Formula Mass the mass of an individual molecule or formula unit also known as molecular mass or molecular weight (MW) sum of the masses of the atoms in a single molecule or formula unit whole = sum of the parts! mass of 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu Tro, Chemistry: A Molecular Approach

Molar Mass of Compounds the relative masses of molecules can be calculated from atomic masses Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu since 1 mole of H2O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g so the Molar Mass of H2O is 18.02 g/mole Tro, Chemistry: A Molecular Approach

Example Example – Find the number of CO2 molecules in 10.8 g of dry ice How many molecules are in 50.0 g of PbO2? (PbO2 = 239.2)

Percent Composition Percentage of each element in a compound By mass Can be determined from the formula of the compound the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding Tro, Chemistry: A Molecular Approach

Example Calculate the mass percent composition of calcium chloride Calculate the mass percent of oxygen in acetic acid (C2H4O2)

Mass Percent as a Conversion Factor the mass percent tells you the mass of a constituent element in 100 g of the compound the fact that CCl2F2 is 58.64% Cl by mass means that 100 g of CCl2F2 contains 58.64 g Cl this can be used as a conversion factor 100 g CCl2F2 : 58.64 g Cl Tro, Chemistry: A Molecular Approach

Example Find the mass of table salt containing 2.4 g of Na Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?

Conversion Factors in Chemical Formulas chemical formulas have inherent in them relationships between numbers of atoms and molecules or moles of atoms and molecules these relationships can be used to convert between amounts of constituent elements and molecules like percent composition E. g Find the mass of hydrogen in 1.00 gal of water Tro, Chemistry: A Molecular Approach

Grams to Grams conversion Grams A  Moles A  Moles B  Grams B How many grams of sodium are in 6.2 g of NaCl? (Na = 22.99; Cl = 35.45)

Empirical Formula simplest, whole-number ratio of the atoms of elements in a compound can be determined from elemental analysis masses of elements formed when decompose or react compound combustion analysis percent composition Tro, Chemistry: A Molecular Approach

Finding an Empirical Formula convert the percentages to grams assume you start with 100 g of the compound skip if already grams convert grams to moles use molar mass of each element write a pseudoformula using moles as subscripts divide all by smallest number of moles if result is within 0.1 of whole number, round to whole number multiply all mole ratios by number to make all whole numbers if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3; if ratio 0.25 or 0.75, multiply all by 4; etc. skip if already whole numbers Tro, Chemistry: A Molecular Approach

Example 3.17 Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53% Tro, Chemistry: A Molecular Approach

Example Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) Determine the empirical formula of hematite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00)

Molecular Formulas The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound Tro, Chemistry: A Molecular Approach

Example Benzopyrene has a molar mass of 252 g/mol and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01) A compound has the emperical formular CH and a molar mass of 78.11 g/mol. Find its molecular formula

Combustion Analysis a common technique for analyzing compounds is to burn a known mass of compound and weigh the amounts of product made generally used for organic compounds containing C, H, O by knowing the mass of the product and composition of constituent element in the product, the original amount of constituent element can be determined all the original C forms CO2, the original H forms H2O, the original mass of O is found by subtraction once the masses of all the constituent elements in the original compound have been determined, the empirical formula can be found Tro, Chemistry: A Molecular Approach

Combustion Analysis Tro, Chemistry: A Molecular Approach

Combustion Analysis Unknown formula: CxHyOx (Oxygen can be replaced with other nonmetal) gCO2  moles CO2  moles C  gC gH2O  moles H2O  moles H  gH g O = g sample – (g H + g C) gO  moles O Follow steps in determine the empirical formula and molecular formula

Example 3.20 Determine the empirical formula of the compound Combustion of a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced the following: CO2 = 2.445 g H2O = 0.6003 g Determine the empirical formula of the compound Tro, Chemistry: A Molecular Approach

Example Upon combustion, a compound containing only carbon and hydrogen produced 1.60g CO2 and 0.819g H2O. Find the empirical formula Tro, Chemistry: A Molecular Approach

Chemical Reactions Reactants  Products Reactions involve chemical changes in matter resulting in new substances Reactions involve rearrangement and exchange of atoms to produce new molecules Elements are not transmuted during a reaction Reactants  Products

Chemical Equations Shorthand way of describing a reaction Provides information about the reaction Formulas of reactants and products States of reactants and products Relative numbers of reactant and product molecules that are required Can be used to determine weights of reactants used and products that can be made Tro, Chemistry: A Molecular Approach

CH4(g) + O2(g) ® CO2(g) + H2O(g) Combustion of Methane methane gas burns to produce carbon dioxide gas and gaseous water whenever something burns it combines with O2(g) CH4(g) + O2(g) ® CO2(g) + H2O(g) O C H C + O H O + 1 C + 4 H + 2 O 1 C + 2 O + 2 H + O 1 C + 2 H + 3 O Tro, Chemistry: A Molecular Approach

Combustion of Methane Balanced to show the reaction obeys the Law of Conservation of Mass, it must be balanced CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) O C + O H + O + H C + 1 C + 4 H + 4 O Tro, Chemistry: A Molecular Approach

Chemical Equations CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) CH4 and O2 are the reactants, and CO2 and H2O are the products the (g) after the formulas tells us the state of the chemical the number in front of each substance tells us the numbers of those molecules in the reaction called the coefficients Tro, Chemistry: A Molecular Approach

Chemical Equations CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) this equation is balanced, meaning that there are equal numbers of atoms of each element on the reactant and product sides to obtain the number of atoms of an element, multiply the subscript by the coefficient 1  C  1 4  H  4 4  O  2 + 2 Tro, Chemistry: A Molecular Approach

Symbols Used in Equations symbols used to indicate state after chemical (g) = gas; (l) = liquid; (s) = solid (aq) = aqueous = dissolved in water energy symbols used above the arrow for decomposition reactions D = heat hn = light shock = mechanical elec = electrical Tro, Chemistry: A Molecular Approach

Example Write a balanced equation for the combustion of butane, C4H10

aluminum(s) + oxygen(g) ® aluminum oxide(s) Practice when aluminum metal reacts with air, it produces a white, powdery compound aluminum oxide reacting with air means reacting with O2 aluminum(s) + oxygen(g) ® aluminum oxide(s) Al(s) + O2(g) ® Al2O3(s) Acetic acid reacts with the metal aluminum to make aqueous aluminum acetate and gaseous hydrogen acids are always aqueous metals are solid except for mercury Tro, Chemistry: A Molecular Approach

Classifying Compounds Organic vs. Inorganic in the18th century, compounds from living things were called organic; compounds from the nonliving environment were called inorganic organic compounds easily decomposed and could not be made in 18th century lab inorganic compounds very difficult to decompose, but able to be synthesized Tro, Chemistry: A Molecular Approach

Modern Classifying Compounds Organic vs. Inorganic today we commonly make organic compounds in the lab and find them all around us organic compounds are mainly made of C and H, sometimes with O, N, P, S, and trace amounts of other elements the main element that is the focus of organic chemistry is carbon Tro, Chemistry: A Molecular Approach

Carbon Bonding carbon atoms bond almost exclusively covalently compounds with ionic bonding C are generally inorganic when C bonds, it forms 4 covalent bonds 4 single bonds, 2 double bonds, 1 triple + 1 single, etc. carbon is unique in that it can form limitless chains of C atoms, both straight and branched, and rings of C atoms Tro, Chemistry: A Molecular Approach

Carbon Bonding Tro, Chemistry: A Molecular Approach

Classifying Organic Compounds there are two main categories of organic compounds, hydrocarbons and functionalized hydrocarbons hydrocarbons contain only C and H most fuels are mixtures of hydrocarbons Tro, Chemistry: A Molecular Approach

Classifying Hydrocarbons hydrocarbons containing only single bonds are called alkanes hydrocarbons containing one or more C=C are called alkenes hydrocarbons containing one or more CC are called alkynes hydrocarbons containing C6 “benzene” ring are called aromatic Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Naming Straight Chain Hydrocarbons consists of a base name to indicate the number of carbons in the chain, with a suffix to indicate the class and position of multiple bonds suffix –ane for alkane, –ene for alkene, –yne for alkyne Base Name No. of C meth- 1 hex- 6 eth- 2 hept- 7 prop- 3 oct- 8 but- 4 non- 9 pent- 5 dec- 10 Tro, Chemistry: A Molecular Approach

Functionalized Hydrocarbons functional groups are non-carbon groups that are on the molecule substitute one or more functional groups replacing H’s on the hydrocarbon chain generally, the chemical reactions of the compound are determined by the kinds of functional groups on the molecule Tro, Chemistry: A Molecular Approach

Functional Groups Tro, Chemistry: A Molecular Approach