3 Stoichiometry Contents

3-1 Law of Conservation of Matter

3-2 Balancing Equation

3-3 Equations on a Macroscopic Scale

3-4 Mass Relationship in Chemical Reactions

3-5 Limiting Reactants

3-6 Theoretical Yield, Actual Yield, and Percent Yield

3.7 Quantitative analysis

3.8 Empirical Formulas from Percent Composition

3.9 Molecular and Structural Formulas

3.10 Percent Composition from Formulas

Stoichiometry is the study of quantitative relationships

between substances involved in chemical changes.

It is a very important subject from the point of view of

both theory and practice.

Analytical chemistry is based on Stoichiometry.

3-1 Law of Conservation of Matter

The Quantity of matter is not changed by chemical

reactions; Matter is neither created nor destroyed by chemical

reactions. The number of atoms of each kind must be the

same after reaction as it was before reaction.

In chemical reactions, the quantity of matter does not

change. The total mass of the products equals the total mass

of the reactants.

Law of Conservation of Matter provided the foundation for

modern chemistry.

H2 + N2 NH3

3H2 + N2 = 2NH3

3-2 Balancing Equation

A chemical reaction is a process in which one set of

substances called reactants is converted to a new set of

substances called products. In other words, a chemical reaction

is the process by which a chemical change occurs.

We need evidence before we can say that a reaction has

occurred. Some of the types of physical evidence to look for

are shown here.

• a color change

• formation of a solid (precipitate) with a clear solution

• evolution of a gas

• evolution or absorption of heat

When none of these signs of a chemical reaction appears,

we need chemical evidence. This requires a detailed chemical

analysis of the reaction mixture to discover whether any new

substances are present.

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.

3.7

2C2H6 NOT C4H12

Balancing Chemical Equations

3. Begin with the compound that has the most atoms

or the most kinds of atoms and use one of these

atoms as a starting point.

C2H6 + O2 CO2 + H2O

3.7

start with C2H6

start with C or H but not O

Balancing Chemical Equations

4. Balance those elements that appear in only once on each side of the arrow first.

C2H6 + O2 CO2 + H2O

3.7

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

start with C or H but not O

Balancing Chemical Equations

5. Balance those elements that appear more than once on a side. Balance free elements last.

3.7

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

The smallest whole-number coefficients are preferred!

Balancing Chemical Equations

6. 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

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)

3.7

Balancing Chemical Equations

3.7

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products

4 C12 H14 O

4 C12 H14 O

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

The system to balance an equation

• Begin with the compound that has the most atoms or

the most kinds of atoms and use one of these atoms as

a starting point.

• Balance elements that appear only once on each side

of the arrow first.

• Then balance elements that appear more than once

on a side.

• Balance free elements last.

Here are some useful strategies for balancing equations

If an element occurs in only one compound on each side of the equation, try balancing this element first.

When one of the reactants or products exists as the free element, balance this element last.

In some reactions, certain groups of atoms(for example, polyatomic ions) remain unchanged. In such cases, balance these groups as a unit.

It is permissible to use fractional as well as integral numbers as coefficients. At times, an equation can be balanced most easily by using one or more fractional coefficients and then, if desired, clearing the fractions by multiplying all coefficients by a common multiplier.

Example

Writing and balancing an equation: The combustion of a carbon-

Hydrogen-Oxygen Compound. Liquid triethylene glycol, C6H14O4,

is used as a solvent and plasticizer for vinyl and polyurethane pla

stics. Write a balanced chemical equation for its complete combu

stion.

Carbon- hydrogen-oxygen compounds, like hydrocarbons, yield carbon dioxide and water when burned in oxygen gas.

Starting expression : C6H14O4 + O2 CO2 + H2O

Balance C : C6H14O4 + O2 6CO2 + H2O

Balance H : C6H14O4 + O2 6CO2 + 7H2O

Solution

At this point, the right side of the expression has 19 O atoms(12 in

six CO2 molecules and 7 in seven H2O molecules). To get 19

atoms on the left, we start with 4 in a molecule of C6H14O4 and

need 15 more. This requires a fractional coefficient of 15/2 for O2.

Balance O : C6H14O4 + (15/2) O2 6CO2 + 7H2O (balanced)

To remove the fractional coefficient, multiply all coefficients by 2,

the denominator of the fractional coefficient, 15/2.

2 C6H14O4 + 15 O2 12CO2 +14H2O (balanced)Check:

Left : (2×6) =12 C; (2 ×14) = 28 H; [(2×4) + (15×2)] = 38 O

Right : (12×1) =12 C; (14 ×2) = 28 H;

[(12×2) + (14×1)] = 38 O

The term stoichiometry means, literally, to measure the elements,

but from a more practical standpoint, it includes all the quantitative

relationships involving atomic and formula masses, chemical form

ulas, and the chemical equation.

The coefficients in the chemical equation

2H2(g) + O2(g) 2H2O(l)

Mean that

2x molecules H2 + 1x molecules O2 2x molecules H2O

Suppose that we let x=6.02214×1023 .

Then the chemical equation also means that

2 mol H2 + 1 mol O2 2 mol H2O

3-3 Equations on a Macroscopic Scale

The coefficients in the chemical equation allow us to make statements, such as:

Two moles of H2O are produced for every two moles of H2

consumed.

Two moles of H2O are produced for every one mole of O2

consumed.

Two moles of H2 are consumed for every one mole of O2

consumed.A stoichiometric factor relates the amounts a mole basis, of any t

wo substances involved in a chemical reaction; thus a stoichiom

etric factor is a mole ratio.

The mole is the key to quantitative relationships between

substances involved in chemical changes on a practical scale.

Example

Relating the Mass of a reactant and a product. What mass of H2O is formed in the reaction of 4.16g H2 with an excess of O2?

Solution

The general strategy for reaction stoichiometry problems, outlined earlier and illustrated in below, suggests these three steps.

1. Convert the quantity of H2 from grams to moles. (use the inverse of molar mass of H2)

2. From the number of moles of H2, calculate the number of moles of H2O formed.

3. Convert the quantity of H2O from moles to grams. (use the inverse of molar mass of H2O)

3-4 Mass Relationship in Chemical Reaction

Grams of H2 GIVENUse inverse of molar mass

as conversion factor:1 mol H2/2.016 g H2

Moles of H2

Use coefficients in balanced chemical Equation to find mole

ratio: 2 mol H2O/2 mol H2

Moles of H2O= (2/2) ×mol H2

Use molar mass as conversion factor: 18.02 g H2O/1 mol H2

Grams of H2O FOUND

( g H2 mol H2 mol H2O g H2O )1 2 3

1 mol H2

2.016 g H2

? g H2O =4.16 g H2 × × × 2 mol H2

2 mol H2O 18.02 g H2O1 mol H2O

= 37.2 g H2O

Always begin solving problem about quantitative relationships

in reaction by writing an equation for the reaction. The

coefficients in the equation describe the relations between moles

of products and moles of reactions. For example:

2H2O = 2H2 + O2

Moles 2 2 1

Formula Mass 18.0 2.02 32.0

Mass 36.0 4.03 32.0

　 When all the reactants are completely and simultaneously co

nsumed in a chemical reaction, the reactants are said to be in stoi

chiometric proportions--in the mole ratios dictated by the coeffic

ients in the balanced equation. At other times, the reactants are n

ot usually present in the proportions shown by the equation. The

quantity of one reactant controls the amount of products that can

be formed. The reactant that is completely consumed: the limiting

reactant--determines the quantities of products formed.

3-5 Limiting Reactants

Solution

1. The first step in a stoichiometric calculation is to writeA balanced equation for the reaction. If the equation

Is not given, you must supply your own.

Determining the Limiting Reactant in a Reaction. Phosphorus trichloride, PCI3, is a commercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products. It is made by the direct combination of phosphorus and chlorine.

P4(s) + 6 Cl2(g) 4 PCl3(l)

What mass of PCl3(l) forms in the reaction of 125 g P4 with 323 g Cl2?

Example

Phosphorus trichloride

? Mol Cl2 = 323g Cl2 × = 4.56 mol Cl2

? Mol P4 = 125g P4 × = 1.01 mol P4

1mol Cl2

70.91 g Cl2

1 mol P4

123.9 g P4

P4(s) + 6 Cl2(g) 4 PCl3(l)

125 g 323 g ?g

2. Note what’s given and asked for the above reaction.

3. Write the formula masses needed below the reaction.

formula masses, u 123.9 70.91 137.3

4. Determine which reactant in limiting. Convert grams to moles.

? g PCl3 = 323 gCl2 × × ×

= 417 g PCl3

1 mol Cl270.91 g Cl2

4 mol PCl3

6 mol Cl2

137.3 g PCl3

1 mol PCl3

We see rather clearly that there is less than 6 mol Cl2 per mole of P4- Chlorine is the limiting reactant.

The remainder of the calculation is to determine the mass of PCl3 formed in the reaction of 323 g Cl2 with

an excess of P4.

4. Calculation and check.

If calculated mole tatio <6/1 chlorine is limiting

If calculated mole tatio >6/1 phosphorus is limiting

Grams of P4

Use inverse of molarMass as conversion factor

: 1 mol P4/123.9g P4

Moles of P4

Grams of Cl2

Use inverse of molarMass as conversion factor

: 1 mol Cl2/70.91g Cl2

Moles of Cl2

Calculate mole ratio of Cl2 to P4

Mole ratio = moles of Cl2/moles of P4

Example

Determining the quantity of excess reactant(s) remaining after a reaction. What mass of P4 remains in excess following the reaction in example above?

Solution

The key to this problem is to calculate the mass of P4 that is consumed, and we can base this calculation either on the mass of Cl2 consumed ：

?g P4 = 323 g Cl2× 1 mol Cl270.91 g Cl2

1 mol P4

6 mol Cl2× ×

123.9 g P4

1 mol P4

= 94.1 g P4

Or on the mass of PCl3 produced.

?g P4 = 417 g PCl3× 1 mol PCl3137.3gPCl3

1 mol P4

4 mol PCl3× ×

123.9 g P4

1 mol P4

= 94.1 g P4

The mass of P4 remaining after the reaction is simply the

difference between what was originally present and what

was consumed; that is,

125 gP4 initially – 94.1 g P4 consumed = 31 gP4 remaining

Theoretical Yield, Actual Yield, and Percent Yield

The theoretical yield of a reaction is the amount of

product calculated to be formed by the reaction.

The amount of product that is actually produced is

called the actual yield. The percent yield is defined

as:

percent yield = actual yield / theoretical yield

× 100%

3-6 Theoretical Yield, Actual Yield, and Percent Yield

In few reactions the actual yield almost exactly equals the theoretical yield, and the reactions are said to be quantitative. On the other hand, in some reactions the actual yield is less than the theoretical yield, and the percent yield is less than 100%. The yield may be less than 100% for many reasons:

(1) The product of a reaction rarely appears in a pure form, and in the necessary purification steps, some product may be lost through handling. This reduces the yield.

(2) In many cases the reactants may participate in reactions other than the one of central interest. These are called side reactions, and the unintended products are called by-products. To the extent that side reactions occur, the yield of the main product is reduced.

(3) Finally, if a reverse reaction occurs, some of the expected product may react to re-form the reactants, and again the yield is less than expected.

Example

Determining Theoretical, Actual, and Percent Yields. Billions

of pounds of urea, CO(NH2)2, are produced annually for use as

a fertilizer. The reaction used is

2 NH3 + CO2 CO(NH2)2 + H2O

The typical starting reaction mixture has a 3: 1 mole ratio of NH3

to CO2. If 47.7 g urea forms per mole of CO2 that reacts, what is

the (a) theoretical yield; (b) actual yield; and (c) percent yield in

this reaction?

60.1 g CO(NH2)2

Solution

(a) The stoichiometric proportions are 2 mol NH3: 1 mol CO2.

Because the mole ratio of NH3 to CO2 used is 3: 1, NH3 is in

excess and CO2 is the limiting reactant. Because the quantity of

urea is given per mole of CO2, we should base the calculation on

1.00 mol CO2.

Theoretical yield = 1.00mol CO2× ×

= 60.1g CO(NH2)2

1 mol CO(NH2)2

1 mol CO2 1 mol CO(NH2)2

(b) Actual yield = 47.7 g CO(NH2)2

(c) %yield = × 100% = 79.4%47.7 g CO(NH2)2

60.1 g CO(NH2)2

3.7 Quantitative analysis

Quantitative analysis is finding out how much of a given

substance is present in a sample.

Qualitative analysis is finding out what substances are

present in a sample.

100sample of mass total

samplein A masssamplein A of massby percent

3.8 Empirical Formulas from Percent Composition

The empirical formulas for a compound is the simplest

formula that shows the ratios of the numbers of atoms of each

kind in the compound.

•Empirical formulas

–give the relative numbers and types of atoms in a

molecule.

–That is, they give the lowest whole number ratio

of atoms in a molecule.

–Examples: H2O, CO2, CO, CH4, HO, CH2.

Example

The composition of a compound is 36.4% Mn, 21.2% S, and 42.4% O. All percents are mass percent. What is the empirical formulas of the compound?

Solution: Suppose there is 100 g sample, then:

Mn: 36.4/54.9=0.663mol; S: 21.1/32.1=0.660mol;

O: 42.4/16.0=2.65mol

0.663:0.660:2.65≈1: 1: 4 MnSO4

• Molecular formulas

– give the actual numbers and types of atoms in a

molecule.

– Examples: H2O, CO2, CO, CH4, H2O2, O2, O3, and

C2H4.

3.9 Molecular and Structural Formulas

• Most molecular substances that we will study in this class contain only nonmetals.

Space-filling models

• Molecular and empirical formulas do not show how

atoms are arranged when bonded together.

Picturing Molecules

• Molecules occupy three dimensional space.

• However, we often represent them in two dimensions.

• The structural formula gives the connectivity between

individual atoms in the molecule.

• The structural formula may or may not be used to

show the three dimensional shape of the molecule.

• If the structural formula does show the shape of the

molecule, then either a perspective drawing, ball-and-

stick model, or space-filling model is used.

Representing Structure in Molecules

Accurately representsthe angles at which molecules are attached.

Class Practice ExerciseThe structural formula of propane and butane is

What is the chemical and empirical formula for thesemolecules?

C C C

H H

H

H

H

HH

H C C C

H H

H

H

H

HH

H C

H

H

3.10 Percent Composition from Formulas

Example

What is the percent by mass of iron (III) to one decimal place of FeCl3?

Solution:

%4.34100335.555.8

55.8 FeClin iron of massby percent

3