A sample problem 3.43) Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4%, H: 6.21%, S: 39.5%, O: 9.86%. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 162 g? empirical formula = molecular formula = C6H10S2O *Note that we arbitrarily chose a 100-g sample to make our math easier. 11/8/2018

Another sample problem 3.52) The empirical formula of a compound is CH. If the molar mass of this compound is about 78g, what is its molecular formula? Figure out molar mass of the empirical formula. Divide the molar mass of the compound by the molar mass of the empirical formula. Multiply each subscript on the empirical formula by this quotient to get the molecular formula. 78/13.01=6 C6H6 11/8/2018

Yet another sample problem Practice exercise, p. 71) A sample of a compound containing boron and hydrogen contains 6.444 g of boron and 1.803 g of hydrogen. The molar mass of the compound is about 30 g. What is its molecular formula? Figure out molar mass of the empirical formula. Divide the molar mass of the compound by the molar mass of the empirical formula. Multiply each subscript in the empirical formula by this whole number to get the molecular formula. B2H6 11/8/2018

A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction 3 ways of representing the reaction of H2 with O2 to form H2O: reactants products 11/8/2018

How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO 2 grams Mg + 1 gram O2 makes 2 g MgO Note: stoichiometric coefficients = atoms or moles… NOT grams! *Note: Stoichiometric coefficients = atoms or moles… NOT grams! 11/8/2018

Balancing Chemical Equations Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane (C2H6) reacts with oxygen to form carbon dioxide & water C2H6 + O2 CO2 + H2O Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C2H6 NOT C4H12 11/8/2018

Balancing Chemical Equations Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not O 1 carbon on right 2 carbon on left multiply CO2 by 2 C2H6 + O2 2CO2 + H2O 6 hydrogen on left 2 hydrogen on right multiply H2O by 3 C2H6 + O2 2CO2 + 3H2O 11/8/2018

Balancing Chemical Equations Balance those elements that appear in two or more reactants or products. multiply O2 by 7 2 C2H6 + O2 2CO2 + 3H2O 2 oxygen on left 4 oxygen (2x2) + 3 oxygen (3x1) = 7 oxygen on right C2H6 + O2 2CO2 + 3H2O 7 2 remove fraction multiply both sides by 2 2C2H6 + 7O2 4CO2 + 6H2O 3.7 11/8/2018

Balancing Chemical Equations 5. Check your balanced eqn to make sure you have the same number of each type of atom on both sides of the eqn arrow. 2C2H6 + 7O2 4CO2 + 6H2O final tally reactants products C 4 C 4 H 12 H 12 O 14 O 14 11/8/2018

Mass Changes in Chemical Reactions Write balanced chemical equation Convert quantities of known substances into moles Use coefficients in balanced equation to calculate the number of moles of the sought quantity Convert moles of sought quantity into desired units 11/8/2018

Balance a combustion reaction Provide a balanced equation for the combustion of methanol (CH3OH). CH4O + O2 CO2 + H2O 11/8/2018

Balance a combustion reaction Provide a balanced equation for the combustion of methanol (CH3OH). 2CH4O + 3O2 2CO2 + 4H2O 11/8/2018

Methanol burns in air according to the equation 2CH3OH + 3O2 2CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH3OH moles CH3OH moles H2O grams H2O molar mass CH3OH coefficients chemical equation molar mass H2O 1 mol CH3OH 32 g CH3OH x 4 mol H2O 2 mol CH3OH x 18 g H2O 1 mol H2O x = 209 g CH3OH 235 g H2O 11/8/2018

Limiting Reagents When a chemist carries out a reaction, the reactants are usually not present in stoichiometric amounts, that is, in the proportions indicated by the balanced equation. It is common to use the cheaper reactant in excess to ensure that all of the expensive reagent is consumed. The more expensive reagent is the limiting reagent and is often completely consumed in a reaction. 11/8/2018

Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3. Which reactant is the limiting reagent? 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Al 27 g Al x 1 mol Fe2O3 2 mol Al x 160 g Fe2O3 1 mol Fe2O3 x = 124 g Al 367 g Fe2O3 Start with 124 g Al need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent 11/8/2018