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Chapter 7 – Chemical Formulas & Compounds

Chapter 7 – Chemical Formulas & Compounds. Chapter 7 – Section 2 – Oxidation Numbers. II. Oxidation Numbers Oxidation numbers or state are assigned to atoms in compounds or ions to indicate their distribution of electrons; they are used to help name compounds and write formulas

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Chapter 7 – Chemical Formulas & Compounds

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  1. Chapter 7 – Chemical Formulas & Compounds

  2. Chapter 7 – Section 2 – Oxidation Numbers • II. Oxidation Numbers • Oxidation numbers or state are assigned to atoms in compounds or ions to indicate their distribution of electrons; they are used to help name compounds and write formulas • Na+1 Mg+2 Al+3 N-3 O-2 Cl-1 Ne0 • A. Assigning Oxidation Numbers • Usually shared electrons belong to the most electronegative atom in the compound

  3. 1. The atoms in any pure element have an oxidation number of zero • 2. In a binary molecular compound, the most electronegative atom is given an oxidation number equal to the charge it would have as an anion; less electronegative atoms are assigned numbers equal to the charge they would have as a cation • 3. Fluorine is the mostelectronegative element, so its oxidation number is always -1

  4. 4. Oxygenis usually a -2, unless it is a peroxide, like H2O2, when it will be -1 or combined with fluorine (OF2), when it will be +2 • 5. Hydrogen is +1 in all compounds where it is the least electronegative, but is -1 in compounds with metals • 6. The sum of all the oxidation numbers in a neutral compound must equal zero • 7. The sum of the oxidation numbers in a polyatomic ion will equal the charge of the polyatomic ion • 8. A monatomic ion’s oxidation number is the same as its charge

  5. B. Using Oxidation Numbers for Formulas & Names • 1. Many nonmetals can have more than one oxidation number • 2. Prefixes can be used to indicate the oxidation numbers (Carbon dioxide or Carbon monoxide) • 3. The Stock system can also be used, which uses Roman numerals instead of prefixes [Carbon (I) oxide or Carbon (II) oxide]

  6. Chapter 7, Section 3 – Using Chemical Formulas • III. A chemical formula can be used to calculate formula mass, molar mass, and percentagecompositionof mass for a compound • A. Formula Masses • 1. The average atomic masses (mass on the periodic table) of the elements in a compound are added together to get the average atomic mass of the molecule or formula unit

  7. 2. The mass of a moleculeis referred to as its molecular mass • 3. The mass of all the atoms in any unit of a formula (molecule, formula unit, or ion) is referred to as the formula mass • B. Molar Masses • 1. A mole of any substance will be 6.022 x 1023 particles of that substance • 2. The molar mass of the substance will be the atomic mass of the substance in grams

  8. 3. The molar mass and the formula mass will be numerically equal (the same), but the formula mass unit will be amu, and the molar mass unit will be grams • C. Molar Mass as a Conversion Factor • 1. Molar mass can be used as a conversionfactor to change moles to grams • 2. Multiply moles by molar mass to get grams • 3. To change grams to moles, divide grams by molar mass

  9. D. Percentage Composition • 1. To find the percentage of an element in a compound, divide the mass of the element by the mass of the compound and multiply by 100 • 2. The percentage of the element will be the same no matter how much of the compound you have • 3. Divide the grams of the element in 1 mole of the compound by the molar mass of the compound and multiply by 100 • 4. The percentage of mass for each element in a compound is known as the percentage composition of the compound

  10. Chp. 7 – Sec. 4 – Determining Chemical Formulas • IV. An empirical formula is made of element symbols and subscripts to show the smallest whole number mole ratio of each atom in the compound • For ioniccompounds, the formula unit is usually the empirical formula • For covalent compounds, the empirical formula does not necessarily tell the actual number of atoms in the molecule • An example is diborane (B2H6), whose empirical formula is BH3, because the empirical formula is a more simple ratio than the molecular formula

  11. A. Calculations of Empirical Formulas • 1. Convert percentage composition to mass • 2. Assume you have 100.00 grams of the compound • 3. Find the mass of each element in the 100 grams • 4. Change the mass of each element to moles • 5. Determine which element has the smallestnumber of moles • 6. Divide both numbers of moles by the smallest one (to find the smallest ratio) • 7. Round the answers to the nearest whole numbers, and these will be the subscripts in the empirical formula

  12. 8. If mass is known instead of percentage composition, then change mass to moles and continue at Step 5 • B. Calculation of Molecular Formulas • 1. The molecular formula shows the actual number of atoms in a molecule • 2. Two different compounds can have the same empirical formula, but have different molecular formulas [Examples: ethene (molecular – C2H4; empirical – CH2) and cyclopropane (molecular – C3H6: empirical – CH2) • 3. To change the empiricalformula to the molecularformula, divide the molecular mass by the empirical mass

  13. 4. When you divide, if your answer does not come out to be a whole number, then follow the 0-9 rule, which states that if the 1st decimal number is a 0, drop it and keep the whole number, but if the 1st decimal number is a 9, round up to the next whole number • 5. If the first decimal number is not 0 or 9, then begin multiplyingyour answer by 2 (or 3 or 4 or 5, etc. if necessary) until you get a number that is 0 or 9 • 6. The answer you get when you divide (and multiply, if necessary) will then be multiplied by each subscript in the empirical formula

  14. 7. After multiplying, this will be the molecular formula • C. Hydrates • 1. A hydrate is a molecule that has one or more watermolecules chemically attached to the compound • 2. To show this chemical attachment, the formula shows a dot between the compound and the water (CuSO4· 5H2O) • 3. The coefficientin front of the water molecule shows how many watermolecules are attached • 4. In calculating the molecular formula for a hydrate, the numbers obtained from dividing and multiplying will be used as coefficients, rather than subscripts

  15. Section Review Questions, pp. 251-252 • 1. a) Ions formed from a single atom • b) Examples include Na+1, Mg+2, and Cl-1 • 2. The nitrate ion, NO3-1, has one more oxygen atom than the nitrite ion, NO2-1, does • 3. a) K+1 b) Ca+2 c) S-2 d) N-3 e) Ba+2 f) Br-1 • 4. a) Na+1 b) Al+3 c) Cl-1 d) N-3 e) Fe+2 f) Fe+3 • 5. a) potassium ion (or potassium cation) • b) magnesium ion (or magnesium cation) • c) Aluminum ion (or aluminum cation) • d) chloride ion (or chloride anion) • e) oxide ion (or oxide anion) • f) calcium ion (or calcium cation)

  16. 6. a) NaI b) CaS c) ZnCl2 d) BaF2 e) Li2O • 7. a) potassium chloride b) calcium bromide • c) lithium oxide d) magnesium chloride • 8. a) CrF4, chromium (II) fluoride • b) NiO, nickel (II) oxide • c) Fe2O3, iron (II) oxide • 9. The less electronegative element is written first • 10. a) carbon dioxide b) carbon tetrachloride • c) phosphorus pentachloride • d) selenium hexafluoride • e) diarsenic pentoxide

  17. 11. a) CBr4 b) SiO2 c) P4O10 d) As2S3 • 12. Binary acids, such as HCl and HBr, contain only two elements, usually hydrogen and a halogen. Oxyacids, such as H2SO4 and HNO3, contain hydrogne, oxugen, and a third element • 13. a) an ionic compound composed of a cation and the anion from an acid • b) Examples include NaCl and MgSO4 • 14. a) hydrofluoric acid b) hydrobromic acid • c) nitric acid d) sulfuric acid • e) phosphoric acid

  18. 15. a) H2SO3 b) HClO3 c) HCl d) HClO • e) HClO4 f) H2CO3 g) HC2H3O2 • 16. a) NaF b) CaO c) K2S d) MgCl2 e) AlBr3 • f) Li3N g) FeO • 17. a) ammonium ion b) chlorate ion • c) hydroxide ion d) sulfate ion e) nitrate ion • f) carbonate ion g) phosphate ion • h) acetate ion i) hydrogen carbonate ion (or • j) chromate ion bicarbonate ion) • 18. a) NH4+1 b) C2H3O2-1 c) OH-1 d) CO3-2 • e) SO4-2 f) PO4-3 g) Cu+2 h) Sn+2 i) Fe+3 • j) Cu+1 k) Hg2+2 l) Hg+2

  19. 19. a) iron (II) ion b) iron (III) ion c) lead (II) ion • d) lead (IV) ion e) tin (II) ion f) tin (IV) ion • 20. a) carbon (IV) bromide b) silicon (IV) oxide • c) phosphorus (V) oxide d) arsenic (III) sulfide • 21. a) PI3 b) SCl2 c) CS2 d) N2O5 • 22. a) numbers assigned to bonded atoms in molecular compounds or polyatomic ions to indicate the general distribution of electrons • b) they aid in naming compounds, writing formulas, balancing chemical equations, and studying chemical reactions

  20. 23. a) sodium chloride b) potassium fluoride • c) calcium sulfide d) cobalt (II) nitrate • e) iron (III) phosphate f) mercury (I) sulfate • g) mercury (II) phosphate • 24. a) +1, -1 b) +3, -1 c) +4, -2 d) +1, -1 • e) +5, -2 f) +1, +5, -2 • 25. a) +5, -2 b) +7, -2 c) +5, -2 d) +6, -2 • e) +4, -2

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