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Chemical Compounds Molecular and Ionic compounds ●Mass percent composition ●Empirical formula determination ●Combustion analysis ●Oxidation states ●Compound Nomenclature ●Examples ● Chemical Compounds Molecular Compounds: Discrete aggregates of atoms. ●Held together by covalent bonds consisting of shared pairs of electrons ● Water Ammonia Methane © 2012 K. Brown Chemical Compounds Ionic Compounds: Groups of oppositely charged atoms or polyatomic species held together by electrostatic forces. No discrete, identifiable molecules. Sodium Chloride © 2012 K. Brown Chemical Compounds Molecular Compounds Empirical Formula: shows the type of atoms in the molecule and their relative numbers. Molecular Formula: shows the type of atoms in the molecule and their actual numbers. Hydrogen peroxide: Hydrazine: empirical: HO molecular: H2O2 empirical: NH2 molecular: N2H4 Chemical Compounds Molecular Compounds Structural Formula: shows the type of atoms in the molecule, their actual numbers and how they are connected. Water: © 2012 K. Brown Chemical Compounds Ratio of number of atoms Exact number of atoms Empirical formula Molecular formula Structural formula © 2012 K. Brown How atoms are connected Chemical Compounds Ionic Compounds Formula unit: shows the types of atoms and their relative numbers very similar to an empirical formula but ... it does not imply the existence of discrete units of ionic molecules. © 2012 K. Brown Chemical Compounds Ions with predictable charges Li + 1 N Al + 3 +1 Na Mg + 2 K +1 +1 Ca Rb Sr Cs +1 +2 +2 Ba −3 −2 F −2 Cl −1 Se−2 Br −1 O S −2 Te +2 © 2012 K. Brown −1 I −1 Chemical Compounds Mixed Molecular/Ionic Compounds Many compounds are a mixture of ions and covalently bonded species. Ammonium hydroxide NH4+ OH- © 2012 K. Brown Chemical Compounds Molar Mass The mass of 1 mol of a substance in grams is defined as it’s molar mass. For a monatomic element it is simply a matter of taking its atomic mass in u/atom and changing the unit to grams/mol. Chemical Compounds Molar Mass For compounds consisting of two or more elements we can find the molar mass (or the formula mass) by using the molecular formula or the formula unit and simply adding up the relative atomic masses of each atom in the formula. Chemical Compounds Molar Mass What is the molar mass of methane (CH4)? Molar mass methane=12.01g/ mol+4×1.008g/ mol =16.04 g/ mol Chemical Compounds Molar Mass What is the molar mass of sodium chloride, NaCl? NaCl is not actually a molecular formula since sodium chloride is an ionic compound, as previously discussed. However, we can treat it as such for the purpose of calculating a formula mass: Formula mass NaCl=22.99g/ mol+35.45g/ mol = 58.44g/ mol Chemical Compounds Mass Percent Composition A relative measure of the masses of the elements in a compound. ● Lavoisier's original estimate of the composition of water was 85% oxygen and 15% hydrogen by mass. Chemical Compounds Percent Composition What is the mass percent composition of carbon dioxide, CO2? Molar mass CO2=12.01g/ mol+2×16.00 g/ mol = 44.01g/ mol 12.01 g/ mol Mass %C= ×100=27.29% 44.01 g/ mol 2×16.00g/ mol Mass % O= ×100=72.71% 44.01g/ mol Chemical Compounds Determining Empirical Formulas Recall that the empirical formula gives us only the ratio of atoms to each other in a compound. The problem in finding empirical formulas is then generally “how do we determine the ratios of atom A to atom B (to atom C ... etc.) in our compound”. Chemical Compounds Determining Empirical Formulas A modern elemental analysis of water gives: %H = 11.19% %O = 88.81% both by mass. What is the empirical formula of water? In100 g of water mass H=11.19g mass O=88.81g Chemical Compounds Determining Empirical Formulas In100 g of water 1 moles H=11.19 g× =11.10mol 1.008 g/ mol 1 moles O=88.81g× =5.551mol 16.00g/ mol Chemical Compounds Determining Empirical Formulas Mol ratio of H to O in water: moles H/ molesO=11.10 mol/ 5.551 mol = 2.000 empirical formula of water is H2 O Chemical Compounds Write empirical formulas for the compounds represented by the molecular formulas. Workbook example 3.1 C4 H8 C H2 B2 H 6 BH 3 CCl 4 CCl 4 Chemical Compounds Determining Empirical Formulas 1. Use the given % composition, or determine what the %composition is and then find the mass of each element in a 100 g sample of the substance. 2. For each elemental mass, determine the number of mols of the element in the 100 g sample using the corresponding atomic mass from the periodic table. Chemical Compounds Determining Empirical Formulas 3. If necessary, divide each mol quantity by the smallest of the numbers of mols to get whole numbers. If any fractions result, multiply all results by the appropriate number to obtain all whole numbers. 4. Use the whole numbers from step 3 to generate an empirical formula. Chemical Compounds Combustion Analysis © 2012 K. Brown Chemical Compounds Combustion Analysis Vitamin C has only elements C, H and O in it. A 6.49 g sample is burned and 9.74 g of CO2 and 2.64 g of H2O are formed. What is the empirical formula of vitamin C? 12.01 g/ mol Fraction of C in CO2 = =0.2729 44.01 g/ mol mass C in CO2 sample=0.2729×9.74 g=2.66 g C 1 mols C in CO2 sample=2.66 g× 12.01g/ mol = 0.221mol C Chemical Compounds Combustion Analysis 2.016g/ mol =0.1119 18.01g/ mol mass H in H2 O sample=0.1119×2.64 g=0.296 g H 1 mols H in H2 O sample=0.296g× 1.008g / mol = 0.293 mol H Fraction of H in H2 O= Chemical Compounds Combustion Analysis Mols C=0.221 mol Mols H=0.293 mol Mols O=? Mass O=mass original sample−massC−mass H = 6.49 g−2.66g−0.296g=3.53 g 1 Mols O=3.53g× 16.00g/ mol = 0.221mol O Chemical Compounds Combustion Analysis Mols C=0.221 mol Mols H=0.293 mol Mols O=0.221 mol Dividing by the smallest: Mols C=1 Mols H=1.33 Mols O=1 Chemical Compounds Combustion Analysis Multiplying by 3: Mols C=3 Mols H=4 Mols O=3 Empirical formula of vitamin C: C3 H4 O3 Chemical Compounds Oxidation States The idea of the oxidation state or oxidation number comes about from considering “what if all compounds were ionic”. There would then be an identifiable charge on each atom in the molecule ... it is this 'what if' charge that is referred to as the oxidation state or oxidation number. Chemical Compounds Oxidation States 1. The oxidation state of an atom in it's elemental form is zero. 2. For ions composed of only one atom the oxidation number is equal to the charge on the ion. 3. The oxidation state of oxygen in most compounds is -2. A notable exception is in peroxides such as hydrogen peroxide in which the oxidation state of oxygen is -1. Chemical Compounds Oxidation States 4. The oxidation state of hydrogen is generally +1 except when it is bonded to metals such as sodium (NaH) in which case it's oxidation number is -1. 5. Fluorine has an oxidation number of -1 in its compounds … always. Group 1 elements have an oxidation number of +1 in their compounds … always. Group 2 elements have an oxidation number of +2 in their compounds … always. Chemical Compounds Oxidation States 6. The sum of the oxidation numbers of all of the atoms in a molecule or ion must equal the net charge on the molecule or ion. Chemical Compounds Oxidation States What is the oxidation number of each atom in the compound KMnO4? K=+1 4Oxygens=−2×4=−8 Mn=? Total=0 Mn=+7 Chemical Compounds Nomenclature of Compounds Metal-containing Binary Compounds Compounds containing only two different elements are binary compounds. If one of the elements is a metal they are named by using the element name of the metal followed by the name of the second element modified with the suffix 'ide'. Chemical Compounds Nomenclature of Compounds Metal-containing Binary Compounds exceptions: NaCl Al2O3 sodium chloride aluminum oxide NaOH KCN sodium hydroxide potassium cyanide Chemical Compounds Nomenclature of Compounds Metal-containing Binary Compounds Many metals can exist in more than one oxidation state and can form two or more compounds with a given anion: Fe+2 Fe+3 iron(II) ion iron(III) ion FeCl2 FeCl3 iron(II) chloride iron(III) chloride Chemical Compounds Nomenclature of Compounds Binary Acids When binary compounds with acidic properties are discussed in the context of their acidity we alter their names by using the prefix 'hydro' and suffix 'ic': HCl HF HCN hydrochloric acid hydrofluoric acid hydrocyanic acid Chemical Compounds Nomenclature of Compounds Polyatomic Ions Most polyatomic ions are anions rather than cations and the most common modifiers used as suffixes are 'ate' and 'ite', depending primarily on oxidation numbers. Also, it is most common for polyatomic anions to have oxygen in them and are called oxoanions. Chemical Compounds Nomenclature of Compounds Polyatomic Ions NH4+ CO3-2 NO3NO2- ammonium carbonate nitrate nitrite The 'ate' suffix is applied to the anion which has the primary atom in the higher oxidation state and 'ite' to the anion with the primary atom in the lower oxidation state. Chemical Compounds Nomenclature of Compounds Oxo Acids H2CO3 H3BO3 carbonic acid boric acid H3PO4 H3PO3 phosphoric acid phosphorous acid H2SO4 H2SO3 sulfuric acid sulfurous acid Chemical Compounds Nomenclature of Compounds Oxo Acids HClO HClO2 HClO3 HClO4 hypochlorous acid chlorous acid chloric acid perchloric acid Chemical Compounds Nomenclature of Compounds Binary Compounds of Nonmetals The nomenclature of these compounds is similar to the binary compounds mentioned earlier except for those in which the primary element can exist in multiple oxidation states. Chemical Compounds Nomenclature of Compounds Binary Compounds of Nonmetals HCl hydrogen chloride PCl5 PCl3 CO2 CO P4O6 phosphorous pentachloride phosphorous trichloride carbon dioxide carbon monoxide tetraphosphorus hexaoxide Chemical Compounds Nomenclature of Compounds Hydrates Some compounds have a fixed number of water molecules associated with them: CuSO4.5H2O copper(II) sulfate pentahydrate Chemical Compounds Nomenclature of Compounds Workbook Examples 3.6, 3.7 and 3.8 Name the compounds: PbCl 4 Lead ( IV ) chloride N I3 Nitrogen Triiodide P 4 S10 Tetraphosphorus DecaSulfide HIO3 ( aq) Iodic acid HF( aq) Hydrofluoric acid Chemical Compounds Chemical Equations The chemical equation is a shorthand description of a reaction that gives the molecular formulas of all of the reactants and products: 2H2(g) +O2(g) → 2H2 O(g) reactants products Chemical Compounds Chemical Equations The equation must be consistent with Lavoisier's law of conservation of mass. The total number of atoms of each element must be the same in the products and reactants. When this condition is satisfied we say the equation is balanced. CH4(g)+O2(g) → CO2(g) +H2 O(g) (unbalanced!) CH4(g)+2O2(g) → CO2(g)+2H2 O(g) (balanced) Chemical Compounds Balancing Chemical Equations 1.Write the unbalanced equation showing each reactant and product compound molecular formula. 2.Balance the equation. Start with the more complex compounds ●Balance atoms in elemental substances last ●Remove fractional coefficients by multiplying with an appropriate factor ● Chemical Compounds Balancing Chemical Equations 3. Indicate the states of matter of the reactants and products. Example: Tin oxide is heated with hydrogen gas to form tin metal and water vapor. Write the balanced equation that describes this reaction. Chemical Compounds Balancing Chemical Equations SnO2+H2 → Sn+H2 O SnO2+H2 → Sn+2H2 O or SnO2+H2 → Sn+H2 O2 ? SnO2+2H2 → Sn+2H2 O SnO2( s)+2H2(g) → Sn(s) +2H2 O(g) Chemical Compounds Valence Valence is the bonding capacity of a particular element. For the second period of the table the valences are: Li – 1 Be – 2 B - 3 C - 4 N - 3 O - 2 F - 1 Ne – 0 Chemical Compounds Valence We can use valences to predict formulas. For example, the combination of nitrogen and fluorine might be expected to give a molecular formula of NF3. Chemical Compounds Example Calculations It was found that 56g of iron combined with 32g of sulfur. Calculate the empirical formula of the compound formed. 1 moles Fe=56g× 55.85g/ mol = 1.0 mol 1 moles S=32g× 32.07g/ mol = 1.0 mol Chemical Compounds Example Calculations 1.0 1.0 empirical formula is FeS mole ratio Fe to S= Chemical Compounds Example Calculations A pure compound was found on analysis to contain 31.9% potassium, 28.9% chlorine and 39.2% oxygen. Calculate its empirical formula. In a 100g sample there will be 31.9g K, 28.9g chlorine and 39.2g oxygen. Chemical Compounds Example Calculations 1 moles K=31.9 g× 39.10g / mol = 0.815mol 1 moles Cl=28.9 g× 35.45g / mol = 0.815mol 1 moles O=39.2 g× 16.00g/ mol =2.45 mol Chemical Compounds Example Calculations Dividing by the smallest number, 0.815, gives a K:Cl:O ratio of 1:1:3 empirical formula: KClO3 Chemical Compounds Example Calculations A metal sulfide, MS2, is 40.06% by mass sulfur. What is the identity of the metal? One mole of MS2 will have two mols of sulfur atoms in it and one mol of the metal, M. Two mols of sulfur has a mass of 32.07 x 2 = 64.14 g, which represents 40.06% of the mass of one mole of MS2. Chemical Compounds Example Calculations We can now determine what the mass of one mol of MS2 will be: 64.14g 40.06 = mol x g/mol 100 100×64.14 x= g/mol 40.06 = 160.1 g/mol Chemical Compounds Example Calculations mass of M in one mole of MS2 =160.1g−64.14 g = 95.96g Referring to the periodic table we find that the element with this molar mass is molybdenum. Chemical Compounds Example Calculations Name the compounds: Cr2O3: MoS2: K2SO3: Fe2(SO4)3: SiCl4: P4O10: KHCO3: chromium(III) oxide molybdenum(IV) sulfide potassium sulfite iron(III) sulfate silicon tetrachloride tetraphosphorus decaoxide potassium hydrogen carbonate Further Reading 1.N.J. Tro, Chemistry: A Molecular Approach, Pearson-Prentice Hall, 2008. (your text!) 2.Extra problems at http://chem4823.usask.ca/chem112