The formula of a chemical compound states how many atoms of each element are in a fundamental unit of the compound. The fundamental unit is called a formula.

The formula of a chemical compound states how many atoms of each element are in a fundamental unit of the compound. The fundamental unit is called a formula unit.

Examples of formula units that you have seen before are NaCl, CaBr2, and Al2O3.

Each formula unit of a compound can be assigned a relative formula mass by adding up the relative masses of all of the atoms in the formula.

The formula mass of H2O would be: (1.008 + 1.008 + 16.00) amu.

For molecular compounds, the formula mass is identical to the quantity called the molecular mass.

A compound’s formula mass may only be calculated if its formula is known.

A quantity of a substance equal to its formula mass in grams will contain the same number of formula units (molecules for a molecular substance) as the formula mass in grams of any other substance.

The following masses all contain the same number of formula units:

22.99 g Na, 35.45 g Cl, 58.44 g NaCl (22.99 + 35.45)

12.01 g C, 1.008 g H, 16.00 g O, 180.2 g glucose (C6H12O6)

(6*12.01 + 12*1.008 + 6*16.00)

The number of atoms contained in the atomic mass (in grams) of an element and the number of formula units contained in the formula mass (in grams) of a compound is called the mole, abbreviated mol.

The mass in grams of one mole of a compound is called its molar mass.

The units of molar mass are grams; the units of atomic mass and formula mass are amu.

The definition of molar mass can be expressed as two conversion factors that can be used to convert grams of a substance to moles and vice-versa.

(8.00 g oxygen atoms) * (1 mole oxygen atoms / 16.00 g oxygen atoms)

= (0.500 mole oxygen atoms)

(0.500 mol nitrogen atoms) * (14.01 g nitrogen atoms / 1 mole nitrogen atoms)

= (7.00 g nitrogen atoms)

An experimental determination of the number of formula units contained in one mole has been performed. This number is 6.022 x 1023 and is called Avogadro’s number.

Avogadro’s number is currently defined as the number of carbon atoms in exactly 12 grams of the pure isotope of carbon, carbon-12.

One mole, or one Avogadro’s number, of dimes would cover the United States to a height of about 10 miles.

One mole of carbon atoms would occupy about a teaspoon of volume.

From the definition of Avogadro’s number we can calculate the mass in grams of 1 amu:

(6.022 x 1023 atoms 12C) x (12 amu / atom 12C) = 12 g

6.022 x 1023 amu = 1 g

1 amu = 1.661 x 10-24 g

From this definition, the masses of an individual atom of an element can be calculated:

The formula for an unknown compound is determined experimentally by a quantitative analysis that determines the identity and percent composition of each element in a sample of the compound.

Using the percent composition data, the simplest formula using whole number subscripts for each of the elements is calculated. This formula is known as the empirical formula.

A sample of calcium chloride is determined to be 36.04% calcium and 63.96% chlorine.

The ratio of calcium to chlorine in the sample (made up of identical formula units) must be the same as that in a single formula unit.

In 100.0 g sample there are 36.04 g calcium and 63.96 g chlorine.

(36.04 g Ca) x (1 mol Ca / 40.08 g Ca) = 0.8992 mol Ca

(63.96 g Cl) x (1 mol Cl /35.45 g Cl) = 1.804 mol Cl

The numbers 0.8992 and 1.804 would have been different had we started with a different mass of sample, but would have still been in the ratio: 0.8992:1.804.

Had we started with (100.0/0.8992) g sample, the two numbers would have been:

(0.8992 / 0.8992) = 1.000 mol Ca

(1.804 / 0.8992) = 2.006 mol Cl

The formula of calcium chloride is CaCl2.

The percent composition of compounds of carbon, hydrogen, and oxygen or of carbon, hydrogen and nitrogen are determined by the analysis of the combustion products of a weighed sample of the compound.

The mass of carbon in the sample is the mass of carbon in the carbon dioxide absorbed.

The mass of hydrogen in the sample is the mass of hydrogen in the water absorbed.

The mass of oxygen or nitrogen in the sample is obtained indirectly by subtracting the masses of carbon and hydrogen from the mass of the sample.

The empirical formula is the simplest formula having the correct ratio of atom types in the substance.

The empirical formula generally corresponds to the actual formula for a salt or other non-molecular substance.

For molecular substances, however, the molecular formula may be an integral multiple of the empirical formula.

The empirical formula of glucose, C6H12O6, is CH2O. The actual molecular formula is exactly 6 times the empirical formula.

The empirical formula of hydrogen peroxide, H2O2, is HO. The actual molecular formula is exactly 2 times the empirical formula.

In determining the actual molecular formula of an unknown molecular compound, the empirical formula is first determined from the percent composition data.

Using an independent method, an approximate value for the molecular weight of the compound is determined.

Comparing the formula mass of the empirical formula to the actual molecular weight allows the molecular formula to be determined.

Compound A: 40.00% C, 6.716% H, 53.29% O.

Approximate molar mass = 180 amu.

Empirical Formula: CH2O

Empirical Formula Mass: 30 amu

(180 amu) / (30 amu) = 6

Molecular formula = C6H12O6 (CH2O) x 6

Alternatively, a table of potential formulas and molar masses can be constructed and compared to the experimentally determined molar mass:

We would select C6H12O6: 180~180.

This equation correctly represents the reactants used and the products formed in a chemical reaction, but is not consistent with the Law of Conservation of Atoms. This reaction is said to be unbalanced.

To balance this reaction, coefficients must be placed in front of some or all of the substances listed so that each atom type occurs the same number of times in the reactants as in the products.

Subscripts within a substance cannot be changed to balance a reaction.

The reaction above can be balanced by first balancing chlorine atoms:

And then balancing sodium atoms:

Balancing: Al + O2 Al2O3

Al + 3 O2 2 Al2O3 Balance oxygen

4 Al + 3 O2 2 Al2O3 Balance aluminum

Balancing: AlCl3 + Na3PO4 AlPO4 + NaCl

AlCl3 + Na3PO4 AlPO4 + 3 NaCl Balance sodium

Balancing: Al(OH)3 + H2SO4 Al2(SO4)3 + H2O

2 Al(OH)3 + H2SO4 Al2(SO4)3 + H2O Balance aluminum

2 Al(OH)3 + 3 H2SO4 Al2(SO4)3 + H2O Balance SO42-

2 Al(OH)3 + 3 H2SO4 Al2(SO4)3 + 3 H2O Balance H and O

Oxidation-reduction reactions occur when there are changes in the valence shells of the participating reactants.

One example already considered was an oxidation reaction involving the addition of oxygen to form a new compound. The new compounds were called oxides.

The oxygen content of an oxide can also be reduced by heating it with hydrogen. In this case, the hydrogen removes the oxygen in the oxide as a molecule of water. This is an example of a reduction.

An oxidation of a compound results from either the addition of oxygen atoms to the compound or the removal of hydrogen atoms from the compound.

2 C2H4O + O2 2 C2H4O2 Oxidation of C2H4O

C2H6 C2H4 + H2 Oxidation of C2H6

The reverse of either reaction would represent a reduction.

Both of these reactions are a subset of a more general definition of oxidation-reduction reactions:

An oxidation is the loss of electrons.

A reduction is the gain of electrons.

Neither an oxidation nor a reduction occurs by itself. An oxidation of one compound is always matched by the reduction of another compound and vice-versa.

Oxidation-reduction reactions can always be separated into two separate half-reactions: an oxidation reaction and a reduction reaction.

2 Na + Cl2 2 NaCl

2 Na 2 Na+ + 2 e- Oxidation

Cl2 + 2 e- 2 Cl- Reduction

Each half-reaction must be balanced both with respect to the atom types on each side of the chemical equation and with respect to the total charge on each side of the chemical equation.

When added together, the electrons on either side of the overall summed reaction must cancel.

Loss and Gain of Electrons

When an oxidation-reduction reaction takes place, a quantity called the oxidation state of one atom increases and the oxidation state of a second atom decreases.

The following rules define the oxidation state of an atom. Each rule takes priority to all rules under it.

Oxidation States

1. Free elements are assigned an oxidation number of zero. The modern usage is to call an oxidation number an oxidation state; so free elements have an oxidation state of zero.

2. The sum of oxidation states of all the atoms in a species (compound or polyatomic ion) must be equal to the net charge of the species.

3. In compounds, the group I metals, such as Na and K, are assigned an oxidation state of one. Note that the charge of an ion is written as one or two. The oxidation states are written one, two, and so on, to differentiate them from ionic charge.

4. In its compounds, fluorine is always assigned an oxidation state of one, because it is the most electronegative element.

5. The group II metals, such as Ca and Mg, are always assigned an oxidation state of two, and group III ions, such as Al and Ga, are assigned an oxidation state of three. These states are identical with the ionic charges of these elements in compounds.

6. Hydrogen in compounds is assigned an oxidation state of one.

7. Oxygen in compounds is assigned an oxidation state of two.

In covalent compounds such as NO2 or CH4 the combined atoms have no real charge, however they are assigned oxidation states that imply an electrical charge.

Determine the oxidation number of carbon in the compound CH3OH.

Hydrogen is assigned an oxidation state of +1.

Oxygen is assigned an oxidation state of -2.

The charge on the molecule is 0.

0 = C + 4*(+1) + 1*(-2)

C = -2

Determine the oxidation number of carbon in the compound HCOOH.

Hydrogen is assigned an oxidation state of +1.

Oxygen is assigned an oxidation state of -2.

The charge on the molecule is 0.

0 = C + 2*(+1) + 2*(-2)

C = +2

Oxidation-reduction reactions can be balanced by the ion-electron method.

Balancing Redox Reactions

1. Separate the oxidation and reduction half-reactions.

2. Make certain that both half-reaction equations are balanced with respect to all elements other than oxygen and hydrogen.

3. Balance the oxygen atoms by adding water.

4. Balance the equations with respect to hydrogen by using protons.

5. Balance the equations with respect to charge.

6. Obtain the complete and balanced equation by adding the two half-reactions and canceling any terms that appear on both sides.

Balance the oxidation of ethanol, C2H5OH, to acetaldehyde, C2H4O, by the dichromate ion, Cr2O7

2- which is reduced to Cr3+.

Cr2O72- 2 Cr3+

Cr2O72- 2 Cr3+ + 7 H2O

14 H+ + Cr2O72- 2 Cr3+ + 7 H2O

6 e- + 14 H+ + Cr2O72- 2 Cr3+ + 7 H2O

C2H5OH C2H4O

C2H5OH C2H4O + 2H+

C2H5OH C2H4O + 2H+ + 2 e-

3 C2H5OH 3 C2H4O + 6H+ + 6 e-

14 H+ + Cr2O72+ + 3 C2H5OH 2 Cr3+ + 7 H2O + 3 C2H4O + 6 H+

8 H+ + Cr2O72+ + 3 C2H5OH 2 Cr3+ + 7 H2O + 3 C2H4O

In each reaction the metal is oxidized and the oxygen is reduced.

Most metals are obtained in elemental form from their oxides by reduction by carbon:

Direct Reaction with Oxygen

Metals react with oxygen to form oxides:

In this reaction, iron is reduced and carbon is oxidized. The oxidation state of oxygen does not change.

The direct reaction of a sample containing carbon, hydrogen, sulfur, and nitrogen with oxygen produces CO2, H2O, SO2, and NO2 as products.

Balance the oxidation of glucose, C6H12O6, with O2.

C6H12O6 + O2 CO2 + H2O

C6H12O6 + O2 6 CO2 + H2O

C6H12O6 + O2 6 CO2 + 6 H2O

C6H12O6 + 6 O2 6 CO2 + 6 H2O

The dehydrogenation (oxidation) of malic acid to oxaloacetic acid occurs in cells while extracting energy from foodstuffs.

Dehydrogenation

The oxidant is nicotinamide adenine dinucleotide, NAD+, which removes a hydride ion, H-, from the malic acid. An additional H+ is also released to the solvent.

The removal of H- and H+ is the equivalent of removing a molecule of H2.

Stoichiometry is the quantitative mass relation of chemical reactions. These relations are always expressed through a balanced chemical equation describing the chemical reaction in question.

The stoichiometric coefficients in a balanced chemical equation provide a set of molar relationships between the amounts of individual reactants and products that are consumed and produced.

The stoichiometric coefficients in this equation tell us that when aluminum hydroxide and sulfuric acid react, the following relationships or conversion factors will be true:

Given the moles of one reactant that disappears, the appropriate conversion factors can be used to calculate the moles of the other reactant that disappears or the moles of either product that appear.

How many moles of Al2(SO4)3 can be produced from 2 moles of H2SO4?

(2 moles H2SO4) x (1 mole Al2(SO4)3 / 3 moles H2SO4)

2/3 mole Al2(SO4)3

Most chemical problems are not expressed in moles of reactants and products, but in grams.

In this case, an addition conversion factor, the molar mass, must be used to convert grams into moles and vice-versa.

How many moles of Al2(SO4)3 can be produced from 196.2 g H2SO4?

(196.2 g H2SO4) x (1 mole H2SO4 / 98.1 g H2SO4) x (1 mole Al2(SO4)3 / 3 moles H2SO4)

2/3 moles H2SO4

The only difference from the previous problem is the first multiplication: converting g H2SO4 to moles H2SO4. The stoichiometric coefficients from the balanced chemical equation are then used to relate moles Al2(SO4)3 to moles H2SO4.

Had the question asked for g of Al2(SO4)3, a final step would be required to convert moles Al2(SO4)3 to g Al2(SO4)3.

… X (342.2 g Al2(SO4)3 / 1 mole Al2(SO4)3)

Consider the question:

Nitrogen gas (N2) and hydrogen gas (H2) can react to form ammonia (NH3). How many grams hydrogen gas are required to react with 14 grams of nitrogen gas.

The steps required to solve this problem are:

•First, balance the equation describing the reaction.

•Using the formula mass as a unit-conversion factor, convert nitrogen’s mass in grams into its equivalent number of moles.

•Using the balancing coefficients as unit-conversion factors, calculate number of moles of hydrogen required to react with the calculated number of moles of nitrogen.

•Using the formula mass as a unit-conversion factor, convert the calculated number of moles of hydrogen into its mass in grams.

Chapter 4 Summary

Chemical Formulas and Formula Masses

• A chemical compound’s formula tells us how many atoms of each element are in a formula unit.

• The sum of the relative masses of the atoms of the formula unit is called the formula mass.

• A compound’s formula mass can be calculated only if its formula is known.

Chapter 4 Summary

The Mole

• The atomic mass (in grams) of an element or the formula mass (in grams) of a compound is called the mole.

• A mole of any element or compound contains the same number of atoms or formula units as a mole of any other element or compound.

• The mole is the factor for converting mass into number of formula units and number of formula units to mass.

Chapter 4 Summary

Avogadro’s Number

• The number of fundamental units in a mole has been determined by experiment to be 6.022 x 1023 and is called Avogadro’s number.

Chapter 4 Summary

Empirical and Molecular Formulas

• The formulas for compounds are determined by quantitative analysis.

• If the numbers of moles obtained by quantitative analysis are fractional, they are transformed into whole numbers by dividing each fractional number by the smallest fractional value. This transformation produces the empirical formula, which gives the number of moles of each atom in the compound.

• The molecular formula gives the actual number of atoms of each element in a molecule.

Chapter 4 Summary

Balancing Chemical Equations

• Chemical reactions are written as equations in which an arrow separates components that react, called reactants, from components resulting from the reaction, called products.

• When an equation has equal numbers of each kind of atom on both sides of the equation, it is called a balanced chemical equation.

• To balance a chemical equation, the correct formulas of reactants and products must be known.

• A chemical equation can be balanced only by inserting whole numbers, called balancing coefficients, in front of the formulas of both reactants and products.

• Each coefficient in a balanced chemical equation multiplies the entire formula unit that it precedes and, therefore, the numbers of all the atoms in it.

Chapter 4 Summary

Oxidation–Reduction Reactions

• Chemical reactions in which electrons are lost by one of the reactants and gained by another are called oxidation–reduction or redox reactions.

• The reactions are written as two separate processes called half-reactions in which electrons are lost in one of the half-reactions and gained in the other.

• In a redox reaction, the number of electrons lost by a compound (the reductant) must be equal to the number of electrons gained by the other (the oxidant).

• To decide when a redox reaction has taken place, we assign a number called its oxidation number to an atom that defines its oxidation state.

• If the oxidation number becomes more positive, an oxidation has taken place. If the oxidation number becomes more negative, a reduction has taken place.

Chapter 4 Summary

Oxidation–Reduction Reactions – cont.

• Rules are used to assign an oxidation number to atoms whether in ionic or in covalent compounds.

• An increase in a compound’s oxygen content, or a decrease in its hydrogen content, or both, also is considered to be an oxidation. Conversely, a decrease in a compound’s oxygen content, or an increase in its hydrogen content, or both, is considered to be a reduction.

Chapter 4 Summary

Stoichiometry

• A balanced chemical equation expresses the quantitative relations between moles of components in the reaction.

• With the use of these relations and the formula masses of the reaction components, moles can be converted into mass and mass into moles.

• This conversion allows the determination of the number of moles of product given the number of moles of reactant, as well as the mass of product in grams given the mass of reactant in grams.