chapter 7 – chemical formulas and chemical compounds

Chapter 7 – Chemical Formulas and Chemical Compounds

Taken from Modern Chemistry written by Davis, Metcalfe, Williams

& Castka

Section 7.1 – Chemical Names and Formulas

Objectives

Students will be able to :• Explain the significance of a chemical formula• Determine the formula of an ionic compound formed between two given ions•Name an ionic compound given its formula•Using prefixes, name a binary molecular compound from its formula.•Write the formula of a binary molecular compound given its name.

HW – Notes on section 7.1 pgs 203-215


Significance of a chemical formula

The chemical formula indicates the relative number of atoms of each element in a chemical compound.

C12H22O11

Elements’ subscripts indicate the number of atoms in the compound.

Al2(SO4)3

Note how in this example parenthesis surround a polyatomic anion and the subscript refers to the entire unit


Monatomic IonsGroup 1 metals lose one e- to give 1+ cations.

Group 2 metals lose two e- to give 2+ cations.

By gaining or losing electrons many main-group elements form ions with stable configurations.

Ions formed from a single atom are known as monatomic ions


Monatomic Ions(continued)

The nonmetals in groups 15, 16 & 17 gain e- to form anions.

Not all main-group elements readily form ions, C and Si form covalent bonds where they share electrons.


Monatomic Ions(continued)

Elements which give up 1 or more e- and take a positive (+) charge are called cations.

Elements which gain 1 or more e- and take a negative (-) charge are called anions.


Naming Monatomic Ions(continued)

Cation naming is simple, element name and the word cation.

For anions you drop the end of the element name and add –ide to the root.

Potassium cation

Magnesium cation

Fluorine fluoride Nitrogen Nitride


Binary Ionic Compounds

Compounds composed of two different elements are known as binary compounds

Cation goes first : Mg2+, Br-, Br-

Balance to become electrically neutral

And you get


Naming Binary Ionic Compounds

The naming system involves combining the names of the compound’s positive and negative ions

And you get

Aluminum cation & oxide

Aluminum Oxide



Some elements form more than one cation, (no elements form more than one monoatomic anion) each with a different charge – add Roman Numerals

Iron(III) oxide

Iron(II) oxide

The Stock system of nomenclature



Oxyanions each is a polyatomic ion that contains oxygen.

Nitrite

Nitrate

Compounds Containing Polyatomic Ions


Naming Binary Molecular Compounds

1 mono-2 di-3 tri-4 tetra-5 penta-6 hexa-7 hepta-8 octa-9 nona-10deca-

1. Less electronegative element given first, prefix only for multiples

2. Second element named with prefix indicating # of atoms, with few exceptions ends with –ide (only 2 elements)

3. The o or a at the end of the prefix is dropped when the word following begins with another vowel.


Naming Binary Molecular Compounds

Example pg 212

Prefix need if more than one

Less-electronegative element

Prefix indicating number

Root name +ide


Covalent-Network CompoundsSimilar to naming molecular compounds


Acids and SaltsAcids are a specific class of compound which usually

refer to a solution of water of one of these special compounds.


Acids and Salts - (continued)An ionic compound composed of a cation and the

anion from an acid is often referred to as a salt.


Practices

PRACTICE – pg 207 Q 1 & 2 all




Practice Naming


Quiz Break Quiz Key

Section 7.1 – Chemical Names and Formulas - POGIL

KEY

Section 7.1 – V2 – Ions and Compounds

Quiz Break Quiz Key

Section 7.1 – Chemical Names and Formulas - POGIL

KEY

Section 7.2 - Oxidation Numbers

Objectives

Students will be able to :• List the rules for assigning oxidation numbers.•Give the oxidation number for each element in the formula of a chemical compound.•Name the binary molecular compounds using oxidation numbers and the Stock sytem.



Oxidation Numbers

To indicate the general distribution of electrons among bonded atoms in molecular compounds , oxidation numbers (or states) are assigned to the atoms that compose the same.

Some are arbitrary, but they are useful in naming compounds, in writing formulas and in balancing equations.


Assigning Oxidation Numbers – the rules The following are guidelines...

1. Atoms of pure elements have an oxidation number of zero.

All zero.


Assigning Oxidation Numbers – the rules

2. The more-electronegative element in a binary compound is assigned the number equal to the negative charge it would have as an anion. The less-electronegative is assigned the number equal to the positive charge it would have as a cation.


Assigning Oxidation Numbers – the rules

3. Fluorine has an oxidation number of -1 as it is the most electronegative element.


Assigning Oxidation Numbers – the rules 4. Oxygen has an oxidation number of -2 in almost all

compounds. Exceptions peroxides is -1, compounds with halogens +2

5. Hydrogen has an oxidation number of +1 in all compounds containing elements that are more- electronegative; it has an oxidation number of -1 in compounds with metals.


Assigning Oxidation Numbers – the rules 6. The algebraic sum of the oxidation numbers of all atoms

in a neutral compound = zero.

7. The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion = the charge of the ion.

8. Rules 1-7 apply to covalently bonded atoms, oxidation numbers can also be assigned to atoms in ionic compounds.

Section 7.2 - Oxidation NumbersUsing Oxidation Numbers for Formulas and Names

Many non-metals have more than one oxidation state. Recall the use of Roman numerals to denote charges.

Formula Prefix system Stock System

PCl3phosphorus trichloride

phosphorus(III) chloride

NO nitrogen monoxide

nitrogen(II)oxide

PbO2lead

dioxide lead (IV) oxide

Section 7.2

Practices

Ba(NO3)2

Is the substance elemental? No, three elements are present.

Is the substance ionic? Yes, metal + non-metal.

Are there any monoatomic ions? Yes, barium ion is monoatomic. Barium ion = Ba2+ Oxidation # for Ba = +2

Which elements have specific rules? Oxygen has a rule....-2 in most compounds Oxidation # for O = -2

Which element does not have a specific rule? N does not have a specific rule. Use rule 8 to find the oxidation # of N

Let N = Oxidation # for nitrogen (# Ba) (Oxid. # of Ba) + (# N) (Oxid. # N) + (# of O) (Oxid. # of O) = 0 1(+2) + 2(N) + 6(-2) = 0 N = +5

Section 7.2

Practices

NF3 Is the substance elemental? No, two elements are present.

Is the substance ionic? No, two non-metals.

Are there any monoatomic ions? Since it is molecular, there are no ions present.

Which elements have specific rules? F = -1

Which element does not have a specific rule? N does not have a specific rule. Use rule 8 to find the oxidation # of N

Let N = oxidation # of N (# N) (Oxid. # N) + (# F) (Oxid. # F) = 0 1(N) + 3(-1) = 0 N = +3

Section 7.2 Practices

(NH4)2SO4 Is the substance elemental? No, four elements are present.

Is the substance ionic? Yes, even though there are no metals present, the ammonium ion is a common polyatomic cation.

Are there any monoatomic ions? No, the cation and anion are both polyatomic.

Which elements have specific rules? H = +1 because it is attached to a non-metal (N) O = -2

Which elements do not have a specific rule? Neither N nor S has a specific rule. You must break the compound into the individual ions that are present and then use rule 9 to find the oxidation numbers of N and S. Notice that if you try to use rule 8, you end up with one equation with two unknowns: 2N + 8(+1) + 1S + 4(-2) = 0 The two ions present are NH4

+ and SO42-.

N + 4(+1) = +1 so N = -3 S + 4(-2) = -2 so S = +6

Section 7.2

Practices

PRACTICE – pg 218 Q 1 all

Practice Sheet

Key

Section 7.2

Quiz

Key

Section 7.3 - Using Chemical Formulas

Objectives

Students will be able to :• Calculate the formula mass or molar mass of any given compound.• Use molar mass to convert between mass in grams and amount in moles of a chemical compound.• Calculate the # of molecules, formula units, or ions in a given molar amount of a chemical compound.•Calculate the % composition of a given chemical compound.


Section 7.3 - Using Chemical FormulasFormula MassesThe formula mass formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all the atoms represented in the formula.

Book exampleAverage atomic mass of H : 1.01 amuAverage atomic mass of O : 16.00 amu

H2O (2 x 1.01 amu) + 16 amu = 18.02 amu


Molar MassesA compound’s molar mass is numerically equal to its formula mass.

Book exampleFormula mass of H2O = 18.02 amu

Which is also the molar mass of water 18.02 g/mol.

Section 7.3 – Using Chemical FormulasMolar Mass as a Conversion Factor – Remember our old friends. . .

There are 3 mole equalities. They are:

1 mol = 6.02 x 1023 particles1 mol = g-formula-mass (periodic table)

1 mol = 22.4 L for a gas at STP*

These become. . .

[-------------] or [-------------] 1 mol

6.02 x 1023 particles 1 mol 6.02 x 1023 particles

[-------------] or [-------------] 1 mol g-formula-mass (periodic table) 1 mol

g-formula-mass (periodic table)

[----] OR [----] 1 mol 22.4 L 1 mol

22.4 L


Percentage CompositionThe percentage by mass of each The percentage by mass of each element in a compound is known as element in a compound is known as the the percentage composition percentage composition of the of the compound.compound.

Mass of X in sample of compound

X 100 % = Mass % X in compoundMass of sample of

compound


Percentage Composition - example

1 Mg = 1 x 24 = 24

2 N = 2 x 14 = 28

6 O = 6 x 16 = 96 148g/mole

% Mg = 24/148 x 100 = 16.2%

% N = 28/148 x 100 = 18.9%

% O = 96/148 x 100 = 64.0%

Percent Composition Example:Calculate the percent composition of Mg(NO3)2

Double check - do they total 100%?


Quiz Break

Key

Percentage Composition - practice

Section 7.4 - Determining Chemical Formulas

Objectives

Students will be able to :•Define empirical formula, and explain how the terms applies to ionic and molecular compounds.•Determine an empirical formula from either a percentage or a mass composition.•Explain the relationship between the empirical formula and the molecular formula of a given compound.•Determine a molecular formula from an empirical formula



Calculation of Empirical Formulas

An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratios of the different atoms in the compound.


Calculation of Empirical Formulas - exampleLet’s Determine the empirical formula for a compound which is 54.09% Ca, 43.18% O, and 2.73% H

1) Divide each percent by that element's atomic weight.

Ca = 54.09/40 = 1.352

O = 43.18/16 = 2.699

H = 2.73/1 = 2.73

2) To get the answers to whole numbers, divide through by the smallest one.

1.352/1.352 = 1

2.699/1.352 = 2

2.73/1.352 = 2

This gives us CaO2H2 better yet Ca(OH)2


Calculations of Molecular Formulas

An empirical formula may or may not be a correct molecular formula.

Book example , diborane1’s empirical formula is BH3, any multiple of that equals the same ratio – B2H6,B3H9,B4H12 etc

It is a colorless gas at room temperature with a repulsively sweet odor. Diborane mixes well with air, easily forming explosive mixtures. Diborane will ignite spontaneously in moist air at room temperature.


Calculations of Molecular Formulas – (continued) The relationship between an empirical formula and a molecular formula is seen below:

X(empirical formula) = molecular formula

X is a whole-number multiple indicating the factor that you need to multiply the empirical

formula by to get the molecular formula.


Calculations of Molecular Formulas – (continued) X = molecular formula

empirical formula

Formula mass of diborane = 27.67 amuEmpirical mass of diborane 13.84 amu

The molecular formula of diborane is therefore B2H6

(BH3)2 = B2H6


PRACTICE

Empirical Formula Practice

Molecular Formula Practice

chapter 7 – chemical formulas and chemical compounds

Documents

chemical names

chemical compoundstaken

chemical formulas

binary compoundscation

different elements

maingroup elements form

gain e

multiplessecond element