Chemistry 1201 Review Preview Chapter Three Review Guide Dr Saundra McGuire Spring 2007 Stoichiometry Calculations with Chemical Formulas and Equations I Balancing Chemical Equations A Memorize the diatomic elements hydrogen oxygen fluorine chlorine bromine iodine B Write the correct chemical formula for all compounds using the charge criss cross method discussed in Chem 1201 Review Preview session C Make the number of each kind of element the same on both sides of the equation by putting in the correct coefficients Never change the subscripts to balance an equation See section 3 1 in the text for more information and examples II Simple Patterns of Chemical Reactivity Three simple types of reactions are frequently encountered in this chapter combination or synthesis reactions decomposition reactions and combustion reactions An example of each type is shown below 1 Combination or synthesis reactions are those in which two or more substances combine to form a larger substance The general representation of a combination reaction is A B C 2 Decomposition reactions are those in which a compound breaks down into two or more simpler substances The general representation of a decomposition reaction is AB A B 3 Combustion reactions are those that produce a flame The combustion reactions most commonly seen in this chapter involve a reactant containing C and H and sometimes O reacting with oxygen gas O2 to form water and carbon dioxide See pages 87 88 of the text for examples III Formula Weights Formula weights are calculated by adding the atomic weights of all of the atoms in a compound e g The formula weight of Ca NO2 3 is 164 1 amu atomic mass units Note that the smallest particles of covalent compounds are called molecules whereas the smallest units of ionic compounds are called formula units Therefore formula weights are often called molecular weights when molecular compounds are involved To get the percentage composition from formulas we calculate the percent of each element present as follows Element X atomic weight of the element X number of atoms of that element X 100 Formula weight of the compound IV Avogadro s Number and the Mole A mole is the amount of substance whose weight in grams is equal to the atomic or formula weight of the substance One mole contains 6 02 X 1023 Avogadro s number particles Interconversions between grams moles and particles can be accomplished as shown below Note that you cannot go directly between grams and particles without first going through moles Atoms Grams Moles Molecules Use the formula wt Use Avogadro s Number Ions See Section 3 4 of the text for a more thorough discussion and examples The molar mass of a substance is the number of grams in one mole 6 02 X 1023 particles of the substance The units of molar mass are grams per mole g mol Note that the molar mass in grams per mole equals the formula weight in atomic mass units amu V Empirical Formula and Empirical Formula Determination A Empirical Formula The simplest formula which shows the ratio of moles of each element in the compound Empirical formulas can be calculated from percent composition or from combustion data To calculate the empirical formula from the composition 1 Assume 100 grams of compound s then become grams 2 Calculate the moles of each element moles grams atomic mass 3 Divide by the smallest number to obtain the subscripts in the empirical formula See Section 3 5 of the text for examples and practice problems To calculate the empirical formula from combustion data 1 Calculate the mass of C in the compound by converting the grams of CO2 to moles of CO2 When we get the moles of CO2 we have the moles of C because the moles of C equals the moles of CO2 We get the grams of C by multiplying the moles of C by 12 0 the atomic mass of C 2 Calculate the mass of H by converting the grams of H2O to moles of H2O When we get the moles of H2O we get the moles of H by multiplying the moles of H2O by two because each mole of H2O contains two moles of H We get the grams of H by multiplying the moles of H by 1 01 the atomic mass of H 3 Calculate the mass of O by finding the sum of the grams of C and H and subtracting this number from the total grams of compound The total grams will be given in the problem Do Sample exercise 3 15 in the text without looking at how they did it VI Quantitative Information from Balanced Equations The coefficients in a balanced equation indicate the numbers of moles or molecules of each reactant and product involved in the reaction For example A 2B 4C means that 1 mole cule of A reactions with 2 mole cule s of B to produce 4 mole cule s of C Note that the number of moles of reactants does not have to equal the number of moles of products The total number of grams must be equal on each side but not the number of moles The numerical coefficients in the balanced equation are used to determine the relationship between moles and grams of substances involved in the reaction Use the diagram in Figure 3 13 of the text to solve stoichiometry problems by the mole method The sequence of steps is as follows moles of given moles of unknown grams of unknown A B C For step A we use molar mass of the given substance for step B we use the ratio of the coefficients in the balanced equation and for step C we use the molar mass of the unknown substance Note that you should NEVER use the molar mass of one substance with the grams of another substance Grams of given VII Limiting Reactants The limiting reactant in a problem is the reactant that runs out before the other is used up For example if we were making tricycles and had 8 frames and 18 wheels we could make only 6 tricycles before we would run out of wheels Therefore the wheels would be the limiting reactant and the frames would be the excess reactant We would have 2 frames left over after we use all of the wheels To do a limiting reactant problem in chemistry Calculate the moles of one reagent A necessary to react with the given number of moles of the other reagent B If the calculated amount of A is larger than the amount of A given in the problem Reagent A is the limiting reagent it will run out first If the calculated amount of A is smaller than the given amount Reagent A is present in excess and Reagent B is the limiting reagent The maximum amount of product possible the theoretical yield is stoichiometrically calculated amount of product possible from the limiting reagent See section 3 7 of the text for specific examples The