Chapter 4Chapter 4

Reaction StoichiometryReaction Stoichiometry

• Multiplying the chemical formulas in a Multiplying the chemical formulas in a balanced chemical equation reflect the fact balanced chemical equation reflect the fact that atoms are neither created nor that atoms are neither created nor destroyed in a chemical reaction.destroyed in a chemical reaction.

• 2H2H2(g) 2(g) +O+O2(g)2(g)------ 2H 2H2 2 OO

• These coefficients can be interpreted in These coefficients can be interpreted in terms of mole. In the above equation we can terms of mole. In the above equation we can say that 1 mole of oxygen reacts with two say that 1 mole of oxygen reacts with two moles of hydrogen to form 2 moles of water.moles of hydrogen to form 2 moles of water.

Mole to mole calculationsMole to mole calculations

• This procedure is based on two This procedure is based on two principles:principles:

• 1. the number of atoms of each element 1. the number of atoms of each element is conserved in a chemical reaction.is conserved in a chemical reaction.

• 2. the coefficients in a given chemical 2. the coefficients in a given chemical equation give the relative numbers of equation give the relative numbers of moles of each reactant and product. moles of each reactant and product. These numbers of moles are used to These numbers of moles are used to construct conversion factors.construct conversion factors.

Class PracticeClass Practice

• Find the number of moles of NFind the number of moles of N2 2 needed to needed to produce 5.0 mol NHproduce 5.0 mol NH3 3 by reaction with Hby reaction with H2.2.

• Step 1. Write the balanced chemical Step 1. Write the balanced chemical equation: Nequation: N2(g) 2(g) +3H+3H2(g)----2(g)---- 2NH2NH3(g)3(g)

• Step 2. It follows from the equation that 1 Step 2. It follows from the equation that 1 mol of Nmol of N2 2 approx. equals to 2mol of NHapprox. equals to 2mol of NH3 3 so so the mol ratio isthe mol ratio is

• Substance required/ substance given Substance required/ substance given =1mol N=1mol N22/2mol NH/2mol NH3.3.

• Step 3. We now apply this conversion Step 3. We now apply this conversion factor to the information given and factor to the information given and obtain the information required:obtain the information required:

• Moles of NMoles of N2 2 =(5.0 mol NH=(5.0 mol NH33))x1mol x1mol NN22/2mol NH/2mol NH3 3 =2.5 mol of nitrogen.=2.5 mol of nitrogen.

• How many moles of NHHow many moles of NH3 3 molecules molecules can be produced from 2.0 mol Hcan be produced from 2.0 mol H2 2 in in the same reaction as in the previous the same reaction as in the previous reaction if all the hydrogen reacts.reaction if all the hydrogen reacts.

• ANS 1.3mol ammoniaANS 1.3mol ammonia

Mass to mass calculationsMass to mass calculations

• By using molar masses we can also find By using molar masses we can also find the mass of one substance that a given the mass of one substance that a given mass of another substance can mass of another substance can produce. We first convert the given produce. We first convert the given mass to the corresponding number of mass to the corresponding number of moles by using the molar mass of that moles by using the molar mass of that substance. We then convert the substance. We then convert the number of moles of the required number of moles of the required substance to mass by using its molar substance to mass by using its molar mass.mass.

STEPSSTEPS

• 1. Convert the given mass of one substance 1. Convert the given mass of one substance (in grams) to number of moles by using its (in grams) to number of moles by using its molar mass.molar mass.

• Number of moles of substance A=mass of A Number of moles of substance A=mass of A in grams/molar mass of A (grams per mole)in grams/molar mass of A (grams per mole)

• 2. Write the balanced equation for the 2. Write the balanced equation for the reaction.reaction.

• 3. Use the mol ratio to convert from the 3. Use the mol ratio to convert from the given number of moles of one substance to given number of moles of one substance to the number of moles of the other substance.the number of moles of the other substance.

• Substance required/substance givenSubstance required/substance given

• 4. Convert from number of moles of 4. Convert from number of moles of the second substance to mass (in the second substance to mass (in grams) by using the molar mass of grams) by using the molar mass of the substance.the substance.

• Mass of B required (grams) =number Mass of B required (grams) =number of moles of B of moles of B x molar mass of B x molar mass of B (grams per mole)(grams per mole)

• mmB B = n= nB B x Mx MBB

• Calculate the mass of oxygen needed Calculate the mass of oxygen needed to react with 0.450 g of hydrogen gas to react with 0.450 g of hydrogen gas in the reaction 2Hin the reaction 2H2(g) 2(g) + O+ O2 (g)2 (g)---- 2H2H22OO(l).(l).

Class PracticeClass Practice

Reaction yieldReaction yield

• Example:Example:

• A major source of atmospheric carbon A major source of atmospheric carbon dioxide is the burning of gasoline in dioxide is the burning of gasoline in automobiles, so we need to know how automobiles, so we need to know how much COmuch CO2 2 is likely to be produced from is likely to be produced from a liter of gasoline.a liter of gasoline.

• 2C2C8 8 HH18(l) 18(l) + 25O+ 25O2(g)2(g)------16CO16CO2(g) 2(g) + 18H+ 18H22OO

• Theoretical yield is the maximum mass Theoretical yield is the maximum mass of product that can be obtained from a of product that can be obtained from a given mass of reactant. The theoretical given mass of reactant. The theoretical yield of carbon dioxide is based on the yield of carbon dioxide is based on the assumptions that this is the only assumptions that this is the only reaction taking place and that every reaction taking place and that every molecule of octane is converted into molecule of octane is converted into carbon dioxide and water.carbon dioxide and water.

Class PracticeClass Practice

• Calculate the theoretical yield of Calculate the theoretical yield of carbon dioxide when 702 g of carbon dioxide when 702 g of octane(Coctane(C88HH1818) is burned.) is burned.

• Steps:Steps:• 1. Convert mass to moles of C1. Convert mass to moles of C88HH18 18

molecules then moles of Cmolecules then moles of C88HH18 18 to to moles of carbon dioxide, then moles of moles of carbon dioxide, then moles of carbon dioxide to mass of carbon carbon dioxide to mass of carbon dioxide.dioxide.

Home workHome work

• Page 166Page 166

• # 4.2,4.3,4.8,4.10,# 4.2,4.3,4.8,4.10,

Percentage yieldPercentage yield

• The percentage yield is the percentage The percentage yield is the percentage of the theoretical yield actually of the theoretical yield actually achieved.achieved.

• When 24 g of potassium nitrate was When 24 g of potassium nitrate was heated with lead, 13.8 g of potassium heated with lead, 13.8 g of potassium nitrite was formed in the reactionnitrite was formed in the reaction

Pb(s) + KNOPb(s) + KNO3 3 (s)---(s)--- PbO(s) +KNO PbO(s) +KNO22(s).(s). Calculate the percentage yield of Calculate the percentage yield of

potassium nitrite. Ans 68.3%potassium nitrite. Ans 68.3%

Limiting reactantLimiting reactant

• Limiting ReactantLimiting Reactant - The reactant in a - The reactant in a chemical reaction that limits the amount of chemical reaction that limits the amount of product that can be formed.  The reaction will product that can be formed.  The reaction will stop when all of the limiting reactant is stop when all of the limiting reactant is consumed. consumed.

• Excess ReactantExcess Reactant - The reactant in a chemical - The reactant in a chemical reaction that remains when a reaction stops reaction that remains when a reaction stops when the limiting reactant is completely when the limiting reactant is completely consumed.  The excess reactant remains consumed.  The excess reactant remains because there is nothing with which it can because there is nothing with which it can react.react.

• 8car bodies + 48 tires --8car bodies + 48 tires -- 8cars with tires 8cars with tires + 16 tires in excess. + 16 tires in excess.

• No matter how many tires there are, if No matter how many tires there are, if there are only 8 car bodies, then only 8 there are only 8 car bodies, then only 8 cars can be made.  cars can be made. 

• Likewise with chemistry, if there is only a Likewise with chemistry, if there is only a certain amount of one reactant available certain amount of one reactant available for a reaction, the reaction must stop when for a reaction, the reaction must stop when that reactant is consumed whether or not that reactant is consumed whether or not the other reactant has been used up. the other reactant has been used up.

• Example Limiting Reactant Calculation:Example Limiting Reactant Calculation: • A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  Which A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen.  Which

is the limiting reactant and how much excess reactant remains after is the limiting reactant and how much excess reactant remains after the reaction has stopped? the reaction has stopped?

• First, we need to create a balanced equation for the reaction:First, we need to create a balanced equation for the reaction:•

4 NH 4 NH 33(g) + 5 O(g) + 5 O22(g) ---(g) ---4 NO (g) + 6 H 4 NO (g) + 6 H 2 2 O (g)O (g)• Next we can use stoichiometry to calculate how much product is Next we can use stoichiometry to calculate how much product is

produced by each reactant. . produced by each reactant. .

• 2.0 g of NH2.0 g of NH3 3 x 1mol of NHx 1mol of NH3 3 / 17.0 g of NH/ 17.0 g of NH3 3 x 4mol NO/4mol NHx 4mol NO/4mol NH3 3 x x 30.0g of NO/1 mol NO =3.53 g30.0g of NO/1 mol NO =3.53 g

• 4.0 g of O4.0 g of O22x 1 mol Ox 1 mol O2 2 /32.0 g of O/32.0 g of O2 2 x 4 mol NO/5 mol Ox 4 mol NO/5 mol O2 2 x 30.0 g x 30.0 g NO/1mol NO=3.00 g of NONO/1mol NO=3.00 g of NO

•

• The reactant that produces the lesser The reactant that produces the lesser amount of product in this case the amount of product in this case the oxygen.oxygen.

• Next, to find the amount of excess Next, to find the amount of excess reactant, we must calculate how reactant, we must calculate how much of the non-limiting reactant much of the non-limiting reactant (oxygen) actually did react with the (oxygen) actually did react with the limiting reactant (ammonia).limiting reactant (ammonia).

• 4.0g of O4.0g of O2 2 x 1 mol of x 1 mol of OO2 2 /32 g of O/32 g of O22 x 1mol x 1mol NHNH3 3 /5 mol O/5 mol O2 2 x 17 g NHx 17 g NH33/1mol NH/1mol NH33=1.70 g of =1.70 g of NHNH33

    1.70 g is the amount of ammonia that 1.70 g is the amount of ammonia that reactedreacted, not what is left over.  , not what is left over. 

• To find the amount of excess reactant To find the amount of excess reactant remaining, subtract the amount that reacted remaining, subtract the amount that reacted from the amount in the original sample. from the amount in the original sample.

• 2.00g of NH2.00g of NH3 3 – 1.70 g (reacted) =0.30 g – 1.70 g (reacted) =0.30 g remaining.remaining.

Class PracticeClass Practice

• In the synthesis of ammonia which is In the synthesis of ammonia which is the limiting reactant when 100 kg of the limiting reactant when 100 kg of hydrogen reacts with 800 kg of hydrogen reacts with 800 kg of nitrogen?nitrogen?

Empirical formulaEmpirical formula

• Empirical FormulaEmpirical Formula - A formula that gives the - A formula that gives the simplest whole-number ratio of atoms in a simplest whole-number ratio of atoms in a compound. compound.

• Steps for Determining an Empirical FormulaSteps for Determining an Empirical Formula • 1. Start with the number of grams of each 1. Start with the number of grams of each

element, given in the problem.  element, given in the problem.  • 2. Convert the mass of each element to moles 2. Convert the mass of each element to moles

using the molar mass from the using the molar mass from the periodic tableperiodic table.  .  • Divide each mole value by the smallest number Divide each mole value by the smallest number

of moles calculated.  of moles calculated.  • Round to the nearest whole number.  This is the Round to the nearest whole number.  This is the

mole ratio of the elements and is represented by mole ratio of the elements and is represented by subscripts in the empirical formula.  subscripts in the empirical formula. 

• If the number is too far to round (x.1 ~ x.9), then multiply If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same  each solution by the same  factor to get the lowest whole number multiple.  factor to get the lowest whole number multiple. 

• e.g.  If one solution is 1.5, then multiply each solution in the e.g.  If one solution is 1.5, then multiply each solution in the problem by 2 to get 3.  problem by 2 to get 3. 

• e.g.  If one solution is 1.25, then multiply each solution in e.g.  If one solution is 1.25, then multiply each solution in the problem by 4 to get 5.  the problem by 4 to get 5. 

• Once the empirical formula is found, the molecular formula Once the empirical formula is found, the molecular formula for a compound can be determined if the molar mass of the for a compound can be determined if the molar mass of the compound is known. compound is known. 

• Simply calculate the mass of the empirical formula and Simply calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the divide the molar mass of the compound by the mass of the empirical formula to find the ratio between the molecular empirical formula to find the ratio between the molecular formula and the empirical formula.  Multiply all the atoms formula and the empirical formula.  Multiply all the atoms (subscripts) by this ratio to find the molecular formula.  (subscripts) by this ratio to find the molecular formula. 

• A compound was analyzed and found to contain A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H.  What is the 13.5 g Ca, 10.8 g O, and 0.675 g H.  What is the empirical formula of the compound?empirical formula of the compound?

• Start with the number of grams of each element, Start with the number of grams of each element, given in the problem. given in the problem.

• Convert the mass of each element to moles using the Convert the mass of each element to moles using the molar mass from the molar mass from the periodic tableperiodic table. .

• Divide each mole value by the smallest number of Divide each mole value by the smallest number of moles calculated.  Round to the nearest whole moles calculated.  Round to the nearest whole number. number.

• In this we divide by the moles of Ca obtained.In this we divide by the moles of Ca obtained.

• This is the mole ratio of the elements This is the mole ratio of the elements and is represented by subscripts in and is represented by subscripts in the empirical formula. the empirical formula.

• CaOCaO22HH2 2 = Ca(OH)= Ca(OH)22

HomeworkHomework

• Page 168Page 168

• # 4.34, 4.36# 4.34, 4.36

SolutionsSolutions

• Molarity : Molarity is the number of Molarity : Molarity is the number of moles of solute dissolved in one liter moles of solute dissolved in one liter of solution. The units, therefore are of solution. The units, therefore are moles per litermoles per liter, specifically it's , specifically it's moles of solutemoles of solute per per liter of liter of solutionsolution..

• molarity = molarity = amount of solute (moles)amount of solute (moles) volume of solution (liter) volume of solution (liter)

Class PracticeClass Practice

• Calculate the molarity of glucose in a Calculate the molarity of glucose in a solution made by dissolving 1.368 g solution made by dissolving 1.368 g of glucose (Cof glucose (C6 6 HH12 12 OO66) in enough water ) in enough water to make 50.00 ml of solution.to make 50.00 ml of solution.

DilutionDilution

• The most important thing to remember concerning dilution The most important thing to remember concerning dilution is that you are only adding is that you are only adding solventsolvent. You are not adding . You are not adding solute when you dilute.solute when you dilute.

• Therefore: Therefore: • moles of solute before dilution = moles of solute after moles of solute before dilution = moles of solute after

dilutiondilution• or Number of millimoles of solute before = millimoles of or Number of millimoles of solute before = millimoles of

solute aftersolute after• or or • Number of grams of solute before dilution = number of Number of grams of solute before dilution = number of

grams of solute after dilutiongrams of solute after dilution• Since the definition for Molarity is: Since the definition for Molarity is: • Molarity = moles of solute / volume of solution in litersMolarity = moles of solute / volume of solution in liters

• Solving for moles of solute gives: Solving for moles of solute gives: • moles of solute = M x V of solution in liters moles of solute = M x V of solution in liters • or or • millimoles of solute = M x V of solution in ml millimoles of solute = M x V of solution in ml • If Moles of solute before dilution = moles of solute after dilution If Moles of solute before dilution = moles of solute after dilution • then M x V in liters before dilution = M x V in liters after dilution then M x V in liters before dilution = M x V in liters after dilution • or or • M1V1 = M2V2 M1V1 = M2V2 • where M1 = Molarity before dilution where M1 = Molarity before dilution • V1 is volume of solution before dilution V1 is volume of solution before dilution • M2 = Molarity of solution after dilution M2 = Molarity of solution after dilution • V2 = Volume of solution after dilution V2 = Volume of solution after dilution • We can use Molarity or mass % as the concentration term so you could We can use Molarity or mass % as the concentration term so you could

have the following alternative where mass % is used: have the following alternative where mass % is used: • mass % x grams solution before dilution = mass % x grams of solution mass % x grams solution before dilution = mass % x grams of solution

after dilution after dilution

HomeworkHomework

• Page 169Page 169

• # 4.40, # 4.40,