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Chapter 9: Chemical Formulas and Chemical Compounds
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Chemical Formulas Represent the type and number of atoms of each element present in a compound –Molecular Formulas represent molecules –Ex: CH 4, O 2 –Formula Units represent salts (ionic) –Ex: NaCl, Ca 3 (PO 3 ) 2 Parenthesis are used when multiple polyatomic ions are present 7.1
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Ions Charged atoms or groups of charged atoms Form Ionic Compounds Represented by Formula Units Cation + Anion – 7.1
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Monatomic ions Charged single atoms (pg. 265-266) –Cations are identified by the elements name Ex. Li + is lithium ion d-block ions may have multiple charges, thus a roman numeral is used to designate the charged-block ions may have multiple charges, thus a roman numeral is used to designate the charge Ex. Fe +2 is Iron (II) and Fe +3 is Iron (III) –Anions drop their ending and replace it with – ide Ex. S -2 is sulfide
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Polyatomic Ions Charged groups of atoms (pg. 268) –Ex: OH -, CrO 4 2- –Oxyanions contain oxygen End in –ite or -ate Each combination has its own name 7.1
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Formula Units (ionic) –The net charge must be zero –Find the proper ratio of cations to anions using their charges Ex. Ca +2 and F - Requires 1 Ca +2 for two F -1 CaF 2 is the formula unit 7.1
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Naming Ionic Compounds Determine the names of the cation and the anion The name of the cation is written first, followed by the anion Ex.Mg(NO 3 ) 2 Magnesium is the cation, while Nitrate is the anion Magnesium Nitrate is the salt Ex.Co 2 (SO 4 ) 3 Cobalt (III) is the cation, while Sulfate is the anion Cobalt (III) Sulfate is the salt 7.1
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Binary Covalent Compounds (Prefix System) Use prefixes to describe the number of atoms of each element Prefixes 1 – mono- 2 – di- 3 – tri- 4 – tetra- 5 – penta- 6 – hexa- 7 – hepta- 8 – octa- 9 – nona- 10 – deca- 7.1
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Binary Covalent Compounds (Prefix System) Least electronegative element comes first If there is one atom for the first element, you may omit the mono- prefix in the name. Second element ends with –ide o and a are dropped from the prefix if the word following starts with a vowel Ex. N 2 O 4 Nitrogen is the least electronegative Di- stands for two and Tetra- stands for four Oxygen becomes oxide Dinitrogen Tetroxide 7.1
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Acids Consist of Hydrogen as a cation and an anion Naming depends on the anion Anion EndingChange -ideic and add hydro- (prefix) -ateic -iteous 7.1
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Acids Ex. HF Fluoride is the anion Add prefix hydro- and change –ide to –ic Hydrofluoric Acid Ex.H 2 CrO 4 Chromate is the anion Change –ate to –ic Chromic Acid Ex.H 2 SO 3 Sulfite is the anion Change –ite to –ous Sulfurous Acid 7.1
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Oxidation Numbers Assigned to atoms that compose a compound or ion Help in naming and writing formulas Rules 1.Atoms of pure elements have oxidation numbers of zero 2.Most electronegative atom has an oxidation number equal to the charge of its anion 3.Fluorine is always –1 4.Oxygen is almost always –2 5.Hydrogen is almost always +1 6.The sum of the oxidation numbers is zero for a neutral compound 7.The sum of the oxidation numbers is equal to the charge of a polyatomic ion 7.2
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Stock System of Naming Alternate method for naming binary covalent compounds The oxidation number follows the name of the first element The second element ends in –ide This can be used for all binary compounds Ex. N 2 O 5 Oxygen is most electronegative Its oxidation number is –2 So Nitrogen must be +5 Nitrogen (V) oxide 7.2
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Formula Mass –Sum of the masses of all the atoms represented in a chemical formula Units of amu Molar Mass –Mass of one mole of particles –Particles may be molecules, formula units or ions –Can be used as a conversion factor between mass and moles –Units of g/mol 7.3
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Percent Composition Percentage by mass of each element in a compound Compare the elements molar mass contribution, to the molar mass of the entire formula molar mass Assume 1 mole of compound %Composition = 100% x Mass of element in sample Mass of sample 7.3
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Percent Composition Find the mass % of water inZnSO 4 · 7H 2 O We have 7 mol of H 2 O per 1 mol of ZnSO 4
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You need molar masses!!! So, molar mass of H 2 O = 7(2x1.01 + 15.99)g = 7 x 18.01g = 126.07 g Total molar mass = molar mass of H 2 O + molar mass of ZnSO 4 = 126.07g + 161.42g = 287.49g So % composition H 2 O = 100% x mass H 2 O = d mass total =100% (126.07g / 287.49g) = 43.9 %
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Empirical Formula –Consists of the elements of a compound with subscripts showing the smallest whole number ratio of atoms that make up the compound 7.4
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Empirical Formula To determine an empirical formula 1.You need a ratio of the elements by mass or by percent composition 2.Divide each mass or percent by the elements molar mass to determine a molar ratio 3.Divide all numbers by the smallest number to get the ratio 4.Convert any decimals to the closest simple fraction 5.Multiply all numbers of the molar ratio by the LCM (lowest common multiple) of all the denominators of any fractions to determine the subscripts of the empirical formula 7.4
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Empirical Formula A Compound is found to contain 63.52% iron and 36.48% sulfur. What is the empirical formula? So 100 g of the sample will have 63.52g of iron and 36.48g of sulfur. Divide by Molar Mass 1. Fe = 63.52/55.847 = 1.137 mol 2. S = 36.48/32.066 = 1.137 mol
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Empirical Formula Divide both by smallest number: 1) 1.137 mol Fe/ 1.137 mol = 1 mol Fe 2) 1. 137 mol S / 1.137 mol = 1 mol S Thus we get a 1:1 ratio The empirical formula is FeS!
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Molecular Formula To determine a molecular formula from an empirical formula 1.Determine the molar mass and empirical mass of the molecule 2.Divide the molar mass by the empirical mass 3.Multiply all subscripts in the empirical formula by the result 7.4
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Molecular Formula Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu. Empirical mass = 12.011+1.008 = 13.019 amu So, 78.110/13.019 = 5.999 ~ 6 So the molecular formula is C 6 H 6
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