Stoichiometry and the Mole Chapter 8 Stoichiometry-What is it? The study of the numerical relationship between chemical quantities in a chemical reaction.

Stoichiometry and the Mole

Chapter 8

Stoichiometry-What is it?

• The study of the numerical relationship between chemical quantities in a chemical reaction is called reaction stoichiometry

• The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction

– Law of Conservation of Mass– balancing equations by balancing atoms

Stoichiometry-What is it?• the number of pancakes you can make depends

on the amount of the ingredients you use

• this relationship can be expressed mathematically1 cu flour 2 eggs ½ tsp baking powder 5 pancakes

1 cup Flour + 2 Eggs + ½ tsp Baking Powder 5 Pancakes

Stoichiometry-What is it?• if you want to make more or less than 5 pancakes

you can use the number of eggs you have to determine the number of pancakes you can make

– assuming you have enough flour and baking powder

pancakes 20 eggs 2

pancakes 5eggs 8

How you measure how much?

• You can measure mass,

• or volume,

• or you can count pieces.

• We measure mass in grams.

• We measure volume in liters.

• We count pieces in MOLES.

The Mole

• The mole is a number.

• A very large number, but still, just a number.

• 6.022 x 1023 of anything is a mole

– A large dozen.

– The number of atoms in exactly 12 grams of carbon-12.

The Mole

The mole is Avogadro’s Number of items (6.02 x 1023).

A mole of atoms weighs the same number of grams as the atomic weight. 1 mole of hydrogen weighs 1.0079g. 1 mole of carbon atoms weighs 12.011g. The atomic weight is not only the number of protons and neutrons but is the grams of 1 mole of atoms.

Using the mole and the atomic weight at grams/mole is stoichiometry.

The Mole• the balanced equation is the “recipe” for a chemical

reaction

• the equation 3 H2(g) + N2(g) 2 NH3(g) tells us that 3 molecules of H2 react with exactly 1 molecule of N2 and make exactly 2 molecules of NH3 or

3 molecules H2 1 molecule N2 2 molecules NH3 – in this reaction

• and since we count molecules by moles

3 moles H2 1 mole N2 2 moles NH3

Mole-to-Mole Conversions

How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2 in the reaction below?

2 Na(s) + Cl2(g) 2 NaCl(s)

How many moles of sodium oxide result from the complete combination of 8.3 mol of O2 with sodium

How many moles of water are formed when 3.6 moles of phosphoric acid react with barium hydroxide

More Practice• 2C2H2 + 5 O2 4CO2 + 2 H2O

• If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?

• How many moles of C2H2 are needed to produce 8.95 mole of H2O?

• If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

Representative particles

• The smallest pieces of a substance.

• For a molecular compound it is a molecule.

• For an ionic compound it is a formula unit.

• For an element it is an atom.

Types of questions

• How many molecules of CO2 are the in 4.56

moles of CO2 ?

• How many moles of water is 5.87 x 1022 molecules?

• How many atoms of carbon are there in 1.23 moles of C6H12O6 ?

• How many moles is 7.78 x 1024 formula units of MgCl2?

Measuring Moles

• The amu was one twelfth the mass of a carbon 12 atom.

• Since the mole is the number of atoms in 12 grams of carbon-12,

• the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

Atomic Mass

• The mass of 1 mole of an element in grams.

• 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron.

• We can right this as 12.01 g C = 1 mole

Examples

• How much would 2.34 moles of carbon weigh?

• How many moles of magnesium in 24.31 g of Mg?

• How many atoms of lithium in 1.00 g of Li?

• How much would 3.45 x 1022 atoms of U weigh?

What about compounds?

• in 1 mole of H2O molecules there are two moles

of H atoms and 1 mole of O atoms

• To find the mass of one mole of a compound – determine the moles of the elements they have– Find out how much they would weigh– add them up

What about compounds?

• What is the mass of one mole of CH4?

• 1 mole of C = 12.01 g• 4 mole of H x 1.01 g = 4.04g

• 1 mole CH4 = 12.01 + 4.04 = 16.05g

• The Gram Molecular mass of CH4 is 16.05g

• The mass of one mole of a molecular compound.

Examples

• Calculate the molar mass of the following.

• Na2S

• N2O4

• C

• Ca(NO3)2

• C6H12O6

• (NH4)3PO4

Molar Mass

• The number of grams of 1 mole of atoms, ions, or molecules.

• We can make conversion factors from these.

• To change grams of a compound to moles of a compound.

For example

• How many moles is 5.69 g of NaOH?

Examples

• How many moles is 4.56 g of CO2 ?

• How many grams is 9.87 moles of H2O?

• How many molecules in 6.8 g of CH4?

• 49 molecules of C6H12O6 weighs how much?

Mole to Mole conversions• How many moles of O2 are produced when 3.34 moles of

Al2O3 decompose?

• 2 Al2O3 Al + 3O2

3.34 moles Al2O3 2 moles Al2O3

3 mole O2 = 5.01 moles O2

Your Turn• 2C2H2 + 5 O2 4CO2 + 2 H2O

• If 3.84 moles of C2H2 are burned, how

many moles of O2 are needed?

• How many moles of C2H2 are needed

to produce 8.95 mole of H2O?

• If 2.47 moles of C2H2 are burned, how

many moles of CO2 are formed?

How do you get good at this?

Mass in Chemical Reactions

How much do you make?

How much do you need?

The Steps in a Stoichiometric Calculation

Mass of substance A

Moles of substance A

Moles of substance B

Mass of substance B

Use molar mass of AUse molar mass of A

Use molar mass of BUse molar mass of B

Use coefficients of A & B in Use coefficients of A & B in balanced eqnbalanced eqn

The equation is :

2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(l)

A certain welding operation, requires that at least 86.0 g of Fe be produced. What is the minimum mass in grams of Fe2O3 that must be used for the operation?

Calculate also how many grams of aluminium are needed.

Strategy:

2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(l)

mass of Fe mol of Fe

mol of Fe mol of Fe2O3

mol of Fe2O3 mass of Fe2O3

For example...• If 10.1 g of Fe are added to a solution of

Copper (II) Sulfate, how much solid copper would form?

• Fe + CuSO4 Fe2(SO4)3 + Cu

• 2Fe + 3CuSO4 Fe2(SO4)3 + Cu

10.1 g Fe

55.85 g Fe

1 mol Fe= 0.181 mol Fe

2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu

0.181 mol Fe2 mol Fe

3 mol Cu= 0.272 mol Cu

0.272 mol Cu1 mol Cu63.55 g Cu

= 17.3 g Cu

Could have done it

10.1 g Fe55.85 g Fe1 mol Fe

2 mol Fe3 mol Cu

1 mol Cu63.55 g Cu

= 17.3 g Cu

More Examples• To make silicon for computer chips they

use this reaction

• SiCl4 + 2Mg 2MgCl2 + Si

• How many grams of Mg are needed to make 9.3 g of Si?

• How many grams of SiCl4 are needed to

make 9.3 g of Si?

• How many grams of MgCl2 are produced

along with 9.3 g of silicon?

For Example• The U. S. Space Shuttle boosters use

this reaction

• 3 Al(s) + 3 NH4ClO4

Al2O3 + AlCl3 + 3 NO + 6H2O

• How much Al must be used to react with 652 g of NH4ClO4 ?

• How much water is produced?

• How much AlCl3?

Gas and Moles

Gases• Many of the chemicals we deal with are

gases.

• They are difficult to weigh.

• Need to know how many moles of gas we have.

• Two things effect the volume of a gas– Temperature and pressure

Standard Temperature and Pressure (STP)

• 0ºC and 1 atm pressure

• At STP 1 mole of gas occupies 22.4 L

• Called the molar volume

• Avogadro's Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

For Example• If 6.45 grams of water are decomposed,

how many liters of oxygen will be produced at STP?

• 2H2O 2H2 + O2

6.45 g H2O 18.02 g H2O1 mol H2O

2 mol H2O1 mol O2

1 mol O2

22.4 L O2

Example• How many liters of CH4 at STP are required

to completely react with 17.5 L of O2 ?

• CH4 + 2O2 CO2 + 2H2O

17.5 L O2 22.4 L O2 1 mol O2

2 mol O2 1 mol CH4

1 mol CH4 22.4 L CH4

= 8.75 L CH4

22.4 L O2 1 mol O2

1 mol CH4 22.4 L CH4

Examples• What is the volume of 4.59 mole of CO2

gas at STP?

• How many moles is 5.67 L of O2 at

STP?

• What is the volume of 8.8g of CH4 gas

at STP?

Example• How many liters of CO2 at STP will be

produced from the complete combustion of 23.2 g C4H10 ?

• What volume of oxygen will be required?

Density of a gas• D = m /V

• for a gas the units will be g / L

• We can determine the density of any gas at STP if we know its formula.

• To find the density we need the mass and the volume.

• If you assume you have 1 mole than the mass is the molar mass (PT)

• At STP the volume is 22.4 L.

Examples

• Find the density of CO2 at STP.

• Find the density of CH4 at STP.

The other way• Given the density, we can find the molar

mass of the gas.• Again, pretend you have a mole at STP, so V

= 22.4 L.• m = D x V• m is the mass of 1 mole, since you have 22.4

L of the stuff.• What is the molar mass of a gas with a

density of 1.964 g/L?• 2.86 g/L?

Limiting Reagent

Limiting Reactants

The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when a reaction goes to completion.

The moles of product are always determined by the starting moles of the limiting reactant.

The Cheese Sandwich Analogy

Which is the limiting reactant?

• Use the relationships from the balanced chemical equation

• You take each reactant in turn and ask how much product would be obtain, if each were totally consumed.

• The reactant that gives the smaller amount of product is the limiting reactant.

Strategy:

Example

• Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?

• If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?

• 2Cu + S Cu2S

10.6 g Cu 63.55g Cu 1 mol Cu

2 mol Cu 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 13.3 g Cu2S

3.83 g S 32.06g S 1 mol S

1 mol S 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 19.0 g Cu2S

= 13.3 g Cu2S

Cu is Limiting Reagent

Limiting Reactant: Example• 10.0g of aluminum reacts with 35.0 grams of

chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?

2 Al + 3 Cl2 2 AlCl3• Start with Al:

• Now Cl2:

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3

27.0 g Al 2 mol Al 1 mol AlCl3

= 49.4g AlCl3

35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3

71.0 g Cl2 3 mol Cl2 1 mol AlCl3

= 43.9g AlCl3

LimitingLimitingReactantReactant

LR Example Continued

• We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete .

Limiting Reactant Practice

• 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

Finding the Amount of Excess

• By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

• Can we find the amount of excess potassium in the previous problem?

Finding Excess Practice• 15.0 g of potassium reacts with 15.0 g of iodine.

2 K + I2 2 KI

• We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced.

15.0 g I2 1 mol I2 2 mol K 39.1 g K

254 g I2 1 mol I2 1 mol K= 4.62 g K USED!

15.0 g K – 4.62 g K = 10.38 g K EXCESS

Given amount of excess reactant

Amount of excess reactant actually used

Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

Limiting Reactant: Recap

1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.

2. Convert ALL of the reactants to the SAME product

3. The lowest answer is the correct answer.

Limiting Reactant: Recap

1. The reactant that gave you the lowest answer is the LIMITING REACTANT.

2. The other reactant (s) are in EXCESS.

3. To find the amount of excess, subtract the amount used from the given amount.

4. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

Your turn

• If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced?

• How many grams of solid?

• How much excess reagent is left?

Your Turn II

• If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced?

• How much excess reagent will remain?

Yield

• The amount of product made in a chemical reaction.

• There are three types

• Actual yield- what you get in the lab when the chemicals are mixed

• Theoretical yield- what the balanced equation tells you you should make.

• Percent yield = Actual x 100 % Theoretical

Example

• 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.

• 2Al + 3 CuSO4 Al2(SO4)3 + 3Cu

• What is the actual yield?

• What is the theoretical yield?

• What is the percent yield?

Empirical Formula

Percent Composition

• Formula:

Mass of one element X 100%

Mass of compound

Tells the relative mass each element contributes to the mass of the whole compound.

Practice

• Find the percent composition of the idicated element in the following:

• Al in aluminum sulfate• O in Tin (IV) phosphate• Iron in Iron (II) nitride• Nitrogen in ammonium phosphite• Hydrogen in perchloric acid• Carbon in glucose

The Empirical Formula• The lowest whole number ratio of elements

in a compound.

• The molecular formula the actual ratio of elements in a compound.

• The two can be the same. –CH2 empirical formula–C2H4 molecular formula–C3H6 molecular formula–H2O both

Calculating Empirical• Just find the lowest whole number ratio

• C6H12O6

• CH4N

• It is not just the ratio of atoms, it is also the ratio of moles of atoms.

• In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.

• In one molecule of CO2 there is 1 atom of C and 2 atoms of O.

Calculating Empirical• We can get ratio from percent composition.

• Assume you have a 100 g.

• The percentages become grams.

• Can turn grams to moles.

• Find lowest whole number ratio by dividing by the smallest.

• Be careful! Do not round off numbers prematurelyBe careful! Do not round off numbers prematurely

Example

• Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.

• Assume 100 g so

• 38.67 g C x 1mol C = 3.220 mole C 12.01 gC

• 16.22 g H x 1mol H = 16.09 mole H 1.01 gH

• 45.11 g N x 1mol N = 3.219 mole N 14.01 gN

Example

• The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N

• The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol N

• C1H5N1

Convert the grams to moles

C mol 3.9 12.01g

C mol 1C 47g

O mol 2.9 16.00g

O mol 1O g 47

Hmol 6.0 1.008g

Hmol 1 Hg 6.0

Determine the Empirical Formula ofAcetic Anhydride if its Percent Composition is

47% Carbon, 47% Oxygen and 6.0% Hydrogen

Divide each by the smallest number of moles

1.3 2.9 C mol 3.9

1 2.9 O mol 2.9

2 2.9 H mol 6.0



If any of the ratios are not a whole number, multiply all the ratios by a factor to make it a whole number– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then multiply by 3;

if ?.25 or ?.75 then multiply by 4

4 3x 1.3 2.9 C mol 3.9

3 3x 1 2.9 O mol 2.9

6 3x 2 2.9 H mol 6.0 Multiply all theRatios by 3 Because C is 1.3

Determine the Empirical Formula ofAcetic Anhydride if its Percent Composition is47% Carbon, 47% Oxygen and 6.0% Hydrogen

° Use the ratios as the subscripts in the empirical formula

4 3x 1.3 2.9 C mol 3.9

3 3x 1 2.9 O mol 2.9

6 3x 2 2.9 H mol 6.0 C4H6O3



Empirical formula from CompositionConsider the following flow-diagram:

Percent composition

Mass Composition

Number of moles of each element

Divide by smallest number of moles to find the molar ratios

Multiply by appropriate number to get whole number subscripts

Practice

• A compound is 43.64 % P and 56.36 % O. What is the empirical formula?

• Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

Molecular Formulas

• The molecular formula is a multiple of the empirical formula

• To determine the molecular formula you need to know the empirical formula and the molar mass of the compound

Determine the empirical formula• May need to calculate it as previous

C5H3

Determine the molar mass of the empirical formula

5 C = 60.05 g, 3 H = 3.024 g

C5H3 = 63.07 g

Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an

empirical formula of C5H3

Divide the given molar mass of the compound by the molar mass of the empirical formula–Round to the nearest whole number

407.63

252 gg



Multiply the empirical formula by the calculated factor to give the molecular formula

(C5H3)4 = C20H12



Example

• A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

Stoichiometry and the Mole Chapter 8 Stoichiometry-What is it? The study of the numerical relationship between chemical quantities in a chemical reaction.

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