STOICHIOMETRYvia ChemLogbyDr. Stephen ThompsonMr. Joe StaleyMs. Mary PeacockThe contents of this module were developed under grant award # P116B-001338 from the Fund for the Improve-ment of Postsecondary Education (FIPSE), United States Department of Education.However, those contents do not necessarily represent the policy of FIPSE and the Department of Education, and you should not assume endorsement by the Federal government.3 Mg (s) + N2 (g)Mg3N2 (s)12345678901234567891010MolesChemLog STOICHIOMETRYChemLog STOICHIOMETRYCONTENTS2Chemical Reactions3Chemical Reactions4Chemical Reactions5Mass Per Cent6Limiting Reactants7Limiting Reactants8Yields9Yields10Molarity11Balancing Chemical Reactions12Mole To Mole Calculations13Avogadro’s NumberChemLog STOICHIOMETRYChemLog STOICHIOMETRYAA2B12345678901234567891010Moles►We can use this ratio to answer the following question: when starting with one mole of A, how many moles of B will you obtain? Let’s set this up mathematically.X=►From the ChemLog, how many moles of B are formed for each mole of A that reacts? To answer this question, fi ll in the appropriate numbers in this sentence: _____ mole(s) of A react(s) to form _____ mole(s) of B.►From the ChemLog, how many moles of A are needed to decompose for each mole of B pro-duced? To answer this question, fi ll in the appro-priate number in this sentence: For every _____ mole(s) of B that is formed, _____ mole(s) of A decomposed.In the following reaction, one mole of reactant A goes to 2 moles of product B.This can also be shown using “blocks” from the ChemLog.A2B►Now try this one, when starting with four moles of A, how many moles of B will you obtain? XAnother way to say this is that the ratio of moles of A to B is 1 to 2.mol Amol BThe ratio of moles of B to A is 2 to 1.mol Bmol Amol Bmol Amol Bmol AYou might be wondering why we chose this ratio (with moles of B on top) rather than the other ratio. The trick is to remember to put the units you want (in our case, we want to get to moles of B) on top. In this example, moles of A cancels, and we’re left with moles of B. ►When starting with one half a mole of A, how many moles of B will you obtain? ►Now try this one, when starting with 3 moles of A, how many moles of B will you obtain? A, how many moles of B will you obtain? mol A=mol Bmol Amol B►We can answer this question simply by us-ing numbers also. Notice that we get the same answer.___ mol AX=___ mol B___ mol A1___ mol A1___ mol A___ mol B2___ mol B2___ mol A1___ mol A2___ mol B2___ mol B►Fill in the appropriate numbers for this reaction.___ mol AX=___ mol B___ mol A___ mol BCHEMICAL REACTIONS2►Here’s a slightly different, but similar question: when you get 2 moles of B from this reaction, how many moles of A did you start with? We’ll go about answering it in the same manner.X=►When you get 6 moles of B from this reaction, how many moles of A did you start with?X=ChemLog STOICHIOMETRYChemLog STOICHIOMETRY►How many moles of B are produced in this reaction when you start with 3 moles of A?►When you get 3 moles of B from this reaction, how many moles of A did you start with?AA2BLet’s continue working with the following reaction.mol Bmol Amol Bmol A►Fill in the appropriate numbers for this reaction.___ mol BX=___ mol A___ mol B___ mol A►Fill in the appropriate numbers for this reaction.___ mol BX=___ mol A___ mol B___ mol ANotice that we’re using a different ratio here. Our answer needs to be in units of “mol A”, so we use the appriopriate ratio with moles of A on top.mol Bmol Amol Bmol ACHEMICAL REACTIONS3CHEMICAL REACTIONSAA3BAA3BChemLog STOICHIOMETRYChemLog STOICHIOMETRY12345678901234567891010Moles►How many moles of A will react to give 6 moles of B? Five moles of B?Here are some more practice questions.2AB2AB5A 2C 5A 2C+ ++ +4DBB4D12345678901234567891010Moles12345678901234567891010Moles►How much of the reactant is left after the reac-tion? When 7 moles of A react, how many moles of product will be obtained? Three moles of B?►How much of the reactant(s) is left after the reaction? When 2 moles of C are formed, how many moles of D are formed?How many moles of C will form from 10 moles of A How many moles of C will form from 10 moles of A and 2 moles of B?Challenge: How many moles of C will form from 10 moles of A and 1 mole of B?mol Amol Bmol Amol Cmol Bmol Amol Cmol Amol Cmol Bmol Bmol Cmol Dmol Amol Amol Dmol Bmol Dmol Dmol Cmol Dmol Bmol Cmol DTo help with answering the following questions, all the possible ratios are given. 4MASS PERCENTDetermining the mass percent composition of a compound refers to the proportion of one element expressed as a percentage of the total mass of the compound.Knowing the mass percent composition of a com-pound can help determine environmental effects from that compound. For example, carbon dioxide (CO2) from burning fossil fuels may contribute to global warming. Methane (CH4) and butane (C4H10) are both fossil fuels that, when burned, produce CO2. Which one will produce less CO2? Well, it’s the one that contains the least amount of carbon as a percentage of the total compound. Let’s use mass percent calculations to determine this.Mass Percent of Carbon (C) Mass of Carbon (C) Total Mass of Compound X 100 % Butane 4 x 12.011 g C (4 x 12.011 g C) + (10 x 1.008 g H) X 100 % C4H10= 48.044 g C 48.044 g C + 10.080 g H X 100 % = 48.044 g C 58.124 g C4H10X 100 % = 48.044 g C 58.124 g C4H10X 100 % 82.66 %= Methane CH4►Use the same technique to calculate the mass percent of carbon in methane.0 10 30 40 20 10 30 40 20 grams Butane ►Which compound, methane or butane, contains a higher mass percent of carbon?Which one will produce more CO2 when burned?0 10 30 40 20 10 grams Methane 30 40 20 First, let us determine the mass percent composi-tion of carbon in butatne.total mass of butane mass of hydrogen in butane mass of hydrogen in butane mass of carbon in butane total mass of methane mass of hydrogen in methanemass of hydrogen in methanemass of carbon in methane Now, let’s consider methane.ChemLog STOICHIOMETRYChemLog STOICHIOMETRY=Mass Percent of Carbon (C) =5LIMITING REACTANTSWhen carrying out a chemical reaction, we may use the exact amount of each reactant needed. Or, we may use an excess of some reactants and a limited amount of others. We may do this if one reactant is very expensive and others are inex-pensive so that we can use