Chapter 3 Stoichiometry. Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance

Chapter 3

Stoichiometry

Mass and Moles of a Substance

• Chemistry requires a method for determining the numbers of molecules in a given mass of a substance.

This allows the chemist to carry out “recipes” for compounds based on the relative numbers of atoms involved.The calculation involving the quantities of reactants and products in a chemical equation is called stoichiometry.

Chemical Stoichiometry

• Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions.

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• Objects behave as though they were all identical.• Atoms are too small to count. • Need average mass of the object.


A pile of marbles weigh 394.80 g. 10 marbles weigh 37.60 g. How many marbles are in the pile?

37.60 gAvg. Mass of 1 Marble = = 3.76 g / marble10 marbles

394.80 g = 105 marbles3.76 g

EXERCISE!EXERCISE!

By definition: 1 atom 12C “weighs” 12 amu

On this scale

1H = 1.008 amu

16O = 16.00 amu

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro Worldatoms & molecules

Macro Worldgrams

• 12C is the standard for atomic mass, with a mass of exactly 12 atomic mass units (amu).

• The masses of all other atoms are given relative to this standard.

• Elements occur in nature as mixtures of isotopes.• Carbon = 98.89% 12C

1.11% 13C < 0.01% 14C


Pg 78b

Average Atomic Mass for Carbon

98.89% of 12 amu + 1.11% of 13.0034 amu =


(0.9889)(12 u) + (0.0111)(13.0034 u) =

12.01 amu

exact number

Average Atomic Mass for Carbon

• Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01.

• This enables us to count atoms of natural carbon by weighing a sample of carbon.


Schematic Diagram of a Mass Spectrometer


Ligh

t

Ligh

t

Hea

vy

Hea

vy

• Isotopes of the same element have identical chemical properties.

• Some isotopes are radioactive.

• Find chlorine on the periodic table.

• What is the atomic number?

17

• What is the mass given?

35.45

• This is not the mass number of an isotope.

• What is this number?

• It is called the atomic mass - the weighed average of the masses of the isotopes that make up chlorine.

• Chlorine consists of chlorine-35 and chlorine-37 in a 3:1 ratio.

• The weighted average is an average corrected by the relative amounts of each isotope present in nature.

Calculate the atomic mass of naturally occurring chlorine if 75.77% of chlorine atoms are chlorine-35 and 24.23% of chlorine atoms are chlorine-37.

Step 1: Convert the percentage to a decimal fraction.

0.7577 chlorine-35

0.2423 chlorine-37

Step 2: Multiply the decimal fraction by the mass of that isotope to obtain the isotope contribution to the atomic mass.

For chlorine-35:0.7577 x 35.00 amu = 26.52 amu

For chlorine-370.2423 x 37.00 amu = 8.965 amu

Step 3: sum to get the weighted average

atomic mass of chlorine = 26.52 amu + 8.965 amu = 35.49 amu

Atomic Weights

Calculate the atomic weight of boron, B, from the following data:

ISOTOPE ISOTOPIC MASS (amu) FRACTIONAL ABUNDANCE

B-10 10.013 0.1978B-11 11.009 0.8022

Atomic Weights

Calculate the atomic weight of boron, B, from the following data:

ISOTOPE ISOTOPIC MASS (amu) FRACTIONAL ABUNDANCE

B-10 10.013 0.1978B-11 11.009 0.8022

B-10: 10.013 x 0.1978 = 1.9805B-11: 11.009 x 0.8022 = 8.8314

10.8119 = 10.812 amu ( = atomic wt.)

Nitrogen consists of two naturally occurring isotopes. 99.63% nitrogen-14 with a mass of 14.003 amu and 0.37% nitrogen-15 with a mass of 15.000 amu. What is the atomic mass of nitrogen?

An element consists of 62.60% of an isotope with mass 186.956 umu and 37.40% of an isotope with mass 184.953 amu.

• Calculate the average atomic mass and identify the element.

186.2 amu Rhenium (Re)


EXERCISE!EXERCISE!

Average atomic mass (6.941)


• The Mole Concept

A mole is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon–12.

The number of atoms in a 12-gram sample of carbon–12 is called Avogadro’s number (to which we give the symbol Na). The value of Avogadro’s number is 6.02 x 1023.

The mole (mol) is the amount of a substance that contains as many elementary entities as there

are atoms in exactly 12.00 grams of 12C

1 mol = NA = 6.0221367 x 1023

Avogadro’s number (NA)

One Mole of:

C S

Cu Fe

Hg

• The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.

• 1 mole of something consists of 6.022 × 1023 units of that substance (Avogadro’s number).

• 1 mole C = 6.022 × 1023 C atoms = 12.01 g C


Calculate the number of iron atoms in a 4.48 mole sample of iron.

2.70×1024 Fe atoms


EXERCISE!EXERCISE!

Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C atom = 12.00 amu

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any element

atomic mass (amu) = molar mass (grams)


• The molar mass of a substance is the mass of one mole of a substance.

For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units.That is, one mole of any element weighs its atomic mass in grams.

• Mass in grams of one mole of the substance:

Molar Mass of N = 14.01 g/mol

Molar Mass of H2O = 18.02 g/mol

(2 × 1.008 g) + 16.00 g

Molar Mass of Ba(NO3)2 = 261.35 g/mol

137.33 g + (2 × 14.01 g) + (6 × 16.00 g)


1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu

1 12C atom12.00 amu

x12.00 g

6.022 x 1023 12C atoms=

1.66 x 10-24 g1 amu

M = molar mass in g/mol

NA = Avogadro’s number

Which of the following is closest to the average mass of one atom of copper?

a) 63.55 g b) 52.00 gc) 58.93 gd) 65.38 g e) 1.055 x 10-22 g


CONCEPT CHECK!CONCEPT CHECK!


• Mole calculationsSuppose we have 100.0 grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent?

g/mol 55.8Fe g 100.0

Fe moles

Fe of moles 1.79


• Mole calculationsConversely, suppose we have 5.75 moles of magnesium (atomic wt. = 24.3 g/mol). What is its mass?

g/mol) (24.3 moles) 75.5( Mg mass

Mg of grams 140

x 6.022 x 1023 atoms K1 mol K

=

Do You Understand Molar Mass?

How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 1023 atoms K

0.551 g K 1 mol K39.10 g K

x

8.49 x 1021 atoms K

1. What is the mass in grams of 2.5 mol Na.

2. Calculate the number of atoms in 1.7 mol B.

3. Calculate the number of atoms in 5.0 g Al.

4. Calculate the mass of 5,000,000 atoms of Au.

Molecular Weight and Formula Weight

• The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not.

For example, one formula unit of NaCl contains 1 sodium atom (23.0 amu) and one chlorine atom (35.5 amu), giving a formula weight of 58.5 amu.

Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound.

1Na 22.99 amu

1Cl + 35.45 amuNaCl 58.44 amu

For any ionic compound

formula mass (amu) = molar mass (grams)

1 formula unit NaCl = 58.44 amu

1 mole NaCl = 58.44 g NaCl

NaCl

Do You Understand Formula Mass?

What is the formula mass of Ca3(PO4)2 ?

1 formula unit of Ca3(PO4)2

3 Ca 3 x 40.08

2 P 2 x 30.97

8 O + 8 x 16.00310.18 amu

Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

SO2

1S 32.07 amu

2O + 2 x 16.00 amu SO2 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2


• Mole calculationsThis same method applies to compounds. Suppose we have 100.0 grams of H2O (molecular weight = 18.0 g/mol). How many moles does this represent?

OH of moles 5.56 2g/mol 18.0

OH g 100.0 OH moles 2

2


• Mole calculationsConversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = 180.0 g/mol). What is its mass?

g/mol) (180.0 moles) 25.3( OHC mass 6126

6126 OHC of grams 585

Do You Understand Molecular Mass?

How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O

1 mol H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O

60 g C3H8Ox

8 mol H atoms

1 mol C3H8Ox

6.022 x 1023 H atoms

1 mol H atomsx =

Calculate the number of copper atoms in a 63.55 g sample of copper.

6.022×1023 Cu atoms



Which of the following 100.0 g samples contains the greatest number of atoms?

a) Magnesiumb) Zincc) Silver



Rank the following according to number of atoms (greatest to least):

a) 107.9 g of silverb) 70.0 g of zincc) 21.0 g of magnesium

b) a) c)


EXERCISE!EXERCISE!

Consider separate 100.0 gram samples of each of the following:

H2O, N2O, C3H6O2, CO2

– Rank them from greatest to least number of oxygen atoms.

H2O, CO2, C3H6O2, N2O


EXERCISE!EXERCISE!

Conceptual Problem Solving

• Where are we going?– Read the problem and decide on the final goal.

• How do we get there?– Work backwards from the final goal to decide

where to start.• Reality check.

– Does my answer make sense? Is it reasonable?


Determining Chemical Formulas

• The percent composition of a compound is the mass percentage of each element in the compound.

We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is,

%100whole the of mass

whole in A"" of mass A"" % mass

Mass Percentages from Formulas

• Let’s calculate the percent composition of butane, C4H10.

First, we need the molecular mass of C4H10.amu 48.0 amu/atom 12.0 @ carbons 4 amu 10.0 amu/atom 1.00 @ hydrogens 10

amu 58.0 HC of molecule 1 104

Now, we can calculate the percents.C%8.82%100 C % total amu 0.58

C amu 48.0 H%2.17%100 H % total amu 0.58

H amu 10.0

Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

• Mass percent of an element:

• For iron in iron(III) oxide, (Fe2O3):


mass of element in compoundmass % = × 100%

mass of compound


• Determining the formula of a compound from the percent composition.

The percent composition of a compound leads directly to its empirical formula.An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts.

Formulas

• Empirical formula = CH– Simplest whole-number ratio

• Molecular formula = (empirical formula)n

[n = integer]

• Molecular formula = C6H6 = (CH)6

– Actual formula of the compound


A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance

An empirical formula shows the simplest whole-number ratio of the atoms in a substance

H2OH2O

molecular empirical

C6H12O6 CH2O

O3 O

N2H4 NH2

Percent Composition and Empirical Formulas

Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent.

nK = 24.75 g K x = 0.6330 mol K1 mol K

39.10 g K

nMn = 34.77 g Mn x = 0.6329 mol Mn1 mol Mn

54.94 g Mn

nO = 40.51 g O x = 2.532 mol O1 mol O

16.00 g O

Percent Composition and Empirical Formulas

K : ~~ 1.00.63300.6329

Mn : 0.63290.6329

= 1.0

O : ~~ 4.02.532

0.6329

nK = 0.6330, nMn = 0.6329, nO = 2.532

KMnO4

g CO2 mol CO2 mol C g C

g H2O mol H2O mol H g H

g of O = g of sample – (g of C + g of H)

Combust 11.5 g ethanol

Collect 22.0 g CO2 and 13.5 g H2O

6.0 g C = 0.5 mol C

1.5 g H = 1.5 mol H

4.0 g O = 0.25 mol O

Empirical formula C0.5H1.5O0.25

Divide by smallest subscript (0.25)

Empirical formula C2H6O


• Determining the empirical formula from the percent composition.Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula?In other words, give the smallest whole-number ratio of the subscripts in the formula

Cx HyOz


Our 100.0 grams of benzoic acid would contain:

C mol )3(73.5g 12.0C mol 1

C g 8.68

H mol 0.5g 1.0H mol 1

C g 0.5

O mol)7(63.1g 16.0O mol 1

O g 2.26

This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might emerge.

Determining the empirical formula from the percent composition.


• Determining the empirical formula from the percent composition.

Our 100.0 grams of benzoic acid would contain:

3.501.63(7)C mol 73.5

3.01.63(7)H mol 0.5

1.001.63(7)O mol )7(63.1

now it’s not too difficult to see that the smallest whole number ratio is 7:6:2.The empirical formula is C7H6O2 .

Analyzing for Carbon and Hydrogen

• Device used to determine the mass percent of each element in a compound.



• Determining the “true” molecular formula from the empirical formula.

An empirical formula gives only the smallest whole-number ratio of atoms in a formula. The “true” molecular formula could be a multiple of the empirical formula (since both would have the same percent composition).To determine the “true” molecular formula, we must know the “true” molecular weight of the compound.


• Determining the “true” molecular formula from the empirical formula.

For example, suppose the empirical formula of a compound is CH2O and its “true” molecular weight is 60.0 g/mol.The molar weight of the empirical formula (the “empirical weight”) is only 30.0 g/mol.This would imply that the “true” molecular formula is actually the empirical formula doubled, or

C2H4O2

The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. – What is the empirical formula?

C3H5O2

– What is the molecular formula?C6H10O4


EXERCISE!EXERCISE!

Determining Chemical Formulas – Ionic Compounds

The percent composition of a compound leads directly to its empirical formula.An empirical formula (or simplest formula) for a ionic compound is the true molecular formula for that compound.

3 ways of representing the reaction of H2 with O2 to form H2O

A process in which one or more substances is changed into one or more new substances is a chemical reaction

A chemical equation uses chemical symbols to show what happens during a chemical reaction

reactants products

How to “Read” Chemical Equations

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

IS NOT

2 grams Mg + 1 gram O2 makes 2 g MgO

Features of a Chemical Equation

)(O )2Hg( )2HgO( 2 gls

Products - written on the right

Reactants- written on the left of arrow

Products and reactants must be specified using chemical symbols

Physical states are shown in parentheses

- means heat is needed.

)(O )2Hg( )2HgO( 2 gls

• The equation must be balanced.

– All the atoms of every reactant must also appear in the products.

• How many Hg’s on left?

• on right?

• How many O’s on left?

• on right?

Coefficient - how many of the substance are in the reaction

• The balanced equation represents an overall ratio of reactants and products, not what actually “happens” during a reaction.

• Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed.


The Experimental Basis of a Chemical Equation

We know that a chemical equation represents a chemical change. The following is evidence for a reaction:

• Release of a gas.

– CO2 is released when acid is placed in a solution containing CO3

2- ions.

– H2 is released when Na is placed in water.

• Formation of a solid (precipitate.)

– A solution containing Ag+ ions is mixed with a solution containing Cl- ions.

• Heat is produced or absorbed (temperature changes)

– Acid and base are mixed together

• The color changes

• Light is absorbed or emitted

• Changes in the way the substances behave in an electrical or magnetic field

• Changes in electrical properties.

Writing and Balancing the Equation for a Chemical Reaction

1. Determine what reaction is occurring. What are the reactants, the products, and the physical states involved?

2. Write the unbalanced equation that summarizes the reaction described in step 1.

3. Balance the equation by inspection, starting with the most complicated molecule(s). The same number of each type of atom needs to appear on both reactant and product sides. Do NOT change the formulas of any of the reactants or products.



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The reactants are starting substances in a chemical reaction. The arrow means “yields.” The formulas on the right side of the arrow represent the products.

A chemical equation is the symbolic representation of a chemical reaction in terms of chemical formulas.

Chemical Reactions: Equations

• Writing chemical equations

NaCl2ClNa2 2

For example, the burning of sodium and chlorine to produce sodium chloride is written

In many cases, it is useful to indicate the states of the substances in the equation.

When you use these labels, the previous equation becomes



)s(NaCl2)g(Cl)s(Na2 2

The law of conservation of mass dictates that the total number of atoms of each element on both sides of a chemical equation must match. The equation is then said to be balanced.



Consider the combustion of methane to produce carbon dioxide and water.

OH CO O CH 2224

For this equation to balance, two molecules of oxygen must be consumed for each molecule of methane, producing one molecule of CO2 and two molecules of water.

Chemical Reactions: Equations• Writing chemical equations

OH CO O CH 2224 Now the equation is “balanced.”

2 2

Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

2C2H6 NOT C4H12


3. Start by balancing those elements that appear in only one reactant and one product.

C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbonon left

1 carbonon right

multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right

multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O


4. Balance those elements that appear in two or more reactants or products.

2 oxygenon left

4 oxygen(2x2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3x1)

multiply O2 by 72

= 7 oxygenon right

C2H6 + O2 2CO2 + 3H2O72

remove fractionmultiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O


5. Check to make sure that you have the same number of each type of atom on both sides of the equation.

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products

4 C12 H14 O

4 C12 H14 O

4 C (2 x 2) 4 C

12 H (2 x 6) 12 H (6 x 2)

14 O (7 x 2) 14 O (4 x 2 + 6)


• Balance the following equations.

332 POCl PCl O

26424 N OP ON P

232232 SO O As O SAs

24243243 )POCa(H POH )(POCa


• Balance the following equations.

332 POCl PCl O

26424 N OP ON P

232232 SO O As O SAs

24243243 )POCa(H POH )(POCa

2 2

6 6

62 29

34

Stoichiometry: Quantitative Relations in Chemical Reactions

• Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction.

It is based on the balanced chemical equation and on the relationship between mass and moles.Such calculations are fundamental to most quantitative work in chemistry.

1. Write balanced chemical equation

2. Convert quantities of known substances into moles

3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity

4. Convert moles of sought quantity into desired units

Amounts of Reactants and Products

Molar Interpretation of a Chemical Equation

• The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as “mole-to-mole” relationships.

For example, the Haber process for producing ammonia involves the reaction of hydrogen and nitrogen.

)g(NH 2)g(H 3)g(N 322


• This balanced chemical equation shows that one mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

Because moles can be converted to mass, you can also give a mass interpretation of a chemical equation.

)g(NH 2 (g)3H (g)N 322

322 NH mol 2 H mol 3 N mol 1 1 molecule N2 + 3 molecules H2 2 molecules NH3

Calculations Using the Chemical Equation

• We will learn in this section to calculate quantities of reactants and products in a chemical reaction.

• Need a balanced chemical equation for the reaction of interest.

• Keep in mind that the coefficients represent the number of moles of each substance in the equation.

7


• Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2.

Because the coefficients in the balanced equation represent mole-to-mole ratios, the calculation is simple.

)g(NH 2 (g)3H (g)N 322

32

32 NH mol 2.3

H mol 3NH mol 2

H mol 8.4

Mass Relationships in Chemical Equations

• Amounts of substances in a chemical reaction by mass.

How many grams of HCl are required to react with 5.00 grams manganese dioxide according to this equation?

)s(MnO HCl(aq) 4 2 )g(Cl (aq)MnCl O(l)H 2 222

First, you write what is given (5.00 g MnO2) and convert this to moles.

Mass Relationships in Chemical Equations

2

22 MnO g9.86

MnO mol 1 MnO g 5.00

Then convert to moles of what is desired.(mol HCl)

2MnO mol 1

HCl mol 4

HCl mol 1

HCl g 36.5

HCl g 40.8

Finally, you convert this to mass (g HCl)

Consider the following reaction:

If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with?

8.07 g O2


EXERCISE!EXERCISE!

Methane (CH4) reacts with the oxygen in the air to produce carbon dioxide and water.

Ammonia (NH3) reacts with the oxygen in the air to produce nitrogen monoxide and water.

– Write balanced equations for each of these reactions.


(Part I)EXERCISE!EXERCISE!

Methane (CH4) reacts with the oxygen in the air to produce carbon dioxide and water.

Ammonia (NH3) reacts with the oxygen in the air to produce nitrogen monoxide and water.

– What mass of ammonia would produce the same amount of water as 1.00 g of methane reacting with excess oxygen?


(Part II)EXERCISE!EXERCISE!

Let’s Think About It

• Where are we going?– To find the mass of ammonia that would produce

the same amount of water as 1.00 g of methane reacting with excess oxygen.

• How do we get there?– We need to know:

• How much water is produced from 1.00 g of methane and excess oxygen.

• How much ammonia is needed to produce the amount of water calculated above.


Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH

32.0 g CH3OHx

4 mol H2O

2 mol CH3OHx

18.0 g H2O

1 mol H2Ox =

235 g H2O

Na + Cl2 NaCl

1. Balance the equation.

2. Calculate the moles of Cl2 which will react with 5.00 mol Na.

3. Calculate the number of grams of NaCl which will be produced when 5.00 mol Na reacts with an excess of Cl2.

4. Calculate the grams of Na which will react with 5.00 g Cl2.

Limiting Reactants

• Limiting reactant – the reactant that runs out first and thus limits the amounts of products that can be formed.

• Determine which reactant is limiting to calculate correctly the amounts of products that will be formed.


Limiting Reagent• The limiting reactant (or limiting reagent) is the

reactant that is entirely consumed when the reaction goes to completion.

The limiting reagent ultimately determines how much product can be obtained.For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, it is clear that the number of frames will determine how many bicycles can be made.

Limiting Reactants


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A. The Concept of Limiting Reactants

Stoichiometric mixture – N2(g) + 3H2(g) 2NH3(g)

Limiting reactant mixture

A. The Concept of Limiting Reactants

N2(g) + 3H2(g) 2NH3(g)

Limiting Reagent

• Zinc metal reacts with hydrochloric acid by the following reaction.

If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced?

)g(H (aq)ZnCl HCl(aq) 2 Zn(s) 22

Limiting ReagentTake each reactant in turn and ask how much product would be obtained if each were totally consumed. The reactant that gives the smaller amount is the limiting reagent.

Since HCl is the limiting reagent, the amount of H2 produced must be 0.26 mol.

22 H mol 0.30

Zn mol 1H mol 1

Zn mol 0.30

22 H mol 0.26

HCl mol 2H mol 1

HCl mol 0.52

Do You Understand Limiting Reagents?

In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

OR

g Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al

27.0 g Alx

1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent

Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3

x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation:

2H2 + O2 2H2O

a) 2 moles of H2 and 2 moles of O2

b) 2 moles of H2 and 3 moles of O2

c) 2 moles of H2 and 1 mole of O2

d) 3 moles of H2 and 1 mole of O2

e) Each produce the same amount of product.



Notice

• We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation.


• You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B. – What information do you need to know

in order to determine the mass of product that will be produced?



Let’s Think About It

• Where are we going?– To determine the mass of product that will be

produced when you react 10.0 g of A with 10.0 g of B.

• How do we get there?– We need to know:

• The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation.

• The molar masses of A, B, and the product they form.


You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation:

A + 3B 2C


EXERCISE!EXERCISE!

Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtainedfrom a reaction.

% Yield = Actual Yield

Theoretical Yieldx 100

Reaction Yield

Theoretical and Percent Yield• The theoretical yield of product is the maximum

amount of product that can be obtained from given amounts of reactants.

The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated).

%100yield ltheoretica

yield actual Yield%

Theoretical and Percent Yield• To illustrate the calculation of percentage yield,

recall that the theoretical yield of H2 in the previous example was 0.26 mol (or 0.52 g) H2.

If the actual yield of the reaction had been 0.22 g H2, then

%42%100H g 0.52H g 0.22

Yield%2

2

Consider the following reaction:

P4(s) + 6F2(g) 4PF3(g)

– What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield?

46.1 g P4


EXERCISE!EXERCISE!

Operational Skills• Calculating the formula weight from a formula.• Calculating the mass of an atom or molecule.• Converting moles of substance to grams and vice versa.• Calculating the number of molecules in a given mass.• Calculating the percentage composition from the formula.• Calculating the mass of an element in a given mass of compound.• Calculating the percentages C and H by combustion.• Determining the empirical formula from percentage composition.• Determining the true molecular formula.• Relating quantities in a chemical equation.• Calculating with a limiting reagent.

Worked Example 3.4

Worked Example 3.6

Worked Example 3.7

Worked Example 3.9

Worked Example 3.11

Worked Example 3.12

Worked Example 3.13a

Worked Example 3.13b

Worked Example 3.15a

Documents

Chapter 3 Stoichiometry. Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance