STOICHIOMETRYGeneral ChemistrySpring 2010

INTRODUCTION

Billions of pounds of chemicals are produced each year across the world.

These chemicals help manufacture: Medicines Computer chips and electronic instruments Fertilizers and pesticides Glass Paper, plastics, synthetic fibers…….

INTRODUCTION

In order to determine the cost of producing such essential items we use each and every day, chemists and chemical engineers perform calculations based on balanced chemical equations

Which chemical do you think is used the most industrially? Sulfuric acid 165 million mega tons (1000 kg)Value of $8,000,000,000 annually…

yes 8 billion dollars…

Uses of H2SO4

Production of fertilizers,. Used in the manufacture of chemicals such as

hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.

It is used in petroleum refining to wash impurities out of gasoline and other refinery products.

Sulfuric acid is used in processing metals. In cleaning iron and steel before plating them with tin or zinc.

Rayon is made with sulfuric acid. It serves as the electrolyte in the lead-acid storage

battery commonly used in motor vehicles (acid for this use, containing about 33% H2SO4).

Stoichiometry? Big Deal…. What are some items you use everyday?

Clothes? Soap and shampoo? CD’s and iPods? Medicine?

Company’s that make these and every other substance do not want the cost to make these items to be higher than the cost they’re sold at. Profit ring a bell? Companies carry out chemical rxns

economically to keep prices down (we hope at least)

Why did I have to learn how to balance an equation?

Chemists use balanced equations as a basis to calculate how much reactant is needed or product is formed in a reaction. If you know the quantity of 1 piece, you can

calculate the quantity of anything else in the reaction

We use grams and molsSometimes L, tons, molecules are used

http://www.theyshoulddothat.com/images/newIpods.jpg

What IS stoichiometry? The calculation of quantities in chemical

reactions We will use a version of the mole road map for

this Equations must be balanced Allows the chemist (you) to keep track of the

amounts of R and P using ratios of moles or particles

http://www.hccbrandon.net/chem1211/mod5/stoichiometryTitle.gif

Still not convinced?

Still not convinced? Air bags must inflate in less than a second. The effectiveness of air bags is based on the

rapid conversion of a small amount of sodium azide into a large volume of gas.

The gas fills up an air bag, preventing the driver and passengers from incurring any life threatening injuries.

2NaN3(s) 2 Na(s) + 3 N2(g)

Sensors trigger an igniter

Fills up the

airbag

INTERPRETING A CHEMICAL EQUATION

What information can you get from a balanced chemical equation?

Nitrogen monoxide is present in car exhaust. UV light catalyzes the reaction of nitrogen monoxide and O2 to produce nitrogen dioxide smog.

2 NO(g) + O2 (g) 2 NO2(g)

UV

INTERPRETING A CHEMICAL EQUATION

• 2 NO(g) + O2 (g) 2 NO2(g) We see that 2 molecules of NO react w/ 1 molecule

of O2 to produce 2 molecules of NO2. These coefficients indicate relative numbers of

reactant and product molecules… it’s a ratio! The coefficients indicate a ratio of moles, or the

MOLE RATIO, of reactants and products in every balanced chemical equation.

UV

VERIFYING THE CONSERVATION OF MASS LAW

“Mass is neither created nor destroyed during a chemical reaction.”

The combined masses of the reactants must equal the combined masses of the products.

NO + O2 NO2

2 NO + O2 2 NO2

Which one obeys the Law of Conservation of Mass? THIS IS WHY YOU BALANCE AN EQUATION

MOLE-MOLE RELATIONSHIPS Remember, the coefficients in a chemical equation

indicate the mole ratio of the reactants and products Consider the synthesis reaction of nitrogen

and oxygen to give nitrogen monoxide:N2(g) + O2(g) 2 NO(g)

We see that 1 mol of nitrogen reacts with 1 mol of oxygen to produce 2 mol of nitrogen monoxide

Mole-Mole

N2(g) + O2(g) 2 NO(g)We can write several mole ratios

for this equation1 mol N2 : 1 mol O2

1 mol N2 : 2 mol NO1 mol O2 : 2 mol NO

Each can become a unit factor to convert between units

MOLE-MOLE RELATIONSHIPS

N2(g) + O2(g) 2 NO(g) How many moles of oxygen react with 2.25 mol of

nitrogen? To cancel units, we will use a mole ratio as our

unit factor: 1 mol O2 / 1 mol N2

2.25 mol N2 1 mol O2 = 2.25 mol O2 1 1 mol N2

Mole-Mole Relationships

N2(g) + O2(g) 2 NO(g) Calculate moles of nitrogen monoxide

produced by the reaction if 2.25 mol of N2 are used.

Plan: 1. Use the mole ratio 2 mol NO/1 mol N2 as the

unit factor.

2.25 mol N2 2 mol NO

= 4.50 mol NO

1 1 mol N2

MOLE- MOLE RELATIONSHIPS

Whenever we have a balanced chemical equation, we can always convert from moles of one substance to moles of another substance using a mole ratio as a unit factor.

MOLE-MOLE RELATIONSHIPS

Carbon monoxide is produced in a furnace by passing oxygen gas over hot coal. The balanced equation is: 2 C(s) + O2(g) 2 CO(g) How many moles of oxygen react with 2.50 mol

of carbon? Mole ratio is 2 mol C / 1 mol O2

2.50 mol C 1 mol O2 = 1.25 mol

O2 1 2 mol C

Mole-Mole relationships

2 C(s) + O2(g) 2 CO(g)How many moles of CO are

produced from 2.50 mol of carbon?Mole ratio is 2 mol C / 2 mol CO

2.50 mol C 2 mol CO= 2.50 mol CO

1 2 mol C

STOICHIOMETRY

We will apply mole ratios in order to relate quantities of reactants and products.

Stoichiometry is a term used to refer to the relationship between quantities in a chemical reaction according to a balanced chemical equation.

We can determine mass-mass relationships, mass-volume relationships (w/ a gas), and volume-volume relationships (w/ 2 gases)

MASS A MASS B PROBLEMS

An unknown mass of substance is calculated from a given mass of reactant or product.

After balancing the equation, proceed as follows:1. Convert the given mass to moles using the

molar mass of the substance as a unit factor2. Convert the moles of the given to moles of the

unknown using the mole ratio (coefficients)3. Convert the moles of unknown to grams using

the molar mass of the unknown as a unit factor.

1) GIVEN MASS MOLES 2) GIVEN MOLES UNKNOWN MOLES 3) UNKNOWN MOLES UNKNOWN MASS

Calculate the mass of tin(IV)chloride produced by the reaction of 1.25 g of metallic tin with yellow chlorine gas.

Sn(s) + 2 Cl2(g) SnCl4(s) First verify the equation is balanced, then calculate the moles

of tin. The molar mass of Sn is 118.71 g/mol.

Next, find the moles of SnCl4 using the mole ratio (1 mol Sn = 1 mol SnCl4)

Last, calculate the mass of product using the molar mass of SnCl4 (260.51 g/mol)

1.25 g Sn 1 mol Sn= 0.0105 mol Sn 1 118.71 g Sn

0.0105 mol Sn 1 mol SnCl4= 0.0105 mol SnCl4

1 1 mol Sn

0.0105 mol SnCl4 260.51 g = 2.74 g SnCl4 1 1 mol

DO YOU HAVE TO COMPLETE THESE IN 3 SEPARATE STEPS?

Once you get comfortable with the calculations, this is what it looks like as one step:

1.25 g Sn x x x = 2.74 g SnCl4

Until then, complete them in three separate steps:

1. Given mass moles (using molar mass)

2. Given moles unknown moles (using mole ratio)

3. Unknown moles unknown mass (using molar mass)

1 mol Sn118.71 g Sn

1 mol SnCl4 1 mol Sn

260.51 g SnCl4 1 mol SnCl4

Mass-Mass

Calculate the mass of potassium iodide, KI required to yield 1.78 g mercury(II)iodide, HgI2.

2 KI + Hg(NO3)2 HgI2 + 2 KNO3

1.78 g HgI2

1 mol HgI2 2 mol KI 166.00g KI= 1.30 g KI 1 454.39 g

HgI21 mol HgI2

1 mol KI

g HgI2 Molar mass Mole ratio Molar massg KI

Mass Mass Concept Map

Mass of Given

Moles of given Moles of Unknown

Mass of unknown

convert to moles using molar mass of given

use molar ratio

convert to mass using molar mass of unknown

MASS-VOLUME PROBLEMS

An unknown volume of gas is calculated from a given mass of reactant or product.

After balancing the equation, proceed as follows:1. Convert the given mass to moles using the molar

mass as a unit factor.2. Convert the moles of the given to moles of the

unknown using the mole ratio (coefficients)3. Convert the moles of unknown to liters using the

molar volume (at STP it’s 22.4 L/mol) as a unit factor.

Of course this can be reversed; we can find the mass of an unknown substance from a given volume of gas.

1) GIVEN MASS MOLE2) GIVEN MOLES UNKNOWN MOLES 3) UNKNOWN MOLES VOLUME

0.165 g of aluminum metal reacts with dilute hydrochloric acid. What is the volume of hydrogen gas produced at STP?

2 Al(s) + 6 HCl(aq) 2 AlCl3 + 3 H2(g) First, calculate the moles of aluminum (26.98 g/mol)

Next, use the mole ratio to find the moles of H2. (2 mol Al = 3 mol H2)

0.165 g Al

1mol Al= 0.00611 mol Al

1 26.98 g Al

0.00611 mol Al

3 mol H2

= 0.00917 mol H2

1 2 mol Al

1) GIVEN MASS MOLE 2) GIVEN MOLES UNKNOWN MOLES 3) UNKNOWN MOLES VOLUME

Last, multiply by the molar volume, 22.4 L/mol, to get the volume of H2 gas produced.

1 step:

0.00917 mol H2 22.4 L H2

= 0.205 L H2

1 1 mol H2

0.165 g Al 1 mol Al 3 mol H2 22.4 L H2

= 0.205 L H2

1 26.98 g Al 2 mol Al 1 mol H2

MASS-VOLUME PRACTICE

Baking soda can be used as a fire extinguisher. When heated, it decomposes to carbon dioxide gas which can smother a fire. If a sample of NaHCO3 (84.01 g/mol) produces 0.500 L of CO2 at STP, what is the mass of the sample?

2 NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

Verify the equation is balanced, find the moles of CO2 by using 22.4 L/mol.

0.500 L CO2 1 mol CO2 2 mol NaHCO3 84.01 g NaHCO3 = 3.75 g

NaHCO3

1 22.4 L CO2 1 mol CO2 1 mol NaHCO3

Mass-Volume Concept Map

Mass of given

Moles of given

Moles of Unknown

Volume of unknown gas

Convert to moles using molar mass of given

Use molar ratio

Convert to volume using molar volume of a gas

VOLUME-VOLUME PROBLEMS

We can convert from a given volume of gas to an unknown volume of gas in a single conversion using the mole ratio.

Sulfuric acid is produced by the conversion of sulfur dioxide to sulfur trioxide using heat and a platinum catalyst. The sulfur trioxide is then passed through water to produce sulfuric acid.

2 SO2(g) + O2(g) 2 SO3(g)

Calculate the liters of sulfur trioxide produced from 37.5 L of SO2. From the balanced equation we see that 2 volumes of SO2 = 2

volumes of SO3.

Pt/

37.5 L SO2 2 L SO3

= 37.5 L SO3 1 2 L SO2

SULFURIC ACID CONTINUED

Calculate how much oxygen gas needs to react with 37.5 L SO2 in order to produce that amount of sulfur trioxide.2 SO2(g) + O2(g) 2 SO3(g)

Summary, for volume-volume problems, you only need to use the volume-ratio.

37.5 L SO2 1 L O2

= 18.8 L O2 1 2 L SO2

Summary

LIMITING REACTANTS

Recipe: ¼ cup butter 1 10.5 oz bag of mini-marshmallows 5 cup crispy rice cereal

Given that we have 1 box of crispy rice cereal (approx. 10 c. cereal), 1 stick of butter (1/2 c. butter per stick) and 1 bag (10.5 oz) of marshmallows, how many recipes can we make? ONE

LIMITING REACTANTS

What did we run out of first?? Marshmallows limited how many rice-crispy

treats we could make. It’s called the limiting reactant.

If we go and get more butter, can we make more recipes?

If we go and get more cereal, can we make more recipes?

If we go and get more marshmallows, can we make more recipes?

NO

NO

YES!

LIMITING REACTANTS

In problems where we are given the amounts of two reactants, we determine the limiting reactant by:

1. Calculate the mass of product that can be produced from the first reactant.

2. Calculate the mass of product that can be produced from the second reactant. Repeat with each reactant… hope that it’s not a big

reaction…

3. State the limiting reactant (the one that produces the least amount of product) and the corresponding mass of product formed.

LIMITING REACTANTS Lets think of this as a balanced equation. Assume the recipe

looks like this:

3 Sugar + Butter + 8 Cereal Rice Crispy Treats Using the above equation, if I had 3 moles of sugar, 2 moles of

butter, and 14 moles of cereal, how many recipes could I make? 3 mol sugar 1 mol treats

= 1 mol treats 1 3 mol sugar

2 mol butter 1 mol treats= 2 mol treats 1 1 mol butter

16 mol cereal 1 mol treats= 1.75 mol treats 1 8 mol cereal

What is the limiting reactant? In other words, what do we run out of first?

SUGAR!!! We can only make one recipe w/ that amount of

sugar. It limits the number of recipes we can make.

LIMITING REACTANTS

3 C6H12O6 + CH3(CH2)14COOH + 8(C6H12O6)12 Rice Crispy Treats

Using the above equation, if I had 986 g of sugar (C6H12O6), 155 g of butter (CH3(CH2)14COOH), and 12,000 g of cereal ((C6H12O6)12), how many recipes (moles) could I make?

Sugar

Butter

Cereal

986 g sugar 1 mol sugar 1 mol treats= 1.83 mol treats

1 180 g sugar 3 mol sugar

155 g butter 1 mol butter 1 mol treats= 0.61 mol treats

1 256 g butter 1 mol butter

12,000 g cereal

1 mol cereal 1 mol treats= 0.69 mol treats 1 2160 g

cereal8 mol cereal

What is the limiting reactant? BUTTER!

Last question… You determined the limiting reactant in the previous problem to be

butter. Calculate how many grams of sugar are needed in order to completely react with that amount of butter (155 g butter). Why? Butter is leaving excess cereal and sugar which is wasted money

($$CHA-CHING!) Start with the grams of your limiting reactant (butter) and do a mass-mass

calculation to determine grams of sugar.

We only need 327 g of sugar to “react” with the butter to make these treats.

155 g butter 1 mol butter 3 mol sugar 180 g sugar= 327 g sugar

1 256 g butter 1 mol butter 1 mol sugar

Ok…one more question

How many grams of sugar would be left over?We have 986 g of sugar.We only need 327 g sugar.

986 g – 327 g = 659 g sugar is left over!

Manufacturers need to do these calculations all the time to be sure they’re not wasting anything.

Percent Yield

When the product from a reaction is less than expected, a percentage is calculated

Theoretical yield The maximum amount of product that could be

formed from given amounts of reactions This is calculated by you on paper!

Actual yield The amount of product that actually forms in

laboratory from an experiment Percent Yield- ratio of the two