Covalent Bonding

Sec. 8.2: Naming Molecules

Objectives

Identify the names of binary molecular compounds from their formulas

Name acidic solutions

Naming Binary Molecular Compounds A binary molecular compound contains

___ different elements, both ________ , bonded through a __________ bond.

Many have common names: – H2O is named water

– NH3 is named ammonia

– See Table 5 p. 251: MUST MEMORIZE!

Naming Binary Molecular Compounds Scientific names reveal a molecular compound’s

composition Rules 1. Name the first element in the formula, using the

entire element name 2. Name the second element in the formula, using the

ROOT of the element name and the SUFFIX -IDE. 3. Add a prefix to each of the names to indicate the

number of atoms of that type that are present. (see pg. 248)

Exceptions The first element name NEVER uses the

prefix mono-. Example: CO is carbon monoxide, NOT monocarbon monoxide.

The final letter in the prefix may be dropped when the element name begins with a vowel. Example: CO is carbon monoxide, NOT carbon monooxide.

Practice: H2O Name of the first element is hydrogen. A prefix

is needed.– There are 2 hydrogen atoms, so the first name

becomes dihydrogen. The root of the second element with -ide is

oxide.– There is 1 oxygen atom, so the second name

becomes monoxide. Dihydrogen monoxide is the chemical name for

water.

Practice problems

1 - CCl42 - As2O3

3 - P2O5

4 - SO2

5 - NF3

Writing Formulas of Molecular Compounds Since the name reveals the

composition, simply translate the information given into a formula. For example, dinitrogen pentaoxide is N2O5.

Practice:– sulfur difluoride– silicon tetrachloride– tetrasulfur tetranitride

Naming Acids

Acids are compounds that produce hydrogen ions in water solutions.

Two types exist: binary acids and oxyacids.

Binary Acids Contain hydrogen and one other element The prefix hydro- is used to name the

hydrogen part of the compound. The rest of first word consists of the root of

the second element with the suffix -ic. The second word is always acid. HCl is hydrochloric acid HBr is ….

Exceptions There are a few acids with that ARE

NOT BINARY that are named according to these rules.

If NO OXYGEN is present in the formula, use the rules for binary acids, except the root of the second name comes from the polyatomic ion name.

Example: HCN is hydrocyanic acid

Naming Oxyacids These acids contain hydrogen and an oxyanion

(anion containing oxygen).1 - Identify the oxyanion present.

Example: HNO3 - the oxyanion is nitrate.

2 - Name the oxyacid using the root of the anion, a suffix, and the word acid.• If the suffix of the anion is -ate, it is replaced by -ic.• If the suffix of the anion is -ite, it is replaced by -ous.

Naming Oxyacids HNO3

HNO2

H3PO4

Anion ends in -ate, so it is named nitric acid

Anion is nitrite, so it is named nitrous acid.

Anion is phosphate, so it is named ...

Practice Problems

1 - HI2 - HClO3

3 - HClO2

4 - H2SO4

5 - H2S

Writing Formulas of Acids

Work backwards from the name, focusing on the rules that were used.

Example: hydroiodic acid– Because “hydro” is used, we know this is a binary

acid.– The acid contains hydrogen and iodine. We know

the second element iodine because of “iodic”.– The formula is derived by criss-crossing H+ and

I-. The acid is HI.

Example: Sulfurous acid

Sulfurous acid was named from a sulfur oxyanion with an -ite ending: sulfite

The oxyacid contains hydrogen and the oxyanion sulfite: H+ and SO3

-2

The formula for this acid would therefore be H2SO3

Practice Problems

hydrofluoric acid bromic acid carbonic acid phosphorous acid chlorous acid

Hydrates As ions in a solution react to form an ionic

compound, water often adheres to the ions as the compound forms.

These molecules of water become part of the structure of the ionic crystal that forms.

The water is called water of hydration and the solid compound formed is called a hydrate.

Hydrates

A hydrate is defined as an ionic compound that has a specific number of water molecules bound to its atoms.

Copper (II) sulfate pentahydrate

(CuSO4 . 5H2O)

The gemstone opal is a hydrate of silicon dioxide.

Naming hydrates In the formula for a hydrate, the number

of water molecules associated with one formula unit is written following a dot:

Na2CO3.10H2O

Name the ionic compound and use a prefix, to indicate the number of water molecules, with the word hydrate: sodium carbonate decahydrate

Practice Problems

CaCl2.2H2O is ____________________

CuSO4.5H2O is

____________________

Ba(OH)2.8H2O is __________________

Writing Formulas for hydrates

Translate the information given in the name into a formula– cobalt (II) chloride hexahydrate

– CoCl2.6H2O

– Practice: iron (III) phosphate tetrahydrate

Uses of hydrates The anhydrous form of a hydrate - the one

in which all the water has been removed through heating- can be used as a dessicant or drying agent. It will absorb water from the air or from a liquid environment. Such substances are called hygroscopic. If the substance absorbs enough water from the air to dissolve and form a liquid solution, it is said to be deliquescent.

Determining % of water in a hydrate Recall the method for determining

percent by mass (pg. 75) For hydrates, the percent by mass of

water will equal mass of water x 100

mass of hydrate

Example: What is the water content in MgSO4

.7H2O? Add up the atomic mass (rounded to the nearest

tenth) for each component of the compound; this is the mass of the hydrate. Mg:24.3, S:32.1, O: 4 x 16.0 = 64.0 H2O: (2 x 1.0) + 16.0 = 18.0

7H2O = 7 x 18.0 = 126.0 (mass of water)

Total mass = 246.4 Divide the mass of water by the mass of hydrate and

multiply by 100 % mass of water = 126.0/246.4 x 100 = 51.14%

Practice Problem: A 4.3 g sample of strontium chloride hexahydrate was heated. The mass of the anhydrous compound that remained after heating was 2.6 g. What percentage of water was in the hydrate?

SrCl2.6H2O

Mass of anhydrous + Mass of water = Mass of hydrate, so Mass of water = 4.3 g - 2.6 g = 1.7 g

% mass of water = mass water/mass hydrate x 100

= 1.7 g/4.3 g x 100 = 39.5%