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Compounds. combination of two or more elements molecular formulas for molecular compounds empirical formulas for ionic compounds. Organic Compounds Organic Chemistry. branch of chemistry in which carbon compounds and their reactions are studied.
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Compounds • combination of two or more elements • molecular formulas for molecular compounds • empirical formulas for ionic compounds
Organic CompoundsOrganic Chemistry • branch of chemistry in which carbon compounds and their reactions are studied. • the chemistry of compounds containing at least one carbon-hydrogen bond
Inorganic Compounds Inorganic Chemistry • field of chemistry in which the chemical reactions and properties of all the chemical elements and their compounds are studied, with the exception of the hydrocarbons (compounds composed of carbon and hydrogen) and their derivatives.
Naming Binary Molecular Compounds • For compounds composed of two non-metallic elements, the more metallic element is listed first. • To designate the multiplicity of an element, Greek prefixes are used: mono 1; di 2; tri 3; tetra 4; penta 5; hexa 6; hepta 7; octa 8
H2O water NH3 ammonia N2O nitrous oxide CO carbon monoxide CS2 carbon disulfide SO3 sulfur trioxide CCl4 carbon tetrachloride PCl5 phosphorus pentachloride SF6 sulfur hexafluoride Common Compounds
Hydrocarbons and Alcohols • alkanes – CnH2n+2 • alkenes – CnH2n • alkynes – CnH2n-2 • alcohols – ROH • where R refers to the hydrocarbon radical backbone • created by substituting an -OH functional group for a H atom in the hydrocarbon
methane – CH4 ethane – C2H6 propane – C3H8 butanes – C4H10 pentanes – C5H12 hexanes – C6H14 heptanes – C7H16 octanes – C8H18 nonanes – C9H20 decanes – C10H22 Alkanes – CnH2n+2
Butane molecules are present in the liquid and gaseous states in the lighter Butane
Naming Branch-Chain Alkanes • select the longest chain alkane as the base name • determine the side chains and give them a number corresponding to the carbon number on the base chain • use Greek prefixes of mono-(1), bi-(2), tri(3), etc. for multiplicity of same side chain
IonicCompounds Characteristics of compounds with ionic bonding: • non-volatile, thus high melting points • solids do not conduct electricity, but melts (liquid state) do • many, but not all, are water soluble
Valance, Charge on Ions • compounds have electrical neutrality • metals form positive monatomic ions • non-metals form negative monatomic ions
Valence of Metal Ions Monatomic Ions Group IA +1 Group IIA +2 Maximum positive valence equals Group A #
Valence of Non-Metal Ions Monatomic Ions Group VIA -2 Group VIIA -1 Maximum negative valence equals (8 – Group A #)
Polyatomic Ions • more than one atom joined together • have negative charge except for NH4+ and its relatives • negative charges range from -1 to -4
ammonium NH4+ perchlorate ClO41- cyanide CN1- hydroxide OH1- nitrate NO31- sulfate SO42- carbonate CO32- phosphate PO43- Polyatomic Ions
Names of Ionic Compounds 1. Name the metal first. If the metal has more than one oxidation state, the oxidation state is specified by Roman numerals in parentheses. 2. Then name the non-metal, changing the ending of the non-metal to -ide.
NaCl sodium chloride Fe2O3 iron(III) oxide N2O4 dinitrogen tetroxide KI potassium iodide Mg3N2 magnesium nitride SO3 sulfur trioxide Nomenclature
Nomenclature NH4NO3 ammonium nitrate KClO4 potassium perchlorate CaCO3 calcium carbonate NaOH sodium hydroxide
Bulk Substances • mainly ionic compounds • such as NaCl and most minerals • empirical formulas • structural formulas
Common Bulk Materials NaCl “table salt”
Molar Mass Sum atomic masses represented by formula atomic masses gaw molar mass MM
Percentage Composition description of a compound based on the relative amounts of each element in the compound
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3 35.453)amu = 119.377amu
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3 35.453)amu = 119.377amu 1(gaw) %C = 100 MM 1(12.011) %C = 100 = 10.061% C 119.377
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3 35.453)amu = 119.377amu 1(1.00797) %H = 100 = 0.844359% H 119.377
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = (12.011 + 1.00797 + 3 35.453)amu = 119.377amu 3(35.453) %Cl = 100 = 89.095% Cl 119.377
EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic? MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl = 119.377amu %C = 10.061% C %H = 0.844359% H %Cl = 89.095% Cl
Simplest (Empirical) Formula • formula describing a substance based on the smallest set of subscripts
EXAMPLE: Phosphorus burns in air to produce a white compound that is 43.7% P and 56.3% O by mass. What is the empirical formula of the compound? % 43.7% P 56.3% O Relative Number of Atoms Multiply by Integer (%/gaw) Divide by Smaller 2 1.00 2 2 2.50 5 43.7/30.97 = 1.41 56.3/15.9994 = 3.52 1.41/1.41 = 1.00 3.52/1.41 = 2.50 Empirical Formula P2O5
EXAMPLE: The burning of fossil fuels in air produces a brown-colored gas, a major air pollutant, that contains 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound? 2.34 %N = 100 = 30.5% N 2.34 + 5.34 5.34 %O = 100 = 69.5% O 2.34 + 5.34
EXAMPLE: The burning of fossil fuels in air produces a brown-colored gas, a major air pollutant, that contains 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound? %N = 30.5% N%O = 69.5% O Relative # Atoms Multiply % (%/gaw) Divide by Smaller by Integer N 30.5 30.5/14.0067 = 2.18 2.18/2.18 = 1.00 11.001 O 69.5 69.5/15.9994 = 4.34 4.34/2.18 = 1.99 11.992 Empirical Formula NO2
Molecular Formula • the exact proportions of the elements that are formed in a molecule
Molecular Formula from Simplest Formula formula mass FM sum of the atomic weights represented by the formula molar mass = MM = X FM
Molecular Formula from Simplest Formula first, knowing MM and FM X = MM/FM then MF = X EF
EXAMPLE: A colorless liquid used in rocket engines, whose empirical formula is NO2, has a molar mass of 92.0. What is the molecular formula? FM = 1(gaw)N + 2(gaw)O = 46.0 MM 92.0 X = = = 2 FM 46.0 thus MF = 2 EF N2O4