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Transcript of Covalent Bonding Sec. 8.2: Naming Molecules. Objectives n Identify the names of binary molecular...
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Covalent Bonding
Sec. 8.2: Naming Molecules
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Objectives
Identify the names of binary molecular compounds from their formulas
Name acidic solutions
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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!
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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)
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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.
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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.
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Practice problems
1 - CCl42 - As2O3
3 - P2O5
4 - SO2
5 - NF3
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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
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Naming Acids
Acids are compounds that produce hydrogen ions in water solutions.
Two types exist: binary acids and oxyacids.
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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 ….
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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
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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.
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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 ...
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Practice Problems
1 - HI2 - HClO3
3 - HClO2
4 - H2SO4
5 - H2S
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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.
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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
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Practice Problems
hydrofluoric acid bromic acid carbonic acid phosphorous acid chlorous acid
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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.
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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)
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The gemstone opal is a hydrate of silicon dioxide.
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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
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Practice Problems
CaCl2.2H2O is ____________________
CuSO4.5H2O is
____________________
Ba(OH)2.8H2O is __________________
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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
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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.
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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
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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%
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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?
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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%