Chemical Formulas and Chemical
Compounds
Chapter 7
Chemical Formulas
• Combinations of symbols are used to
represent compounds of two or more
elements.
• Also indicate the ratio of the number of
atoms of each type of element in the
compound.
• H2O – means that there are 2 hydrogen atoms
for every oxygen atom.
• No subscript on Chemistry
O – means
there is 1
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Chemical Formulas
• Show either one molecule or one formula
unit
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Organic Compounds
• Written differently than other formulas
• The shorthand shows how the atoms are
joined, not just the number present.
• Example –
• CH3COOH, not C2H4O2
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Ions
• Ion – charged atom or group of atoms
• Monatomic Ions – single atom
• Polyatomic Ions – more than one atom
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Monatomic Ions
• Can be anions or cations
• Transition elements can form more than one
kind of ion
• See table 7-1 on page 205
• You must memorize this table.
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Naming monatomic ions
• Cations
• Element’s name
• Roman numerals are used when there are
multiple ions
• Anions
• Drop the element name ending
• Add -ide
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Binary compounds
• Contain two different elements
• When we write chemical formula for a
compound, the charges must add up to zero.
• Write the positive ion first.
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Example
• Write a formula for a compound of tin (II)
and Iodine.
• Tin (II) is 2+
• Iodine is 1• We need two iodines to cancel out the
charge on the tin (II).
• SnI2
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Nomenclature
• Naming system
• Works for most compounds
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Naming binary compounds
• Write the name of the positive cation first.
• Add the name of the negative anion
• AlN – Aluminum nitride
• KCl – potassium chloride
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The stock system
• Elements with more than one possible
charge
• Cu2S – copper (I) sulfide
• CuS – copper (II) sulfide
• Note – in an older naming system the above
could be written as cuprous sulfide and
cupric sulfide
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Oxyanions
• Polyatomic ions that contain oxygen
• When there are two or more oxyanions
formed from the same two elements, the
most common has the ending –ate
• The ion with one less oxygen than –ate ends in
–ite
• The ion with one less oxygen than –ite adds the
prefix hypo• The ion with one more oxygen than –ate adds
the prefix perChemistry chapter 7
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Compounds with polyatomic ions
• See table 7-2 on page 210
• They are written like binary compounds.
• Except the ending isn’t changed to end in –ide
• CuSO4 – copper (II) sulfate
• Sn(SO4)2 – tin (IV) sulfate
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Discuss
• Practice problems 7-1, 7-2, and 7-3 on
pages 207, 209, and 211
• Practice
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Polyatomic ions you must memorize
•
•
•
•
•
•
•
•
Ammonium
Acetate
Chlorate
Chlorite
Hydroxide
Hypochlorite
Nitrate
Nitrite
•
•
•
•
•
•
•
Perchlorate
Permanganate
Carbonate
Peroxide
Sulfate
Sulfite
Phosphate
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Naming binary molecular
compounds
• Two systems – one will be covered in
section 7-2
• Older system
•
•
•
•
•
Prefixes used – see table 7-3 on page 212
CO – carbon monoxide
CO2 – carbon dioxide
SO2 – sulfur dioxide
SO3 sulfur trioxide
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Rules
• List the less-electronegative element first.
• Only has a prefix if there is more than one.
• The second element
• Has a prefix
• Root of the element name
• -ide ending
• If the word begins with a vowel, drop the o or a at
the end of the prefix (monoxide, not monooxide)
• Order: C, P, N, H, S, I, Br, Cl, O, F
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Examples
• PF5
• Phosphorus pentafluoride
• N2O5
• Dinitrogen pentoxide
• OF2
• Oxygen difluoride
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Acids
• Have a different naming rules.
• Some common ones are listed in table 7-5
on page 214
• You should know
• Hydrochloric acid (HCl)
• Sulfuric acid (H2SO4)
• Acetic acid (CH3COOH) (vinegar)
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Salts
• An ionic compound composed of a cation
and the anion from an acid
• Sometimes the salt keeps one or more
hydrogen atoms from the acid
• The prefix bi- or the word hydrogen is added to
the anion name
• HCO3• Hydrogen carbonate ion or bicarbonate ion
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Discuss
• Sample problem 7-4 on page 213
• Practice
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Discuss
• www.dhmo.org/facts.html
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Oxidation numbers
• Also called oxidation states
• Assigned to atoms in molecules
• Indicate the general distribution of electrons
among the bonded atoms
• Sort of like ionic charge
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Pure elements
• Have oxidation numbers of zero
• Single atoms – Na
• Molecules of a pure substance
• O2
• P4
• S8
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Like charges on ions
• Shared electrons are assumed to belong to
the more-electronegative atom
• The more electronegative element gets a
number equal to the negative charge it
would have as an anion.
• The less electronegative element gets a
number equal to the positive charge it
would have as a cation.
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Fluorine
• Oxidation number of -1
• The most electronegative element
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Oxygen
• Usually -2
• In peroxides, -1
• H2O2
• In compounds with halogens, +2
• OF2
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Hydrogen
• +1 with more electronegative elements
• -1 with metals
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Sum of oxidation numbers
• In a neutral compound must be zero
• In a polyatomic ion must equal the charge
on the ion
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Ion
• Can be assigned an oxidation number equal
to the charge on the ion
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Example
• Assign oxidation numbers to each atom in
the following compound:
• KClO4
• O is -2, which gives -8, since there are 4.
• The charge on perchlorate is 1-, so Cl must be
+7
• K must be +1 to cancel out the 1-
• +1, +7, -2
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Example
• Assign oxidation numbers to each atom in
the following compound:
• SO32• O is -2, which gives -6, since there are 3.
• The charge on sulfite is 2-, so S must be +4
• +4, -2
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You try
• Assign oxidation numbers to each atom in
the following compound:
• CO2
• O is -2, which gives -4, since there are 2.
• The charge is 0, so C must be +4
• +4, -2
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You try
• Assign oxidation numbers to each atom in
the following compound:
• NO3• O is -2, which gives -6, since there are 3.
• The charge is 1-, so N must be +5
• +5, -2
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More oxidation numbers
•
•
•
•
•
•
•
•
•
See Appendix Table A-15
There is also a pattern on the periodic table
Group 1 is usually +1
Group 2 is usually +2
Group 13 is usually +3
Group 14 is usually +2 or +4
Group 15 is usually -3
Group 16 is usually -2
Group 17 is usually -1
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The stock system
• Can be used instead of prefixes for molecular
compounds
• Use the oxidation number
• SO2
• Sulfur dioxide
• Sulfur (IV) oxide
• SO3
• Sulfur trioxide
• Sulfur (VI) oxide
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Discuss
• Name each of the following binary
molecular compounds according to the
stock system
• CI4
• SO3
• As2S3
• NCl3
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Formula mass
• The sum of the average atomic masses of all
the atoms in a formula
• For ionic compounds or molecules
• Can also be called molecular mass for
molecules
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Example
• Find the formula mass of Na2SO3
• 126.05 amu
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Example
• Find the formula mass of HClO3
• 84.46 amu
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You try
• Find the formula mass of MnO4• 118.94 amu
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You try
• Find the formula mass of C2H6O
• 46.08 amu
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Molar Mass
• Chapter 3
• The mass in grams of one mole (6.022 x
1023 particles) of a substance
• Example: H2O
• The mass of two moles of hydrogen atoms and
one mole of oxygen atoms
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Example
• Find the molar mass of K2SO4
• 174.27 g/mol
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You try
• Find the molar mass of (NH4)2CrO4
• 152.10 g/mol
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Formula mass and molar mass
• Numerically equal
• Only the units are different
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Discuss
• How many moles of atoms of each element
are there in one mole of ammonium
carbonate, (NH4)2CO3
• 2 mol N, 8 mol H, 1 mol C, 3 mol O
• Determine both the formula mass and the
molar mass of ammonium carbonate
• 96.11 amu, 96.11 g/mol
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Converting with molar mass
• Relate mass in grams to number of moles
• Relate mass in grams to number of particles
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Example
• What is the mass in grams of 3.04 mol of
ammonia vapor, NH3?
• 51.8 g
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You try
• What is the mass in grams of 0.257 mol of
calcium nitrate, Ca(NO3)2?
• 42.2 g
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Example
• How many moles of SO2 are in 3.82 g?
• 0.0596 mol
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You try
• How many moles of Cl2 are there in 77.1 g?
• 1.09 mol
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Example
• How many molecules are there in 77.1 g
Cl2?
• 6.55 x 1023 molecules
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You try
• How many molecules are in 4.15 x 10-3 g of
C6H12O6?
• 1.39 x 1019 molecules
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Percentage composition
• Percentage by mass of each element in a
compound
• Example: gum
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Example
• Find the percentage composition of sodium
nitrate NaNO3.
• 27.05% Na
• 16.48% N
• 56.47% O
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You try
• Find the percentage composition of silver
sulfate, Ag2SO4.
• 69.19% Ag
• 10.29% S
• 20.53% O
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Discuss
• Zinc chloride, ZnCl2 is 52.02% chlorine by
mass. What mass of chlorine is contained
in 80.3 g of ZnCl2?
• How many moles of Cl is this?
• 41.8 g
• 1.18 mol
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Empirical formula
• The symbols for the elements combined in a
compound
• Subscripts show the smallest whole-number
mole ratio of the atoms
• Determined from the percent composition
of a substance
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Empirical formula
• Usually the same as an ionic compound’s
formula unit
• Not always the same as the molecular
formula
• Diborane’s molecular formula is B2H6
• The empirical formula is BH3
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Example
• A compound is analyzed and found to
contain 36.70% potassium, 33.27%
chlorine, and 30.03% oxygen. What is the
empirical formula of the compound?
• KClO2
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You try
• Determine the empirical formula of the
compound that contains 17.15% carbon,
1.44% hydrogen, and 81.41% fluorine.
• CHF3
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Example
• A 60.0 g sample of tetraethylead, a gasoline
additive, is found to contain 38.43 g lead,
17.83 g carbon, and 3.74 g hydrogen. Find
its empirical formula
• PbC8H20
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You try
• A 170.00 g sample of an unidentified
compound contains 29.84 g sodium, 67.49 g
chromium, and 72.67 g oxygen. What is its
empirical formula?
• Na2Cr2O7
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Discuss
• Find the empirical formula of a compound
that contains 53.70% iron and 46.30%
sulfur.
• Fe2S3
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Molecular formula
• Show how many atoms are in each
molecule
• Related to empirical formula
xempiricalformula  molecularformula
• x is the whole number the subscripts must
be multiplied by
• It might be 1
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Mass relationship
xempiricalformulamass  molecularformulamass
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Example
• The empirical formula for
trichloroisocyanuric acid is OCNCl. The
molar mass of this compound is 232.41
g/mol. What is its molecular formula?
• O3C3N3Cl3
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Example
• Determine the molecular formula of a
compound with an empirical formula of
NH2 and a formula mass of 32.06 amu.
• N2H4
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You try
• Determine the molecular formula of the
compound with an empirical formula of CH
and a formula mass of 78.110 amu.
• C6H6
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Example
• If 4.04 g of N combine with 11.46 g of O to
produce a compound with a formula mass
of 108.0 amu, what is the molecular formula
of this compound?
• N2O5
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You try
• The molar mass of a compound is 92 g/mol.
Analysis of a sample of the compound
indicates that it contains 0.606 g N and
1.390 g O. Find its molecular formula.
• N2O4
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