9 Holt Stoichiometry: Calculating Quantities in Reactions

• Stoichiometry is a branch of chemistry dealing with the quantities of substances in chemical reactions.

Start-Up Activity: All Used Up

• This activity uses nuts and bolts to demonstrate the concept of stoichiometry.
• Nuts (N) and bolts (B) are used to create nut-bolt (NB) and nut-nut-bolt (N2B) models.
• The goal is to find the mass of 8 nuts and 5 bolts using a balance.
• Attach 1 nut to 1 bolt to make as many NB models as possible and record the amount of models were formed.
• Attach 2 nuts to 1 bolt to assemble the most N2B models as possible and record this amount.
• By calculating the masses of nuts and bolts, explain if it is possible to predict which material would be used up.
• Balanced equations can help predict which component runs out for the "reaction" that forms NB and N2B.
• With 18 bolts and 26 nuts, calculating possible NB and N2B models is possible

Section 1: Calculating Quantities in Reactions

• Stoichiometry is the Key Term for this section
• Proportional reasoning is used to determine mole ratios from balanced chemical equations.
• Mole ratios are central to solving stoichiometry problems.
• Stoichiometry problems involving mass need molar mass to be solved.
• Stoichiometry problems involving volume use density.
• Stoichiometry problems involving particle number use Avogadro's number.
• Balanced chemical equations show the proportions of reactants and products involved in a reaction.
• Coefficients in balanced equations represent the number of particles or moles of each substance.
• Calculations use proportions from balanced chemical equations to find the quantity of each reactant and product.
• Assumed reactions go to completion, meaning the initial reactant changes into product.
• Sufficient other reactants are assumed, allowing complete reaction of the given reactant.
• Reactions are assumed to happen perfectly, and there is no product loss during collection.

Relative Amounts in Equations

• Equations can be interpreted in terms of particles or moles.
• Coefficients in balanced equations represent the moles of each substance.
• 2C8H18 + 25O2 → 16CO2 + 18H2O shows 2 molecules of C8H18 react with 25 molecules of O2.
• The equation shows 2 mol C8H18 react with 25 mol O2 to form 16 mol CO2 and 18 mol H2O.
• Stoichiometry determines how much reactant is needed or how much product is formed.

The Mole Ratio Is the Key

• The mole is unit that bridges the gap between substances
• Coefficients in balanced equations show relative numbers of moles.
• Mole ratios are conversion factors that convert between moles of substances.

Skills Toolkit 1: Converting Between Amounts in Moles

• First, the amount that is known in moles from the problem needs to be identified.
• Second, using the coefficients from the balanced equation the mole ratio must be set up with the known substance on the bottom and the unknown on top.
• Finally, the original amount is multiplied by the equation.

Sample Problem A: Using Mole Ratios

• Consider N2 + 3H2 → 2NH3, for the commercial preparation of ammonia: the key question is how many moles of hydrogen are needed to prepare 312 moles of ammonia.
  • amount of NH3 is 312 mol
  • the amount of H2 is unknown
  • 3 mol H2 = 2 mol NH3
• The mole ratio must cancel out the units of mol NH3 given in the problem and leave the units of mol H2. The mole ratio is 3 mol H2/2 mol NH3
• Problem is solved by: ? mol H2 = 312 mol NH3 × 3 mol H2/2 mol NH3 = 468 mol H2
• Answer is larger than the initial number of moles of ammonia so there needs to be further investigation.
• The number of significant figures must be correct because the coefficients 3 and 2 are considered to be exact numbers.

Getting into Moles and Getting out of Moles

• Substances are measured by mass or volume, so units for mass and volume need to be converted using the unit "mol" before using the mole ratio.
• Each stoichiometry problem has the step in which moles of one substance are converted into moles of a second substance using the mole ratio from the balanced chemical equation.
• Three basic steps include:
  • changing the units that are given into moles.
  • use the mole ratio to determine moles of the desired substance.
  • changing out of moles to whatever unit finalize the answer in.

Skills Toolkit 2: Solving Stoichiometry Problems

• All types of stoichiometry problems can be solved by using these steps:

1. Gather information
   • If given, balance the equation. If no equation is given, write a balanced equation for the reaction described. write the information provided for the given substance. If an amount in moles is missing, determine what is needed to change any given units into moles, and write it down. Report the units being asked for in the unknown substance, and what info is needed to change moles into the desired units and write it down. create an equality using substances and coefficients that shows the relative amount from the balanced equation. Plan the work Think through the three basic steps used to solving stoichiometry problems: change to moles, the mole ratio and change out of moles, know which conversion factors to use in each part. Make sure you write the mole ratio in the form: (moles of unknown substance)/(moles of given substance)
2. Calculate
   • Write a question mark with units of the answer, followed by an equal sign plus quantity of the given substance. Write the conversion factors (including the mole ratio in the write order so that the units match what is needed for the answer. Make sure the units are the correct, required units of the unknown substance. When through, perform the calculations and round off that answer to the correct number of figures and report the final answer.
3. Verify your Calculations
   • Use estimation by finding the approximated value in step 3 or by comparing the conversion factors in the setup to decide if the result makes sense for the actual problem. Verify that the answer is reasonable, for instance make sure that the quantity calculated makes sense. Always double check the work

Problems Involving Mass, Volume and Particles

• Includes how to use the molar mass of each substance involved in a stoichiometry problem
• The molar mass is the sum of atomic masses from the periodic table for the atoms in a substance.
• It is a three step process:
  • the mass in grams of the given substance is converted into moles.
  • Next, the mole ratio is then converted into moles of the desired substance.
  • Finally, convert the amount in moles into grams.

Skills Toolkit 3: Solving Mass-Mass Problems

• Amount of Known (g) is multiplied by (1 mol/g)
• Amount of Unknown is multiplied by (2 mol unknown/2 mol known)
• Mass of Unknown is multiplied by (2 g/1 mol)

Sample Problem B: Problems Involving Mass

• Use the following equation: N2+3H2 → 2NH3

1. Gather information:
   • mass of H2 = 1221 g H2
   • molar mass of H2 = 2.02 g/mol
   • mass of NH3 = ? g NH3
   • molar mass of NH3 = 17.04 g/mol
   • balanced equation: 3 mol H2 = 2 mol NH3.
2. Plan the work (change of grams):
   • H2 to moles needs H2's molar mass.
   • The mole ratio must cancel out the units of mol H2 given in the problem and leave the units of mol NH3, which makes it 2 mol NH3/3 mol H2
   • To change the moles of NH3 to grams, use the molar mass of NH3.
3. Calculate:
   • ? g NH3 = 1221 g H2 x 1 mol H2/2.02 g H2 X 2 mol NH3/3 mol H2 X 17.04 g NH3/1 mol NH3 = 6867 g NH3
4. Verify the result.
   • The units cancel to provide the correct units for the answer and Estimating shows that the answer should be about 6 times the original mass.

Skill Toolkit 4: Solving Volume-Volume Problems

• For volume, Density and Molar Mass might have to be used
• Involves adding two more steps to the sequence of mass-mass problems—the conversions of volume to mass and of mass to volume.
• Consists of Five conversion factors: two densities, two molar masses, and a mole ratio
• To convert from volume to mass or from mass to volume of a substance, the density of the substance should be used as the conversion factor. The units you want to cancel should be on the bottom of your conversion factor.

Sample Problem C: Problems Involving Volume

• Calculate the amount that ? mL H3PO4 forms when 56 mL POCl3 completely react:
• Density of POCl3 = 1.67 g/mL, Density of H3PO4 = 1.83 g/mL
• POCl3(l) + 3H2O(l) → H3PO4(l) + 3HCl(g)
• First Gather the Information*
  • Volume of POCl3 = 56 mL POCl3 , density of POCl3 = 1.67 g/mL, molar mass POCl3 = 153.32 g/mol, volume H3PO4 = ? , density H3PO4 = 1.83 g/mL , molar mass H3PO4 = 98.00 g/mol. With equality 1 mol POCl3 = 1 mol H3PO4.
• From the above use the Plan to map out the work*
  • The volume needed will be used milliliters (mL) and start with the density of POCl3 for moles which will then followed by its own molar mass to balance the equations. The mole ratio has to cancel out the units completely like normal. Then to move out of the mL use the molar mass of its density.
  • ? mL H3PO4 = 56 mL POCl3 × 1.67 g POCl3 / 1 mL POCl3 x 1 mol POCl3/ 153.32 g POCl3 X 1 mol H3PO4/1 mol POCl3 x 98.00 g H3PO4/ 1 mol H3PO4 x 1 mL H3PO4/ 1.83 g H3PO4 = 33 mL H3PO4.
  • From this the units of the answer are correct and Estimating shows results should be 2/3's of the original volume to match

Skill Toolkit 5: Solving Particle Problems

- Avogadro's number is used when solving for the # of particles: 6.022 × 10^23 particles/mol, in stoichiometry problems and if particles and the goal to find # of particles then no need to use/ Avogadro's number cancels out thus need only the coefficients from the balanced equation again where you are interpreting it again in terms of number of particles.

Sample Problem D: Problems Involving Particles

• Determine how many grams of C5H8 form from 1.89 × 10^24 molecules C5H12 with the base equation being: C5H12(l) → C5H8(l) + 2H2(g)

1. Starting gathering the info needed: for solving

• quantity of C5H12 = 1.89 × 1024 molecules
• Avogadro's number which holds its value; = 6.022 × 1023 molecules/mol
• mass of C5H8 = ? g C5H8 to look for
• Molar mass of C5H8 = 68.13 g/mol known
• base/ basic equation to = 1 mol C5H12 = 1 mol C5H8.

2. The Plan Set up problem using Avogadro's number to change to moles, then use the mole ratio, and finally use the molar mass of to change to grams. like so:

• ? g C5H8 = 1.89 × 10^24 molecules C5H12 × 1 mol C5H12 / 6.022 × 10^23 molecules C5H12 x 1 mol C5H8/ 1 mol C5H12 X 68.13 g C5H8/ 1 mol C5H8=214 g C5H8

3. Verify Result

• results should cancel correctly and by estimating the value it makes sense that 210 is the correct estimate value

Practice Problems

• The equation provided, Br2(l) + 5F2(g) → 2BrF5(l), can be used to answer the following questions:
  • Find out how many BrF5 molecules form when 384 g Br2 react with excess F2.
  • Find out How many molecules of Br2 react with 1.11 × 10^20 molecules F2?

Many Problems, Just One Solution

Although you could be given many different problems, the solution boils down to just three steps:

• Take the "giver" and find a way to change it into moles.
• Then, use a mole ratio from the balanced equation to get moles of the second substance.
• Finally, find a way to convert the moles into the units that you need for your final answer.

Section 1 Review

• There are a few ways to fully determine what key the answers in these sections with different perspectives.
• For the chemical equation Br2 + Cl2 → 2BrCl, determine:
  • How many moles of BrCl form when 2.74 mol Cl2 react with excess Br2
  • How many grams of BrCl form when 239.7 g Cl2 react with excess Br2
  • How many grams of Br2 are needed to react with 4.53 × 10^25 molecules Cl2
• Consider the equation for burning C2H2: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g) then:
  • At STP where 1 mol = 22.41 L for any gas, how many moles of CO2 are produced upon burning of 15.9 L C2H2?
  • upon burning of 59.3 mL O2 where CO2 weighs 1.977 g/L and O2 weighs 1.429 g/L, what is the equivalent amount in milliliters of CO2 formed?

Critical Thinking

• Moles is what what needed in amount to perform and solve stoichiometry.
• Mass and Mass can not be just a solution or simple calculation
• Some of key reactions can help form air in craft that's sustainable

Section 2: Limiting Reactants and Percentage Yield

• The Key Terms for this section are:
• limiting reactant
• excess reactant
• actual yield
• By using the the limiting reactant for a reaction you can identify and calculate the theoretical yield
• The ability to perform calculations involving percentage yield is beneficial

Limiting Reactants and Theoretical Yield

• Just like you need both gasoline and oxygen to drive, chemical reactions need all reactants to make it work
• Limiting Reactant = Limits the distance/amount you can travel/use because it stops the engine/reaction.
• Theoretical aspect = all reactants change into products BUT in reality, other factors limit yield for this particular result ex; amounts of all reactants, completeness of the reaction and possible product loss.
  • Analogy is used for at home coming/ fun raisers to help understand supply quantity vs amount of product made
• Short amount = limited amount of product made( most directly impacts overall outcome

The Limiting Reactant Forms the Least Product

• When assembling items they may have a list materials for one individual item ex one helmet, one flier, 8 blue ribbons, 6 white ribbons, 2 bells. As a result no assembly can go past one item is all out
• Just the reactants of a reaction are never in equal ratios like in balanced/ basic equation so most reactants are used up before any other
• Example: to to make H2 is Zn + 2HCl → ZnCl2 + H2
• From this if combine 0.23 mol Zn and 0.60 mol HCl, would they react completely?
• 0.23 mol Zn = 0.23 mol H2 possible to make while 0.60 mol HCl = 0.30 mol H2 to see - Zinc is the limiting agent here because of the limit in Zinc to = limit H2 product -HCI is the excess agent because of its abundance and how left overs are possible
• Looking at the mum picture, flowers are limiting agent due to low short supply
• to solve figure out which reactant is needed, calculate the amount of product for them to then see the limiting reactant to match result.
• whatever reactant produces lower estimate then that is limiting

Determine Theoretical Yield from the Limiting Reactant

• In the past reactions were assumed perfect but in reality; max quantity or product in reaction that happens in perfect conditions is theoretical yield. The amount of a product is limited. And depends on the theoretical yield. The "yield" in the reaction should come with limiting reactant. Example: the (theoretical yield) of Zn w/ HCI equals = 0.23 mol H2 if theoretical yield is 0.30 mol H2 w/ HCI. This is the base example"

Sample problem E: Limiting Reactants and Theoretical Yield

• key focus and goal is to identify: the Limiting Reactant/ reagent(LR) + what is/are the theoretical yields as a result if 225 g of PCI3 is added/ mixed into 125 g H2O to react
• First Step " Gathering basic information to start solving".
• Equation: PCI3+3H2O<-->H3PO3 + 3HCI
• Mass: 225g of PCI3 being added in , H2O=125 g
• Molecular Masses for all of the starting pieces
• PCI3<== 137.32g/mol *H2O<== 18.02 g/mol *(looking for answer) H3PO3 mass = ??

2. Plan out your work strategy to help solve

• Need to determine and know the mass of the H3PO3 you were would expect
• Solving* g H3PO3 = 225 g * PCI3 * (1mol PCI3 / 137.32g ) * (1mol H3PO3 / 1mol PCI3) * (82 g H3PO3/ 1mol H3PO3) = 134 g H3PO3 to see
• Next H2O*g H3PO3 = 123 g H2O * (1mol H2O / 18.02g) (1mol H3PO3 /3mol H2O ) * (82 g / 1mol) = 187 g H3PO3 to see Now as a result PCI3 is limiting and theory yield is 134 g H3PO3
• Verifying = units should be correct and estimates match 128

The Food You Eat

• Cheapest reactant used in industries due to it helps to uses the expensive reactants

• Also in a controlled setting it is beneficially ex; making apple cider vinegar.

• Juices are kept with (No oxygen) in a controlled environment to cause micro-organisms to break sugars down such as glucose to release CO2 + ethanol and the result is a "Hard Cider" solution

• in next step the High Oxygen allows organisms from change thanol into ethanoic acid to become cider. With this because air is free to take Ethanol becomes a " Limiting reactant and runs out

• cost also choses too, like when flavoring banana flavoring" (Isopenttyl acetate in banana flavor).

• Acetic acid becomes/gets used in excess due its cheap amount >" amount needed for its Alcohol counterpart + the equation between them is Equation ==(CH3OOH + C5H11OH---> CH3COOC5H11 + H2O)==

With this and also knowing its less expensive " alcohol is only twice more expensive as acid" when there's an abundance of acetic acid present, and a lot reacts.

choosing this by value makes sense in some scenarios

Actual Yield and Percentage Yield

• Equation/ balanced equations show and tell you exactly what should happen- it does NOT account for what would actually happened in terms of real world conditions
• Because reactions aren't fully a complete reflection of the expected results from calculations or the base equation.
• Thus this is why it can be be: the( actual yield= is mass of made result that happens) "In some times has a mass that's less than theoretical"
• Imagine a factory were 500.0 gold.
• (isopenttyl alcohol is mixed with 1.25 × 103 g acetic acid. in the same factory*)
• In that scenario the theory and actual results do in fact not equal or is less.

there are multiple reasons " reasons why "actual" yield is less/ or not expected:

• Reactions not fully going to completion-
• Products get turned back to the reactants- a constant change in state until equilibrium happens
• Also needs 2nd steps needed to purify the resulting product because in the real world their mixed with extras materials. Ex; needing distilling process to separate " banana flavors or solid sugar. can then use the re-crystallize with sugars in solid.
• Product is always potentially in the process*

also by chance because they had to be side other reactions not accounted for in formula- and not the desired result for side product." Side reactions"

Determining % yield

-"Relating the Actual Yield to the % and its theoretical is the %Y "Calculation method of similar of how a person who bats averages their stats"

• A batter could potentially in theory make 100 percent shots when in real situations they don't
• SO to look look at example A player/ (Batter= 41 hits in 126 times)
• Then a batter makes 41/(divides) 126= 0.325 average ( as real stats and reflection)
• Back to yield: * (theor result of reaction/ is 738 to what's real)=591 g what is percent yield?
• Equation* %Y= (591g < what you earn in real result>) / (738g theor math result= expected) * Multiplies by 100 percent = (80. 1 percent yield)<- expected to make

Sample problem F: Calculating % yields steps/ equations + information/ theore Yield

1. Find percentage yield + determine the limiting reactant. If 16.1 g NH3 result if 14.0 gold N2 are mixed + combined/reacted to with 9.0 g H2

Gather

• what basic information do i i need?*
• Mass* =
• KnowN2 = 14.0 g
• KnowH2 = 9.0 g un -Known: what's theor yield of NH3 + What is %

from above we also know ( the given or basic equality from base:)"" equation "" _1 mol nitrogen equals / reacts = 2 "" NH3 and also "" 3 mol hydrogen = equals / reacts = 2 ""(mol:) of NH3

==Plan out work+ set up the problem==We use + set up (what we know from calculations). "to also look for results from NH3 and what its expected to also form"

• what info does it give me that i dont know?*= what is percent equation to use=? g NH3= what we know(14.0 gold N2 as baseline) x *(1 mol N2 / 28.02 value of gold from its form is )*x multiply also the side by 1 to **(2mol NH3/ all divided by 1mol also of baseline)*x(**17.0.04 all gold of NH3 all over 1 mol)value as baseline-=***17 g NH3< =to result from equationsThen we use baseline by H2 mass what equals

to ==?g NH3 == (9 all baseline gold H2 value) / (2.02 what it cost also from equation baseline))* / ( the mol x ratio that is given ( 2 mol baseline/ all of baseline the ( 3mol value H2* / **also the baseline baseline mass 1 is ( also the formula given = 17.04 value we get or what the mass is is of) (divided by the amount of 1 from formula) Baseline== 5 NH3 to reflect
==Result== 51 g of ""

• (Note from calculation = the less value = (17.0 Gold NH3) = is its theory due value with agent name "" NITROGEN"**
• Equation*
• % to determine from yield ** =< (what we saw happen is + the given or "Actual = result " 16.1)/ what we math found theor as limited (17 gold NH3) x 100< what it all equals->>"" 94.7*

14. Verify if your result is accurate** and test that values

• make sure if % is< 100. then correct

Designing to calculate how much or estimate better if/ how long

• estimating in most industries better by use % with in reactions if you want " more estimate close than its needs to found in EXP/lab
• equation or design of how it will work < % =< how it looks vs theory x100=>>

Sample problem G: Actual yield of what % of what is possible

• With the example of : How many grams of CH3COOC5H11 (should with form. when mass = 48 all theoretical) + yields

==Gather** to find- theory all up of the total: for to create results) " also+ what we see happen"" CH3COOC5H11 = <<4.808 value for it + what = what do we expect after or after percent + yield + for the" actual """"" all CH""""

• Plan*
• Equation*: 80 value= "" actual"" x110 mass:""<>

what actually what what to get or happen" =mass the

Calculating what to Do: " what the actual yields are vs what it means if there pollution"

• note that: yield needs < Less for it the be safe than theoretical and is potentially reverse as estimate otherwise
• equation or design of how it will work < % =< how it looks vs theory x100=>>

Air Bag Stoichiometry

• Assume when 65.1L N2 inflates in air the bag its at proper measurements + Density is ( gold mass/volume) at: 0. mass0.60 g/l ==what MASS needs to used from NaN3 + and to implement?==

1st== " gathering values:

• what has high significance to start with Equation 1
• = 2(NaN3) =<<-->> 2(Na) + 3() is a

2. known density

• what is known+ significance (mass/ value=what all matters to create" formula +
• volume+of N all over ""2 is "65.1 = mass / gold -* value that exists ( N)is 28.02

3. to create better map work use strategy in solving

• equation

##air " for "bags

? <<

• formula value of equation/ given* *

(1 over the amount in to = the # as 9.03 mass: over it the over *mass by gold is "" = ""gold""

to see mass = for better numbers use correct for the number

Air Quality Design and Stoic Precision

• The better the more results of

Air to fuel ration

• for more perfect ration of air it helps for " combustion" what = .79/mill

Stoichiometry deals with measuring substances in chemical reactions. Using nuts and bolts demonstrates the concept of stoichiometry. Balanced equations predict which component is used up in reactions forming NB and N2B models.