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Covalent Bonding

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.

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Covalent Bonding

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  1. Covalent Bonding Sec. 8.2: Naming Molecules

  2. Objectives • Identify the names of binary molecular compounds from their formulas • Name acidic solutions

  3. 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!

  4. 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)

  5. 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.

  6. 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.

  7. - CCl4 - As2O3 - P2O5 - SO2 - NF3 Practice problems

  8. 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

  9. Naming Acids • Acids are compounds that produce hydrogen ions in water solutions. • Two types exist: binary acids and oxyacids.

  10. 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 ….

  11. 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

  12. Naming Oxyacids • These acids contain hydrogen and an oxyanion (anion containing oxygen). • - Identify the oxyanion present. Example: HNO3 - the oxyanion is nitrate. • - 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.

  13. 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 ... Naming Oxyacids

  14. Practice Problems • - HI • - HClO3 • - HClO2 • - H2SO4 • - H2S

  15. 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.

  16. 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

  17. Practice Problems • hydrofluoric acid • bromic acid • carbonic acid • phosphorous acid • chlorous acid

  18. 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.

  19. 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)

  20. The gemstone opal is a hydrate of silicon dioxide.

  21. 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

  22. Practice Problems • CaCl2.2H2O is ____________________ • CuSO4.5H2O is ____________________ • Ba(OH)2.8H2O is __________________

  23. 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

  24. 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.

  25. 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

  26. 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%

  27. 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?

  28. 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%

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