Slide 1:Chapter 04 STOICHIOMETRY


Slide 3:The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = NA = 6.0221367 x 1023 Avogadro’s number (NA)


Slide 4:Each would have 602,213,670,000,000,000,000,000 atoms


Slide 5:Let’s convert from mass of an element to number of atoms of an element How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 0.551 g K 8.49 x 1021 atoms K


Slide 6:Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2


Slide 7:How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol H = 6.022 x 1023 atoms H 5.82 x 1024 atoms H 1 mol C3H8O molecules = 8 mol H atoms 72.5 g C3H8O Let’s convert from mass of a compound to number of atoms of a compound


Slide 8:Percent Composition Percent composition of an element in a compound = n is the number of moles of the element in 1 mole of the compound 52.14% + 13.13% + 34.73% = 100.0%


Slide 9:How to use Percent Composition to get Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent.


Slide 10:How to use Percent Composition to get Empirical Formulas nK = 0.6330, nMn = 0.6329, nO = 2.532 KMnO4


Slide 11:g of O = g of sample – (g of C + g of H) Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O How to use Combustion Analysis to get Empirical Formulas


Slide 12: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


Slide 13:2 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 How to “Read” Chemical Equations


Slide 14: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 reacts with oxygen to form carbon dioxide and water 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.


Slide 15:Start by balancing those elements that appear in only one reactant and one product. start with C or H but not O multiply CO2 by 2 multiply H2O by 3 Balancing Chemical Equations


Slide 16:Balance those elements that appear in two or more reactants or products. = 7 oxygen on right remove fraction multiply both sides by 2 Balancing Chemical Equations


Slide 17:Check to make sure that you have the same number of each type of atom on both sides of the equation. Balancing Chemical Equations


Slide 18: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 Amounts of Reactants and Products


Slide 19:Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? molar mass CH3OH coefficients chemical equation molar mass H2O 209 g CH3OH 235 g H2O Getting the mass of a product from mass of a reactant


Slide 20:NO is the limiting reagent O2 is the excess reagent Concept of Limiting Reagents


Slide 21:In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. 124 g Al 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent Getting the mass of a product from mass of a reactant


Slide 22:Use limiting reagent (Al) to calculate amount of product that can be formed. 124 g Al 234 g Al2O3


Slide 23:Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. Reaction Yield