lose electrons ---> cations

nonmetals(F) gain electrons ---> anions


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METALS

M ---> n e- + Mn+

where n = periodic group

Na+ sodium ion

Mg2+ magnesium ion

Al3+ aluminum ion

Transition metals --> M2+or M3+

are common

Fe2+

iron(II) ionFe3+ iron(III) ion


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NONMETALS

NONMETAL + n e- ------> Xn-

where n = 8 - Group no.

C4-

,carbide

N3-, nitride

O2-, oxide

S2-, sulfide

F-, fluoride

Cl-, chloride

Group 7AGroup 6AGroup 4A Group 5A

Br-, bromide

I-, iodide

Name derived

by adding -ideto stem


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POLYATOMIC IONS

Note: many Ocontaining anions

have names ending

in ate (or -ite).


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Naming binary ionic compounds

The name of the cation is the same as the name of the metal.

Many metal names end in -ium.

The name of the anion takes the root of the nonmetal name

and adds the suffix -ide.

Calcium and bromine form calcium bromide.

The name of the cation is written first, followed by that of theanion.


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Naming Acids

1) Binary acids solutions form when certain gaseous compoundsdissolve in water.

For example, when gaseous hydrogen chloride(HCl) dissolves in

water, it forms a solution called hydrochloric acid. Prefix hydro-+

anion nonmetal root+ suffix -ic+ the word acid - hydrochloric

acid

2) Oxoacid names are similar to those of the oxoanions, except for

two suffix changes:

Anion -atesuffix becomes an -icsuffix in the acid.

Anion -ite suffix becomes an -oussuffix in the acid.

The oxoanion prefixes hypo-and per-are retained. Thus,

BrO4

-isperbromate, and HBrO4isperbromicacid; IO

2

-is iodite,

and HIO2is iodousacid.


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Properties of Ionic CompoundsForming NaCl from Na and Cl2

A metal atom cantransfer an electron to

a nonmetal.

The resulting cation

and anion are attractedto each other by

electrostaticforces.


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Electrostatic Forces

The oppositely charged ions in ionic compounds are attracted to one

another by ELECTROSTATIC FORCES.

These forces are governed by COULOMBS LAW.


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Electrostatic Forces

COULOMBS LAW

As ion charge increases, the attractive force_______________.

As the distance between ions increases, the

attractive force ________________.This idea is important and will come upmany times in future discussions!


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Molecular CompoundsCompounds without Ions

CH4 methane

CO2 Carbon dioxide

BCl3boron trichloride

All are

formed from

two or morenonmetals.

Ionic

compounds

generally

involve a metal

and nonmetal(NaCl)


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Empirical & Molecular

FormulasA pure compound always consists of the same

elements combined in the same proportions by

weight.

Therefore, we can express molecular composition asPERCENT BY WEIGHT

Ethanol, C2H6O

52.13% C13.15% H

34.72% O


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Percent CompositionConsider some of the family of nitrogen-oxygen compounds:

NO2, nitrogen dioxide and closely related, NO, nitrogen

monoxide (or nitric oxide)

Structure of NO2

Chemistry of NO,nitrogen monoxide


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Percent CompositionConsider NO2, Molar mass = ?

What is the weight percent of N and of O?

Wt. % O = 2 (16 .0 g O per mole )

46 .0 gx 100 % =69 .6%

Wt. % N =

14.0 g N

46.0 g NO2 100% = 30.4 %

What are the weight percentages ofN and O in NO?


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How to Determine a Formula?

Mass spectrometer


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Mass Spectrum of EthanolMass Spectrum of Ethanol(from the NIST site)

46

45

CH3CH2OH+

CH3CH2O+

H2C=OH+31

CH2O+

30


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DETERMINE THE FORMULA OF A

COMPOUND OF Sn AND I

Sn(s) + some I2(s) ---> SnIx


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Data to Determine the formula of a Sn

I Compound

Reaction of Sn and I2is done using excess Sn.

Mass of Sn in the beginning = 1.056 g

Mass of iodine (I2) used = 1.947 g Mass of Sn remaining = 0.601 g

Mass of Sn used = 0.455 g

Find moles of Sn used

0.455 g Sn 1 mol

118.7 g = 3.83 x 10-3 mol Sn


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Tin and Iodine Compound

Now find the number of moles of I2that combinedwith 3.83 x 10-3mol Sn. Mass of I2used was

1.947 g.

1.947 g I2 1 mol

253.81 g

= 7.671 x 10-3 mol I2

How many mol of iodine atoms?

= 1.534 x 10-2mol I atoms

7.671 x 10-3

mol I22 mol I atoms

1 mol I2

$

%


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Tin and Iodine CompoundNow find the ratio of number of moles of moles of I and Sn

that combined.

1.534 x 10-2

mol I

3.83 x 10-3

mol Sn

=4.01 mol I

1.00 mol Sn

Empirical formula is SnI4


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Chemical Equations

Depict the kind of reactantsand productsand their relative amountsin a reaction.

4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)

The numbers in the front are called stoichiometric coefficientsThe letters (s), (g), and (l) are the physical states of compounds.


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Chemical Equations

Because the same atoms are present in

a reaction at the beginning and at theend, the amount of matter in a system

does not change.

The Law of the Conservation of Matter

Demo of conservation of matter, See

Screen 4.3.

2HgO(s) ---> 2 Hg(liq) + O2(g)Lavoisier, 1788


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BalancingEquations

____C3H8(g) + _____ O2(g) ---->

_____CO2(g) + _____ H2O(g)

____B4H10(g) + _____ O2(g) ---->

___ B2O3(g) + _____ H2O(g)


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STOICHIOMETRY

- the study of thequantitative aspects ofchemical reactions.


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PROBLEM: If 454 g of NH4NO3decomposes, how much N2Oand H2O are formed? What is the theoretical yield of

products?

STEP 1

Write the balanced chemical

equation

NH4NO3 ---> N2O + 2 H2O


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454 g of NH4NO3--> N2O + 2 H2O

STEP 2 Convert mass reactant(454 g) --> moles

454 g

1 mol

80.04 g = 5.68 mol NH4NO3

STEP 3 Convert moles reactant

(5.68 mol) --> moles product


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454 g of NH4NO3--> N2O + 2 H2O

STEP 3 Convert moles reactant --> moles product

Relate moles NH4NO3to moles product expected.

1 mol NH4NO3--> 2 mol H2O

Express this relation as the STOICHIOMETRICFACTOR.

2 mol H2O produced

1 mol NH4NO3 used

5.68 mol NH4NO3 2 mol H2O produced

1 mol NH4NO3used= 11.4 mol H2O produced


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454 g of NH4NO3--> N2O + 2 H2O

11.4 mol H2O 18.02 g

1 mol= 204 g H2O

STEP 4 Convert moles product(11.4 mol) --> mass product

Called the

THEORETICAL YIELD

ALWAYS FOLLOW THESE STEPS INSOLVING STOICHIOMETRY PROBLEMS!


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Reactions Involving aLIMITING REACTANT

In a given reaction, there is not enough of one

reagent to use up the other reagent completely.

The reagent in short supply LIMITSthequantity of product that can be formed.


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An ice cream sundae analogy for limiting reactants.


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LIMITING REACTANTS

Demo of limiting reactantsMethanol combustion


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Rxn 1: Balloon inflates fully, some Zn left

* More than enough Zn to use up the 0.100 mol HCl

Rxn 2: Balloon inflates fully, no Zn left* Right amount of each (HCl and Zn)

Rxn 3: Balloon does not inflate fully, no Zn left.* Not enough Zn to use up 0.100 mol HCl

LIMITING REACTANTS

React solid Zn with 0.100

mol HCl (aq)

Zn + 2 HCl ---> ZnCl2+ H2

1 2 3


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Rxn 1 Rxn 2 Rxn 3

mass Zn (g) 7.00 3.27 1.31

mol Zn 0.107 0.050 0.020

mol HCl 0.100 0.100 0.100

mol HCl/mol Zn 0.93/1 2.00/1 5.00/1

Lim Reactant LR = HCl no LR LR = Zn

LIMITING REACTANTS

React solid Zn with 0.100 molHCl (aq)

Zn + 2 HCl ---> ZnCl2+ H2

0.10 mol HCl [1 mol Zn/2 mol HCl]= 0.050 mol Zn


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PROBLEM: Mix 5.40 g of Al with8.10 g of Cl

2

. What mass of Al2

Cl6

can form?

Massreactant

StoichiometricfactorMoles

reactantMolesproduct

Massproduct


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Step 1 of LR problem:compare actualmole ratioof reactants totheoreticalmole ratio.


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2 Al + 3 Cl2 --->

Al2Cl6Reactants must be in the mole ratio

Step 1 of LR problem:compare actual mole ratio of

reactants to theoreticalmole ratio.

mol Cl2

mol Al =

3

2


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Deciding on the Limiting Reactant

If

There is not enough Al to use up all

the Cl2

2 Al + 3 Cl2 ---> Al2Cl6

mol Cl2

mol Al

>3

2

Lim reag = Al


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If

There is not enough Cl2to use

up all the Al

2 Al + 3 Cl2 ---> Al2Cl6

mol Cl2

mol Al Al2

Cl6


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CALCULATIONS: calculate mass ofAl2Cl6expected.

Step 1: Calculate moles of Al2Cl6expected based on LR.

0.114 mol Cl2 1 mol Al2Cl6

3 mol Cl2

= 0.0380 mol Al2Cl6

0.0380 mol Al2Cl6

266.4 g Al2Cl6

mol = 10.1 g Al2Cl6

Step 2: Calculate mass of Al2Cl6expectedbased on LR.


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Cl2was the limiting reactant.

Therefore, Al was present in

excess. But how much?

First find how much Al was required.

Then find how much Al is in excess.

How much of which reactant willremain when reaction is complete?


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2 Al + 3 Cl2 products

0.200 mol 0.114 mol = LR

Calculating Excess Al

Excess Al = Al available - Al required

0.114 mol Cl2 2 mol Al

3 mol Cl2 = 0.0760 mol Al req'd

= 0.200 mol - 0.0760 mol

= 0.124 mol Al in excess


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Determining the Formula of a Hydrocarbonby Combustion

Active Figure 4.9


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Using Stoichiometry toDetermine a Formula

Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g

of CO2and 0.1035 g of H2O.

CxHy + some oxygen --->

0.379 g CO2+ 0.1035 g H2O

What is the empirical formula of CxHy?


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First, recognize that all C in CO2and all H in H2O is from

CxHy.


0.379 g CO2+ 0.1035 g H2O

Puddle of CxHy

0.115 g

0.379 g CO2+O2

+O2 0.1035 g H2O

1 H2O molecule forms for

each 2 H atoms in CxHy

1 CO2molecule forms for

each C atom in CxHy


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First, recognize that all C in CO2and all H in H2O is from CxHy.

1. Calculate amount of C in CO2

8.61 x 10-3 mol CO2--> 8.61 x 10-3 mol C2. Calculate amount of H in H2O

5.744 x 10-3mol H2O -- >1.149 x 10-2 mol H


0.379 g CO2+ 0.1035 g H2O


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Now find ratio of mol H/mol C to find values of x and y in CxHy.

1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C

= 1.33 mol H/ 1.00 mol C= 4 mol H/ 3 mol C

Empirical formula = C3H4


0.379 g CO2+ 0.1035 g H2O

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2. Molecules Ions Stoic