Bond Breaking and Bond Making in Chemical Reactions

When a chemical reaction has taken place it is usual that the products have either less stored energy than the reactants (overall exothermic) or more stored energy than the reactants (overall endothermic).

That is, with regard to the balanced chemical equation, the total molar enthalpy of the products is less (overall exothermic) or more (overall endothermic) than that of the reactants.

DH = SH_products - SH_reactants

Enthalpy level diagrams illustrate this:

Bond breaking and bond making...

When chemical reactions take place, chemical bonds are broken in the reactants and new chemical bonds are formed in the products. It is as a result of these processes that a reaction is overall exothermic (energy is given out to the surroundigs) or overall endothermic (energy is absorbed from the surroundings).

Bond breaking is obviously an endothermic process.

It is not quite so obvious that...

Bond making is an exothermic process.

But, if a particular bond in a molecule is broken and then reformed the same amount of energy must be involved, because the First Law of Thermodynamics (also known as the 'law of conservation of energy') must apply.

Thermodynamics in the study of transforming energy from one form to another.
The First Law of Thermodynamics states that: Energy can be transformed from one form to another , but cannot be created or destroyed. The total amount of energy in the universe never changes. Hess's Law (G.H. Hess, 1840) of 'constant heat summation' follows from the First Law.

F₂(g)    ®    2F(g)      DH° = +158 kJ mol^-1
2F(g)    ®    F₂(g)     DH° = -158 kJ mol^-1

Bond Enthalpy...

The molar bond enthalpy is the energy required to break one mole of bonds between pairs of atoms in the 'gaseous molecule'. Bond enthalpy values are usually expressed in kJ mol^-1 of bonds broken.

Molar bond enthalpy is also called bond dissociation enthalpy, but there are other terms too that refer to this.

Bond enthalpies are often averaged values...

However, for more complicated molecules a precise bond enthalpy for a particular chemical bond depends to some extent on the environment in the molecule where the bond exists. That is, a precise bond enthalpy depends on what other atoms are attached to the two atoms of the bond to be broken. For this reason, tables of bond enthalpy values are averaged over those found in a large number of different compounds.
Also...

Bond enthalpy values apply to molecules in the gaseous state.

Because bond enthalpy values are averaged and also apply to molecules in the gaseous state, when they are used in calculations the answer will be only approximate. However, the approximation is often very good and such calculations are useful in predicting the overall enthalpy change, DH, for a chemical reaction. There is often close agreement with the standard value for an enthalpy change given in a chemical data book.

Using bond enthalpies...

Bond enthalpies can be used to calculate an overall enthalpy change, DH, for a chemical reaction. It is an over-simplification, but bonds are broken in the reactants and new bonds are formed in the products.

If it takes more energy to break bonds in the reactants than is released when new bonds form in the products then the reaction will be overall endothermic. A reaction is overall exothermic if more energy is released when new bonds form in products than is used when bonds in the reactants are broken.

This idea is illustrated below for the combustion of ethanol:

CH₃CH₂OH(g)  +  3O₂(g)  ®  2CO₂(g)  +  3H₂O(g)

Note that all reactants and products are gaseous.

In the method above all the chemical bonds are broken in the reactants and new bonds formed in the products. There are very few reactions in which the reactant molecules are completely broken up into atoms, but this does not matter here because Hess's Law applies.

Calculating an Enthalpy Change of Reaction, DH...

This is quite straigtforward. First write out the balanced equation for the reaction showing full structural formulae. Now simply add together the molar bond enthalpies involved for the reactants to obtain a total endothermic value. Do the same for the products to obtain a total exothermic value. Finally, add the total endothermic and exothermic values together to find DH for the reaction.

Bond	Number broken	Number formed	Average Molar Bond Enthalpy kJ mol-1
C-C	1	0	+347
C-H	5	0	+413
C-O	1	0	+358
O-H	1	6	+464
O=O	3	0	+498
C=O	0	4	+805
Here bond enthalpies are defined endothermically.

Bond breaking:
Total endothermic value = (+347 x 1) + (+413 x 5) + (+358 x 1) + (+464 x 1) + (+498 x 3) = +4728 kJ
Bond making:
Total exothermic value = (-464 x 6) + (-805 x 4) = -6004 kJ
Sum total of bond breaking and bond making:
DHc = +4728 + - 6004 = -1276 kJ mol^-1

Now compare this with the value of -1368 kJ mol^-1 given in a chemical data book. The data book value is more exothermic by 92 kJ mol^-1. This is explained by the use of averaged molar bond enthalpy values and that all reactants and products are gaseous, in the calculation of DH using bond enthalpies. For example, when water vapour forms liquid water the process is exothermic.

Now check your own calculations below...

Bond	Number broken	Number formed	Enthalpy Change, DH = kJ mol-1
C-C
C-H
C-O
O-H
O=O
C=O

Bond Enthalpies and Bond Lengths

Bond Bond Length nm  Bond Enthalpy kJ mol-1 C - C 0.154 +347 C = C 0.134 +612 C º C 0.120 +838 Bond Length and Bond Enthalpy are averaged values. F - F 0.142 +158.0 Cl - Cl 0.199 +243.4 Br - Br 0.228 +192.9 I - I 0.267 +151.2 H - F 0.092 568.0 H - Cl 0.127 +432.0 H - Br 0.141 +366.3 H - I 0.161 +298.3

The molar bond enthalpies and bond lengths in the table above show that in general...

The longer a chemical bond the weaker it is.

For example, hydrogen chloride shows little tendencey to decompose into its constituent elements when heated. Strong heating of hydrogen bromide produces a brown colour of bromine vapour, while copious violet fumes of iodine form when a hot glass rod is plunged into a gas jar of hydrogen iodide.

HI(g)    ®    ½H₂(g)   +   ½I₂(g)

The stability of the hydrogen halides to thermal decomposition therefore decreases in the order:

HCl   >   HBr   >   HI

and this is due to the progressive decrease in the H ¾ X bond enthalpy.

Hot
glass rod

Hydrogen Iodide
HI(g)

With regard to the diatomic halogen molecules (X₂), fluorine has an abnormally low molar bond enthalpy. This low value is in part explained by the repulsion between the non-bonding electrons on the fluorine atoms, the other halogen molecules having longer bond lengths making this repulsive force is less significant.

chemical equation

=> one product?syntesist

In chemistry, chemical synthesis is purposeful execution of chemical reactions to get a product, or several products. This happens by physical and chemical manipulations usually involving one or more reactions. In modern laboratory usage, this tends to imply that the process is reproducible, reliable, and established to work in multiple laboratories.
A chemical synthesis begins by selection of compounds that are known as reagents or reactants. Various reaction types can be applied to these to synthesize the product, or an intermediate product. This requires mixing the compounds in a reaction vessel such as a chemical reactor or a simple round-bottom flask. Many reactions require some form of work-up procedure before the final product is isolated.^[1] The amount of product in a chemical synthesis is the reaction yield. Typically, chemical yields are expressed as a weight in grams or as a percentage of the total theoretical quantity of product that could be produced.A side reaction is an unwanted chemical reaction taking place that diminishes the yield of the desired product.


Chemical reactions convert reactants to products, whose properties differ from those of the reactants. Chemical equations are a compact and convenient way to represent chemical reactions. They have the general form
Reactant(s) → Product(s)
The arrow in the equation means "changes to" or "forms." The reaction of gaseous nitrogen with hydrogen to produce ammonia, NH ₃ , is represented by the chemical equation
Although there are thousands of chemical reactions, a significant number of them, especially those that are not organic reactions, can be classified according to four general patterns: combination, decomposition, displacement, and exchange.

1. Combination. A combination reaction is one in which two or more substances (the reactants) are combined directly to form a single product (the product). An example is the reaction in which sodium (Na) combines with chlorine (Cl ₂ ) to form sodium chloride, or table salt (NaCl).

A reaction of sodium with chlorine to produce sodium chloride is an example of a combination reaction.

2 Na + Cl ₂ → 2 NaCl
The physical states of reactants and products are included where necessary. The symbols used are: ( s ) for solid, ( l ) for liquid, ( g ) for gas, and ( aq ) for aqueous (water) solutions. In the case of sodium chloride formation, the equation is modified accordingly.
2 Na ( s ) + Cl ₂ ( g ) → 2 NaCl ( s )
2. Decomposition. A decomposition reaction can be considered to be the reverse of a combination reaction. In a decomposition reaction, one substance (the reactant) decomposes to form two or more products. For example, calcium carbonate (limestone) decomposes at high temperatures to calcium oxide (lime) and carbon dioxide. This reaction is used industrially to produce large quantities of lime.
3. Displacement. A displacement reaction (also called a single replacement reaction) occurs when an element reacts with a compound to form a new compound and release a different element. An example is the reaction that releases silicon (Si) from silicon dioxide (sand), SiO ₂ , via its reaction with carbon. Carbon monoxide, CO, is the reaction's other product. When further purified, the silicon can be used in computer chips.
SiO ₂ ( s ) + 2 C ( s ) → Si ( s ) + 2 CO ( g )
4. Exchange. During an exchange reaction, "partners" in compounds exchange their partners. One type of exchange reaction is called a neutralization reaction, the reaction between an acid and a base. The reaction of sodium hydroxide (lye), NaOH, with hydrochloric acid, HCl, to produce NaCl and water is such a reaction. In this case, Na ⁺ switches partners from OH ⁻ to Cl ⁻ , and H ⁺ from Cl ⁻ to OH ⁻ .
NaOH ( aq ) + HCl ( aq ) → NaCl ( aq ) + H ₂ O ( l )
Organic chemical reactions, those in which carbon plays a predominant role, are very important in biochemical systems and industrial processes. These reactions can also be represented by balanced chemical equations, a few examples of which are given.
The fermentation of glucose to produce ethyl alcohol (ethanol)
The synthesis of acetylsalicylic acid (aspirin) from the reaction of salicylic acid with acetic anhydride
The formation of a triglyceride (a fat), such as the biochemical synthesis of tristearin via the reaction of stearic acid with glycerol:
Matter is conserved in chemical reactions: The total mass of the products equals the total mass of the reactants. Chemical equations reflect this conservation. It is why chemical equations must be balanced. Atoms have mass, and the numbers of each kind of atom on each side of the equation must be the same. Coefficients, the numbers to the left of the formulas, are used to balance equations. Many equations can be balanced directly by simply adjusting the coefficients, as illustrated in the equations given above. Other equations are more difficult to balance, such as that for the decomposition of nitroglycerine (an explosive)
4 C ₃ H ₅ (NO ₃ ) ₃ ( l ) → 12 CO ₂ ( g ) + 10 H ₂ O ( l ) + 6 N ₂ ( g ) + O ₂ ( g )
and this complicated reaction involving several reactants and products
4 CuSCN + 7 KIO ₃ + 14 HCl → 4 CuSO ₄ + 7 KCl + 4 HCN + 7 ICl + 5 H ₂ O
Balanced chemical equations provide a significant amount of information. Consider the equation for photosynthesis, the natural process by which green plants form glucose, C ₆ H ₁₂ O ₆ , and oxygen from the reaction of carbon dioxide with water.
This balanced equation and its coefficients can be interpreted as indicating that six carbon dioxide molecules and six water molecules react to form one molecule of glucose and six oxygen molecules, each containing two oxygen atoms. A coefficient multiplies the term following it. The "6 CO ₂ " denotes six CO ₂ molecules containing a total of six carbon atoms and twelve oxygen atoms.
Applying these concepts to the remainder of the balanced equation yields information that confirms that the equation is balanced—the atom counts for both sides of the equation are the same.

Reactants	Products
Carbon atoms = 6	Carbon atoms = 6
Hydrogen atoms = 12	Hydrogen atoms = 12
Oxygen atoms = 12 + 6 = 18	Oxygen atoms = 6 + 12 = 18

Coefficients also apply to a larger scale, in which the counting unit is the mole (there are 6.02 × 10 ²³ molecules per mole of a compound), rather than individual molecules. Thus, this balanced equation also represents the reaction of six moles of glucose with six moles of water to produce one mole of glucose and six moles of oxygen.
Oxidation-reduction (redox) reactions are an important, general kind of reaction, one involving the transfer of electrons. Oxidation is the loss of an electron or electrons from an element, ion, or compound. Reduction is the gain of an electron or electrons from an element, ion, or compound. The two processes occur simultaneously; electrons released during oxidation are gained in a reduction process. In every redox reaction, a reactant is oxidized (loses electrons) and a reactant is reduced (gains electrons). During a redox reaction there is a change in oxidation numbers—evidence of a redox reaction. An oxidation number compares the charge of an uncombined atom, one not in a compound, with its actual or relative charge when it is part of a compound. Oxidation numbers are zero, positive, or negative.
These guidelines are used to determine oxidation numbers.

• Atoms of pure elements, that is, atoms not combined with any other element, have an oxidation number of zero. For example, sodium in metallic sodium, Na; oxygen in molecular oxygen, O ₂ ; and chlorine in molecular chlorine, Cl ₂ , each have an oxidation number of 0.
• Monatomic ions have an oxidation number equal to the charge of the ion. Thus, a sodium ion, Na ⁺ , has an oxidation number of +1; that of chlorine in a chloride ion, Cl ⁻ , is −1.
• Generally, hydrogen atoms in compounds have an oxidation number of +1; oxygen atoms in compounds are typically −2.
• The sum of oxidation numbers in a neutral compound is zero. Water, H ₂ O, is an example. Hydrogen: 2 H × (+1/H) = +2; oxygen: 1 O × (−2/O) = −2; (+2) + (−2) = 0
• The sum of oxidation numbers of the atoms in a polyatomic ion equals the charge on the ion. For example, the sulfate ion, SO ₄ ⁻² , a polyatomic ion, has a net charge of −2. Each oxygen in a sulfate ion has an oxidation number of −2, and four oxygens add up to −8. For the sulfate ion to have a net −2 charge, sulfur must have a +6 oxidation number: −2 = 4(−2) +6.

Oxidation numbers and their changes can be used to identify the reaction of sodium with chlorine to form NaCl as a redox reaction.
2 Na + Cl ₂ → 2 NaCl

Reactant	Products
Oxidation number	Oxidation number
Na = 0	Na + = +1
Cl = 0	Cl − = −1

During this reaction, reactant sodium atoms (oxidation number 0) are converted to sodium ions (oxidation number +1); reactant chlorine atoms (oxidation number 0) are transformed to chloride ions (oxidation number −1). Because there is a change in the oxidation numbers of the reactants during the reaction, this is a redox reaction. The definitions of oxidation and reduction can be broadened a bit using oxidation numbers: Oxidation is an increase in oxidation number; reduction is a decrease in oxidation number. The gain in oxidation number occurs because electrons are lost during oxidation; the gain of electrons during reduction causes a decrease in the oxidation number. This can be shown by using so-called half-reactions for each process.
Oxidation half-reaction: Na → Na ⁺ + e ⁻
Reduction half-reaction: Cl ₂ + 2 e ⁻ → 2 Cl ⁻
Notice in the balanced equation that two moles of Na were used to react with the two moles of chlorine atoms in one mole of Cl ₂ . Each mole of Na lost one mole of electrons; each mole of chlorine atoms gained a mole of electrons. Two moles of electrons were transferred to form two moles of NaCl. The overall reaction is the sum of the two half-reactions; the moles of electrons cancel, and the sodium ions and chloride ions combine to form sodium chloride. Note that the sum of the oxidation numbers in sodium chloride is zero: (+1) + (−1) = 0.
Oxidation half-reaction: 2 Na → 2 Na ⁺ + 2 e ⁻
Reduction half-reaction: Cl ₂ + 2 e ⁻ → 2 Cl ⁻
Overall reaction: 2 Na + Cl ₂ → 2 NaCl
Oxidation-reduction reactions, even complex ones, can be balanced using either the half-reaction method or the oxidation number method. The half-reaction method will be discussed first, using the reaction of iron with chlorine to produce iron chloride.
Fe + Cl ₂ → FeCl ₃ (unbalanced equation)

Half-Reaction Method

Step 1. Divide the reaction into two half-reactions; one corresponding to oxidation, the other, reduction.
Oxidation: Fe → Fe ³⁺
Reduction: Cl ₂ + → Cl ⁻
Step 2. Balance each half-reaction for mass and then charge. The iron half-reaction is balanced with respect to mass because there is one iron on each side. However, the charge is not balanced; the left side has a charge of zero, the right side has a charge of +3. Charge is balanced by adding three electrons to the right side.
Fe → Fe ³⁺ + 3 e ⁻
The chlorine half-reaction is unbalanced in terms of mass and charge. Mass balance is achieved by using a coefficient of 2 on the right side.
Cl ₂ → 2 Cl ⁻
Charge is then balanced by adding two electrons to the left side.
Cl ₂ + 2 e ⁻ → 2 Cl ⁻
The two half-reactions indicate that three electrons are lost per Fe atom during oxidation, and that two electrons are gained as each Cl ₂ molecule is reduced.
Step 3. Combine the two half-reactions in such a way as to balance the electrons lost and gained. The oxidation half-reaction lost three electrons; the reduction half-reaction gained two electrons. Therefore, to balance electrons lost and gained, multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3. Add the resulting half-reactions to get the final balanced equation for the formation of FeCl ₃ . Note that, in doing so, the electrons cancel (as they should if the final equation is balanced).
2 [Fe → Fe ³⁺ + 3 e ⁻ ] → 2 Fe → 2 Fe ³⁺ + 6 e ⁻
3 [Cl ₂ + 2 e ⁻ → 2 Cl ⁻ ] → 3 Cl ₂ + 6 e ⁻ → 6Cl ⁻
The Fe ³⁺ and Cl ⁻ ions combine to form FeCl ₃ and the overall balanced equation is
2 Fe + 3 Cl ₂ → 2 FeCl ₃
The half-reaction method can be applied to more complex redox reactions, such as the reaction of permanganate ion, MnO ₄ ⁻ , with Fe ²⁺ in acidic solution.
MnO ₄ ⁻ ( aq ) + Fe ²⁺ ( aq ) → Mn ²⁺ ( aq ) Fe ³⁺ ( aq ) (unbalanced equation)
Step 1. Oxidation: Fe ²⁺ ( aq ) → Fe ³⁺ ( aq )
Reduction: MnO ₄ ⁻ ( aq ) → Mn ²⁺ ( aq )
Step 2. Mass and charge balance are achieved this way:
The iron is balanced by adding one electron on the right
Fe ²⁺ ( aq ) → Fe ³⁺ ( aq ) e ⁻
To balance oxygen, we use H ₂ O on the right side; to balance hydrogen, we use H ⁺ on the left side (recall that the reaction is taking place in acidic solution)
8 H ⁺ ( aq ) + MnO ₄ ⁻ ( aq ) → Mn ²⁺ ( aq ) + 4 H ₂ O ( l )
The reduction half-reaction has a net charge of +7 on the left [(8+) + (−1)] and +2 on the right [(+2) + 0]. Adding 5 electrons to the left side balances the charge.
8 H ⁺ ( aq ) + MnO ₄ ⁻ ( aq ) + 5 e ⁻ → Mn ²⁺ ( aq ) + 4 H ₂ O ( l )
Step 3. Equalize the electrons transferred. Multiply the oxidation half-reaction by 5. Add the half-reactions, canceling the electrons.
5 Fe ²⁺ ( aq ) → 5 Fe ³⁺ ( aq ) + 5 e ⁻
8 H ⁺ ( aq ) + MnO ₄ ⁻ ( aq ) + 5 e ⁻ → Mn ²⁺ ( aq ) + 4 H ₂ O( l )
Balanced equation: 5 Fe ²⁺ ( aq ) + 8 H ⁺ ( aq ) + MnO ₄ ⁻ ( aq ) → 5 Fe ³⁺ ( aq ) + Mn ²⁺ ( aq ) + 4 H ₂ O ( l )

Oxidation Number Method

MnO ₄ ⁻ ( aq ) + Fe ²⁺ ( aq ) + H ⁺ ( aq ) → Mn ²⁺ ( aq ) + Fe ³⁺ ( aq ) H ₂ O ( l ) (unbalanced equation)
As in the half-reaction method, H ₂ O is used to balance oxygen, and H ⁺ is used to balance hydrogen.
Step 1. Identify the oxidation number of each element on each side of the equation. Determine which has undergone oxidation and which has undergone reduction. This is indicated in Table 1.

Table 1.

Element	Oxidation Number as Reactant	Oxidation Number as Product	Change in Oxidation Number	Oxidation or Reduction
Mn	+7	+2	Decrease by 5	Reduction
O	−2	−2	0	Neither
Fe	+2	+3	Increase by 1	Oxidation
H	+1	+1	0	Neither

Step 2. Use coefficients so that the total increase in oxidation number equals the total decrease. In this case, the total decrease is 5 (Mn ⁺⁷ becomes Mn ²⁺ ), and the total increase must also be 5; each iron must be multiplied by 5: (5 Fe ²⁺ becomes 5 Fe ³⁺ ). Balance hydrogen and oxygen in the usual manner. The balanced equation is
MnO ₄ ⁻ ( aq ) + 5 Fe ²⁺ ( aq ) + 8 H ⁺ ( aq ) → Mn ²⁺ ( aq ) + 5 Fe ³⁺ ( aq ) + 4 H ₂ O ( l )

STOICHIOMETRY

Mole Concept and Avogadro’s Constant

 Describe the mole concept and apply it to substances.

The mole concept applies to all kinds of particles: atoms, molecules, ions, formula units etc.  The amount of substance is measured in units of moles.  The approximate value of Avogadro’s constant (L), 6.02 x 10²³ mol^-1, should be known.

A mole (often abbreviated mol) is the number equal to the number of carbon atoms in exactly 12 grams of pure ¹²C.  Techniques such as mass spectrometry, which count atoms very precisely, have been used to determine this number as 6.02214 x 10²³ (6.02 x 10²³ is good enough for IB).  This number is called Avogadro’s number to honor his contributions to chemistry.  One mole of something consists of 6.02 x 10²³ units of that substance.  Just as a dozen eggs is 12 eggs, a mole of eggs is 6.02 x 10²³ eggs.

The magnitude of the number 6.02 x 10²³ is difficult to imagine.  To give you some idea, 1 mole of seconds represents a span of time 4 million times as long as the earth has already existed, and 1 mole of marbles is enough to cover the earth to a depth of 50 miles!  However, since atoms are so tiny, a mole of atoms or molecules is a perfectly manageable quantity to use in a reaction.

The mole is also defined as such that a sample of a natural element with a mass equal to the element’s atomic mass expressed in grams contains 1 mole of atoms.  This definition also fixes the relationship between the atomic mass unit and the gram.  Since 6.02 x 10²³ atoms of carbon (each with a mass of 12 amu) have a mass of 12 g, then

(6.02 x 10²³ atoms)(12 amu/atom)= 12 g

6.02 x 10²³ amu=1 g

   Calculate the number of particles and the amount of substance (in moles).

Calculate between the amount of substance (in moles) and the number of atoms, molecules or formula units.

In order to convert from x moles of anything to how many actual atoms, multiply x by Avagadro’s constant to find how many actual particles you have.

For example, say you are given 2.3 moles of hydrogen.  In order to find how many actual particles of hydrogen you have, you do the following calculations…

2.3 mol H x (6.02 x 10²³ particles H/1 mol H) = 1.38 x 10²⁴

1.2 Formulas

 Define the term molar mass (M) and calculate the mass of one mole of a species.

The molar mass of a substance is the mass in grams of one mole of the compound.

How can we calculate the mass of one mole of a substance?  Let’s take methane for example.  Methane is CH₄, which means in a molecule of methane there is one carbon atom and four hydrogen atoms.  So, the molar mass of CH₄ would be, what is the mass in grams of one mole of methane, or what is the mass of 6.02 x 10²³ CH₄ molecules?  The mass of 1 mole of methane can be found by summing the masses of carbon and hydrogen.

Mass of 1 mol C: 12.01 g (This can be found on the periodic table)

Mass of 4 mol H: 4 x 1.008 g (If you don’t understand where the four came from, learn what subscripts mean)

Add them.

Mass of 1 mol CH4: 16.04 g

Traditionally, the term molecular weight has been used for this quantity.  The molar mass of a known substance is obtained by summing the masses of the component atoms as we did for methane.

Distinguish between atomic mass, molecular mass and formula mass.

The term molar mass (in g mol^-1) can be used for all of these.

Molecular Mass: The mass in grams of one mole of molecules or formula units of a substance; the same as molar mass.

Formula Mass: The mass of a formula, including ionic compounds.  Technically, molecular mass only applies to molecules (which are defined by covalent bonding).  Formula mass can also include ionic compounds, etc.

Atomic Mass: Atomic mass is somewhat difficult to define.  According to Zumdahl, atomic mass is technically the average atomic mass of an element, which is determined by finding the different isotopes of an element present in nature, and how much percent of the total amount of that element in the world they make up.  For example, in nature there are two primary isotopes of Carbon, ¹²C and ¹³C (there are other isotopes but they are so rare they are insignificant at this level of precision.  98.89% of Carbon is ¹²C, but 1.11% is ¹³C.  So to find the atomic mass of Carbon, we do the following…

(.9889)(12 amu (mass of ¹²C by definition)) + (.0111)(13.0034 amu (mass of ¹³C)) = 12.01 amu

This is the atomic mass of carbon. 

Define the terms relative molecular mass (M_r) and relative atomic mass (A_r).

The terms have no units.

The relative atomic mass is the mass of one atom of an element compared to the mass of Carbon 12.  The relative atomic mass has no units because it is a ratio of masses and the units cancel out.  The relative molecular mass is the mass of the relative atomic masses of all the atoms in a molecule added up.  It’s basically the molecular mass without units as far as I can tell.

 State the relationship between the amount of substance (in moles) and mass, and carry out calculations involving amount of substances, mass and molar mass.

The relationship between moles and mass is demonstrated in the mass formula.

Number of Moles (N)= mass (m)/Molecular Mass (M)…

N=m/M

So, say you are given 40 grams of Carbon Dioxide, how many moles do you have? The molecular mass of Carbon Dioxide (CO2) is 44 g.  So, the way to solve this would be to divide 40 grams by 44 g.

Number of Moles: 40g/44= .91 mol

Note: I am not sure how the units work out for this.  It would appear that the grams cancel each other out leaving no value, but for some reason you are left with moles.  This may be something to ask your teacher about.

 Define the terms empirical formula and molecular formula.

The molecular formula is a multiple of the empirical formula.

The empirical formula is best understood by knowing its usefulness.  One way scientists find the formulas for molecules is by determining the percent of the mass each element in the molecule takes up.  For example, let us take CH₅N.  A scientist is attempting to find the formula but doesn’t know what it is.  He breaks it up and finds that 38.67% Carbon, 16.22% Hydrogen, and 45.11% Nitrogen.  By assuming the compound weighs 100 grams, each of these percentages now becomes gram units.  Then by changing them to moles and setting them up in ratio to each other, the scientist finds that the ratio of Nitrogen to Hydrogen to Carbon is 1:5:1.  So, he knows that the formula can be CH₅N.  However, the formula could also be C₂H₁₀N₂, or C₃H₁₅N₃.  That is because all he has is a ratio.  The empirical formula is CH₅N.  This is the simplest form.  The exact formula is called the molecular formula, and that will be (CH₅N)n where n is an integer.  Sometimes the empirical and molecular formula will be the same, other times they will not be.  If you did not understand the whole percent composition and ratios, don’t worry, those will be explained later.  The important part is understanding the difference between empirical and molecular formula and how they are related.  The empirical formula is the formula for the compound that has the lowest values possible for it to still work.  The molecular formula is the formula with the actual values for the compound, and it will always be the empirical formula multiplied by some integer.

  

Determine the empirical formula and/or the molecular formula of a given compound.

Determine the:

• Empirical formula from the percentage composition or from other suitable experimental data.
• Percentage composition from the formula of a compound.
• Molecular formula when given both the empirical formula and the molar mass.

The steps taken in the previous objective will be examined in much more detail here.  In order to do these problems we will actually work the problems out, and taking the same steps with any similar step should work out the same way.

1.  Empirical formula from the percentage composition or from other suitable experimental data.

When a new compound is prepared, one of the first items of interest is the formula of the compound.  This is most often determined by taking a weighed sample of the compound and either decomposing it into its component elements or reacting it with oxygen to produce substances such as CO₂, H₂O, and N₂, which are then collected and weighed.  We will see how information of this type can be used to compute the formula of a compound.  Suppose a substance has been prepared that is composed of carbon, hydrogen, and nitrogen.  When 0.1156 gram of this compound is reacted with oxygen, 0.1638 gram of carbon dioxide (CO₂) and 0.1676 gram of water (H₂O) are collected.  Assuming that all the carbon in the compound is converted to CO₂, we can determine the mass of carbon originally present in the 0.1156-gram sample.  To do this, we must use the fraction (by mass) of carbon in CO₂.  The molar mass of CO₂ is

C: 1 mol x 12.01g/mol=12.01 g

O: 2 mol x 16.00 g/mol=32.00 g

Added together, the molar mass of CO₂ = 44.01 g/mol.

The fraction of carbon present by mass is

Mass of C/Total mass of CO₂ = 12.01 g C/ 44.01 g CO₂

This factor can now be used to determine the mass of carbon in 0.1638 gram of CO2:

0.1638 g CO₂ x 12.01 g C/44.01 g CO₂ = 0.04470 g C

Remember that this carbon originally came from the 0.1156-gram sample of unknown compound.  Thus the mass percent of carbon in this compound is

(0.04470 g C/0.1156 g compound) x 100= 38.67% C

The same procedure can be used to find the mass percent of hydrogen in the unknown compound.  We assume that all the hydrogen present in the original 0.1156 gram of compound was converted to H₂O.  The molar mass of H₂O is 18.02 grams, and the fraction of hydrogen by mass of H₂O is

Mass of H/Mass of H₂O = 2.016 g H/18.02 g H₂O

Therefore, the mass of hydrogen in 0.1676 gram of H₂O is

0.1676 g H₂O x (2.016 g H/18.02 g H₂O)= 0.01875 g H

And the mass percent of hydrogen in the compound is

(0.01875 g H/0.1156 g) compound x 100 = 16.22%

The unknown compound contains only carbon, hydrogen, and nitrogen.  So far we have determined that is is 38.67% carbon and 16.22% hydrogen.  The remainder must be nitrogen.

100.00% - (38.67% + 16.22%) = 45.11% N

We have determined that the compound contains 38.67% carbon, 16.22% hydrogen, and 45.11% nitrogen.  Next we use these data to obtain the formula.  Since the formula of a compound indicated the numbers of atoms in the compound, we must convert the masses of the elements to numbers of atoms.  The easiest way to do this is to work with 100.00 grams of the compound in the present case, 38.67% carbon by mass means 38.67 grams of carbon per 100.00 grams of compound, etc.  To determine the formula, we must calculate the number of carbon atoms in 38.67 grams of carbon, the number of hydrogen atoms in 16.22 grams of hydrogen, etc.  You can do that using the mass formula (found in 1.2.4.).  The answers are 3.220mol C, 16.09 mol H, 3.219 mol N.  Thus 100.00 grams of this compound contains 3.220 moles of carbon atoms, 16.09 moles of hydrogen atoms, and 3.219 moles of nitrogen atoms.  We can find the smallest whole number ratio (empirical formula) of atoms in this compound by dividing each of the mole values above by the smallest of the three:

C: 3.220/3.219 = 1.000 = 1

H: 16.09/3.219 = 4.998 = 5

N: 3.219/3.219 = 1.000 = 1

So, the empirical formula for the unknown compound is CH₅N.  Now, if you want to know the molecular formula, you have to know the formula mass of the unknown compound.  If the formula mass of the unknown is the same as the formula mass of the empirical formula, then the empirical formula is the molecular formula, but if the formula mass of the unknown is say n times more then the formula mass of the empirical formula, then the molecular formula is n times the empirical formula.

2. Percentage composition from the formula of a compound.

Say you are given the formula H₂O, and you want to find out the percentage composition of hydrogen.  How you do this is you find the formula mass of water, and then find the atomic mass of H₂, and divide the latter by the former.  Then multiply by 100 to get a percent value.  So, formula mass of H₂O is 18, H₂ is 2, so 2 g H₂/ 18 g H₂O=.11, x 100=11.1%. 

3.  Molecular formula given both the empirical formula and the molar mass. 

Say you are given the empirical formula CH₅N, and are told that the molar mass is 62.12 g, and are told to find the molecular formula.  First you find the formula mass of CH₅N, which ends up being 31.06 g.  You know that that multiplied by n equals the molecular mass of the formula you want, and the n in this is case is 2 (you could find this by dividing 62.12 g unknown/ 31.06 g empirical formula).

 Chemical Equations

Balance chemical equations when all reactants and products are given.

Distinguish between coefficients and subscripts.

Say you are given the reaction C₆H₁₄ + O₂à CO₂ + H₂O and are told to balance it.  The main idea behind balancing an equation is you have to have the same amount of each element on both sides of the equation, and the only thing you can edit are the coefficients, not the subscripts.  So, for example, in this equation, in the hexane (reactant), you have 6 carbons, but in the products you only have one carbon.  So, to balance the carbons you have to place a 6 in front of the CO2.  You now have 6 carbons on both sides.  Next look at the hydrogen (a good rule of thumb is whenever you have an element that is by itself, such as the oxygen in the reactants, leave it for last).  You have 14 hydrogen on one side and only 2 on the other, so you have to multiply the water on the right by 7 (because the hydrogen has a subscript of 2, which means there are two hydrogen in the water compound.)  You now have fourteen hydrogen on both sides.  You now move onto the oxygen.  There is 2 oxygen in the reactants and 19 in the products.  There is a rule when balancing equations that you are not really supposed to use anything but integers (although when you get into more advanced stuff you will occasionally use fractions to make it easier on yourself, but you’re not supposed too).  So, there is no way to balance it as is.  So, you’ll have to go back and change your original coefficients.  If you put a 2 in front of the hexane and a 12 in front of the carbon dioxide you’ll have balanced the carbons.  Then to balance the hydrogen put a 14 in front of the water.  Now, you have 28 hydrogen on both sides.  You now have two oxygen in the reactants and 26 in the products.  So, by multiplying the oxygen in the reactants by 13, you have 26 oxygen on both sides and you have balanced the equation.  

2C₆H₁₄ + 13O₂à 12CO₂ + 14H₂O

In reality, practice makes perfect when it comes to balancing equations.  The only way to learn is to keep trying, and after a few you’ll get pretty good and fast at it.  Some other rules are that if all of the coefficients have a common denominator, you can divide everything by that common denominator (so if the reactants had coefficients of 2 and 4 and the products had coefficients of 6 and 10, you could divide all of them by 2).  And remember that you cannot change a subscript, and when you multiply a compound by a coefficient the whole compound is multiplied, not just the front element.

 Identify the mole ratios of any two species in a balanced chemical equation.

Use balanced chemical equations to obtain information about the amounts of reactants and products.

Using the reaction from above, 2C₆H₁₄ + 13O₂à 12CO₂ + 14H₂O.

The mole ration from hexane to Oxygen is 2 mol hexane/13 mol O₂.  Assuming stoichiometric equilibrium (there are no limiting or excess reagents) it takes 2 moles of hexane to react with every 13 moles of oxygen.  This can also be done from reactants to products.  Another mole ratio is 13 mol O₂/12 mol CO₂.  Once again assuming stoichiometric equilibrium, for every 13 mol O₂ you get 12 mol CO₂.  This can also be done for products and products (it can be done for any two species in a reaction.)

Apply the state symbols of (s), (l), (g), and (aq).

Encourage the use of state symbols in chemical equations.

The state symbols are placed after each species to specify what state each each species in a reaction is in.  The symbol (s) stands for the solid state, (l) stands for liquid, (g) stands for gaseous, and (aq) stands for in aqueous solution (dissolved in water).  Often times you don’t have to write the state symbols and they won’t have much importance, but when they are needed it is really bad to miss them.  For that reason I HIGHLY recommend you get used to using them every time you know what they are, it’ll save you later on from making the stupid mistake of forgetting them when they are needed.

Mass and Gaseous Volume Relationships in Chemical Reactions

 Calculate stoichiometric quantities and use these to determine experimental and theoretical yields.

Mass is conserved in all chemical reactions.  Given a chemical equation and the mass or amount (in moles) of one species, calculate the mass or amount of another species.

The first rule to remember is mass is always conserved (at least at this point in the syllabus).  So the total mass of one side should always add up on another side.  Now, once again we will use the chemical equation from above.

2C₆H_{14 (g)} + 13O_{2 (g)}à 12CO_{2 (g)} + 14H₂O _(l)

Say you are told that you have 40 grams of hexane, how many grams of oxygen do you need to completely react with all the hexane?  Well, in order to solve this you use mole rations.

2 moles hexane/13 moles O2 = 40 g hexane/x moles O2.  You then solve this like a normal ratio problem (multiply diagonally then solve for x).  You answer is 260 grams of oxygen is needed to react with 40 g of hexane. Note: This is not the best example because you normally don’t use the measurement grams for gases, but it would be the same way for liters.

  

Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given.

Given a chemical equation and the initial amounts of two or more reactants:

• Identify the limiting reactant
• Calculate the theoretical yield of a product
• Calculate the amount(s) of the reactant(s) in excess remaining after the reaction is complete.

1.  Identify the limiting reactant.

There are many ways to solve for the limiting reactant, and everyone tends to like their own way.  I will show mine below but if you don’t like it feel free to ask your teacher or someone else for another way to do it.  Say you have the equation

4FeCr₂O_4(s) + 8K₂CO_3(s) + 7O_2(g) à 8K₂CrO_4(s) + 2Fe₂O_3(s) + 8CO_2(g)

The masses are 169 kg, 298 kg, and 75 kg respectively for the reactants, and you want to find out which one is the limiting reagent.  In order to do this, you first have to find out how many moles of each of the reactants you have (using the mass formula), we find their to be 7.55 x 10² mol chromite ore, 2.16 x 10³ mol potassium hydroxide, and 2.34 x 10³ mol O₂.

You then set up the mole ratios you need to react correctly and the mole ratios you have.

So, to react correctly you should 4 moles of chromite ore for every 8 moles of potassium carbonate.  That is, 4 mol FeCr2O4/8 mol K₂CO_3(s), which equals .5.  You then take the values that you actually have, which are 7.55 x 10² mol chromite ore/2.16 x 10³ mol potassium hydroxide, which equals .35.  This is smaller then .5, and that means that the numerator is too small (if the answer was bigger then .5, the denominator would be too small).  Whichever value is too small, that number is the limiting reagent FOR THAT PAIR.  You then have to take that reagent and compare it with the other reactant, or the rest of the reactants if there are more then three.  So, taking chromite ore we do the same thing comparing it with oxygen.  We have 7.55 x 10² mol chromite ore/ 2.34 x 10³ mol O₂ which then equals .32.  The ratio we want is 4/7, or .57, which is bigger then the actual value we got, which means O₂ is the limiting reagent of the entire reaction (you don’t have to then compare it to the first reactant tested).  So, oxygen is the limiting reagent of the formula.

2 and 3.  Calculate the theoretical yield of a product and amount of excess reagent remaining after reaction is complete.

Theoretical yield of a product is the amount of a product formed when the limiting reactant is completely consumed.  In order to obtain it, you go through several steps, that we will go through with the reaction 2C₂H₆ + 7O₂ à 4CO₂ + 6H₂0 starting with 20 g of ethane and 40 grams of oxygen.  So, the first step is to identify the limiting reagent.

2 mol ethane/7 mol oxygen=.29  .66 mol ethane/1.25 mol oxygen=.528.  Since the number is too big, that means the denominator is the limiting reagent, or oxygen is the limiting reagent.  We then have to find how much ethane is used up before the oxygen is consumed.  We do this by multiplying the actual amount of oxygen by the mole ratio of the two reactants.

1.25 mol oxygen x (2 mol ethane/ 7 mol oxygen)= .36 mol ethane. 

This is how much ethane is used up in the reaction, so now we know how much ethane was used and we know how much ethane we started with, so to find how much ethane is left over we do a simple subtraction.

.66 mol ethane started with - .36 mol ethane used = .3 mol ethane left over.

So, we have .3 mol ethane left over after the reaction.  Then, to find theoretical yield, you have two options.   You can use how much oxygen use, or you can use how much ethane you use in the reaction, either should gain you the right answer.  I would recommend using the value given to you by the problem, so that way if you made a mistake in finding the excess reagent left over you won’t mess up your second answer.  You know you used 1.25 mol oxygen, so you do another mole ratio.

1.25 mol oxygen x (4 mol carbon dioxide/7 mol oxygen)=.714 mol carbon dioxide.

So, .714 mol carbon dioxide is your theoretical yield for that product.

Note: Rarely do you ever calculate the theoretical yield, because side reactions normally occur that decrease from the theoretical yield.  The actual yield is sometimes compared to the actual yield by doing the following calculations: actual yield/theoretical yield x 100= Percent Yield.

Apply Avagadro’s law to calculate reacting volumes of gases.

Avagadro’s law is V=an, where V stands for volume, a stands for proportionality constant, and n is the number of moles of gas particles.  Basically, what this equation states is that for constant temperature and pressure, the volume is directly proportional to the number of moles of gas.  This relationship is obeyed closely by gases at low pressures.  An example of this can be exemplified by the following reaction 3O_2(g) à 2O_3(g) .  The temperature and pressure of this reaction are constant (25 degrees C and 1 atm).  Say we start with a 12.2-L sample containing 0.50 mol oxygen gas.  What would be the volume of the ozone?

0.50 mol oxygen x 2 mol ozone/3 mol oxygen= 0.33 mol ozone.

Avagadro’s law can be rearranged to show V/n=a.  Since as is a constant, an alternative representation is V₁/n₁= a =V₂/n₂

So, for this reaction n₁ equals 0.50 mol, V₁ equals 12.2 L, n₂=0.33 mol, and V₂=?  By setting up the above rearrangement of avagadro’s law, 12.2/0.5=?/0.33, so the volume is 8.052 L.

Solutions

 Define the terms solute, solvent, solution and concentration (g dm^-3 and mol dm^-3).

Concentration in mol dm^-3 is often represented by square brackets around the substance under consideration, eg. [CH₃COOH].

Solute: A substance dissolved in a  liquid to form a solution.

Solvent: The dissolving medium in a solution.

Molarity: moles of solute per volume of solution in liters.  This often times written by IB with units mol dm^-3, which means moles/dm³, and a dm³ equals a liter.  Same meaning, just different ways of writing it.  I recommend writing it mol dm^-3 whenever your in IB class because it gets you well practiced, but if you are working with people outside of the IB program write it per liters, because I have found they have a hard time understanding IB notation.  This is also often times written short hand in square brackets such as [CH₃COOH].  That means the molarity of that compound.  This could also be written as g dm^-3,(molality) however I have never seen it written that way, even by IB.  If it is ever written that way, I would recommend converting it to moles (by just using Avagadro’s number), however I could see how having it in grams may help calculate it at some point.  So, if you dissolve 5 moles of HCl in 3 Liters of water, your molarity [HCl] is 5 mol/3 L, or 1.66 mol L^-1, or 1.66 mol dm^-3.

 Carry out calculations involving concentration, amount of solute and volume of solution.

These are fairly easy.  Just remember that concentration equals amount of solute/volume of solution.  So say you know the concentration of a solution is 4 mol dm-3, and you know the volume is 5 L.  That means x mol/ 5 L=5 mol/L.  That means x must equal 20 mol.

Note:  A good rule to remember is the equation m₁L₁=m₂L₂.  m stands for the molarity, and L stands for volume.  So, if you had a 2 L concentration of 5 M HCl solution (M is often used to stand for molarity) and you wanted to dilute it to a 2 M HCl, how much water would you need to add?  Well, m1=5, L1=2, m2=2, and L2=?.  So, (5 mol/L)(2 L)=(2 mol/L)(? L).  By solving for the unknown, you find your final solution will be a volume of 5 liters, and since you started with 2 liters, you need to add 3 liters to get to 5 liters.

Solve solution stoichiometry problems.

Given the quantity of one species in a chemical reaction in solution (in grams, moles or in terms of concentration), determine the quantity of another species.

When it is given to you in grams or moles, it’s just like a regular stoichiometric problem.  However, if it is given to you in terms of concentration, work your way around it to find moles or grams (so, if you have a concentration of 5 M, then you have 5 moles per liter, and if you have 3 liters, that means you have 15 moles.)