Download Summer Assignment: Some Review / Basic Prep

NAME _________________________________REVIEW OF BASIC CHEMISTRY FOR AP CHEMISTRY
DIRECTIONS: Hello! The following pages should help you review your work from your last chemistry class. Some of these pages are
informative. Some of the pages are practice problems. You should be able to do most of these problems with a little bit of effort …. A few of the
problems MAY CHALLENG YOU. Do not despair …Rather get researching …. Scan this packet over, during the next few weeks. I will NOT
COLLECT this work … You do not need to do it – assuming you have these ideas mastered. If not, do as much as you need to do. PREP
AT HOME, BY READING/ TRYING, researching other sources… There is a quiz on this material during the second week of the new school
year. We will finalize a date at the start of the year. You can contact me via my school email, [email protected], I will try
to respond in a timely fashion – but I may not be able to get back to you quickly. So, consider using the internet or a text from the library…
The mastery of this work, is your responsibility … So, work with peers, ask questions … seek answers … try …seek again …
The following pages cover review topics on
I.
II.
Matter: Compounds
A) Ionic
B) Molecular
C) Organic
D) Inorganic
E) Matter: Mixtures
Nomenclature & Formulae
A) Inorganic nomenclature
B) Writing inorganic formulae
III.
Significant Figures
A) Identification
B) Adding/Subtracting
C) Multiplying/Dividing
D) Determination of the sig figs for the answer to a calculation
IV.
Stoichiometry
***********************************************
I) Types of Compounds

Essentially there are binary compounds and ternary compounds of:
Ionic Compounds of Main Group metals
Ionic Compounds of Transition metals
Inorganic Molecular (Covalent) Compounds
and then there are the:
Organic Compounds (the terms binary and ternary are of far less importance)

Types of compounds can be further subdivided into examples of acids / bases and salts

Organic compounds exhibit acid/base activity in terms of a variety of classes of organic
compounds with functional groups; such as, but not limited to: carboxylic acids, ketones, and
amines. While these compounds ultimately become important to any student of chemistry, they
are not within the scope of this review (Phew!!!!)
A) Ionic Compounds
1) Ionic Compounds tend to be classified as: any compound exhibiting an ionic bond (as the
predominant bond responsible for the chemistry of the compound). The ionic bond is
essentially due to a strong columbic attraction between a cation and anion.
1
2) Ionic compounds (as a very general rule) can be identified as
a) a metal cation bonded to a nonmetal anion
i) Note that this is a general schema only. For instance the compounds BeF2,
AlBr3, and AlI3 are considered to exist as molecules.
3) Ionic compounds can contain metal ions of the Main Group elements and the Transition
metals.
a) Main Group metals would be those metal species found in groups 1,2, (12), 13-15
i) Main Group metals, generally have singular, stable oxidation states.
e.g.) Group 1 alkali metal cations = + 1
Group 2 alkaline earth cations = +2
Group 13 metal cations = +3 commonly (Ga, In, Tl may also be +1 or +2)
Group 14 metal cations = +4 commonly (Sn & Pb may also be +2)
Group 15 metal cations = + 3 commonly (Bi may also be +5)
b) Transition metals are those found in groups 3 – 11, (12)
i) As a rule, transition metals may have cations of multiple positive oxidation
states. This fact is the reason for including the exact whole number oxidation
state in the names of ionic compounds containing transition metal ions.
c) You’ll note that there is some sort of overlap regarding group 12 metals. According to
IUPAC group 12 metals are not transition metals. Yet, they are not really classified
as main group metals.
i) IUPAC defines the transition metals as any element with an incomplete d
sublevel &/OR any element which may form stable ions only with an
incomplete d sublevel.
This definition leaves Zn, Cd, and Hg out of the transition metal family.
Consider the electron configuration of a zinc atom: [Ar] 3d10 4s2.
The atom has a complete d sublevel, thus it is not technically transitional.
The only stable ion of zinc is +2, leaving the ion with a complete 3d sublevel,
thus it is not technically transitional, either.
Traditionally the Group 12 metals have been lumped in with the transition
metals … Yet, the chemistry of the Group 12 metals appears to exhibit a
blend of properties characteristic of both Main Group and Transition Metals.
My best advice is to treat any compound composed of Group 12 metal ions as
ionic compounds and to name them WITHOUT using a roman numeral
indicating their oxidation state. Hence, for ZnCl2 zinc chloride is the
preferred name, (not, zinc(II) chloride) You may infer then, that I am looking
at the Grp 12 as “not being transition metals” … for I feel on firmer ground,
siding with IUPAC, than with a text book author or internet source.
2
4) Generally, ionic compounds are considered to be inorganic compounds.
a) An ionic compound may contain carbon, but the carbon is not the “central atom”.
The carbon atom may express some sort of hybridization (for the polyatomic ion), but
the primary bond which creates the compound, is ionic.
B) Molecular Compounds
1) Molecular Compounds tend to be classified as compounds exhibiting a preponderance of
covalent bonding between nonmetal atoms.
2) While the physical properties of ionic compounds are due, essentially to the ionic bond, the
physical properties of the molecular compounds are due to the strength of the intermolecular
forces of attraction between the molecules.
a)
b)
c)
d)
Dipole-Dipole Attractions
Hydrogen Bonding
London Dispersion Forces
Induced Dipole Attractions
3) Molecular compounds may be inorganic compounds, or organic compounds.
a) An organic compound can be loosely defined as any compound with at least one
carbon atom as the central atom, exhibiting sp3 hybridization (or perhaps one with
can demonstrate sp3 hybridization via halogenation or hydrogenation e.g. alkenes).
But this is a very loose definition. It has proven quite difficult to provide students
a straightforward definition, due to the exceptions of compounds such as; urea,
mellitic acid, ketomalonic acid, oxalic acid, carbon tetrachloride, sodium lauryl
sulfate …etc…. So be flexible. A nice rule of thumb re: organic compounds is
that they are generally molecular compounds of carbon, with C to C or C to H bonds,
excluding oxides, carbonates, and hydrogen carbonates.
b) Organic compounds include, but are not limited to; the hydrocarbons (alkanes,
alkenes, alkynes, arenes), alcohols, ketones, carboxylic acids, aldehydes, esters,
halocarbons, ethers, amines, amides … etc…
4) The chemical and physical characteristics of molecular materials are in large part dependent
upon the
 molecular geometry (pyramidal, tetrahedral, linear, bent, trigonal planar)
 bond lengths / bond energies / bond angles
AND THE
 distribution of partial charges across the molecule
3
A Comparison of Ionic and Molecular Compound Properties
Physical Property
Phase at STP
Ionic Compound
Solid/Hard Crystals. Exist in crystal
lattice structures of formula units (not
molecules)
Melting Point
Relatively high (arbitrarily about
150°C or greater)
Very High … since the melting point
is relatively high, it makes sense that
the normal boiling point is also high.
Normal Boiling
Point
Vapor Pressure
(in a closed
system)
Very Low. Most ionic compounds do
not produce a gas (thus they tend
NOT to exert a vapor pressure)
Solubility in
Water at 20°C
Generally soluble – to some extent.
In the absolute sense some ionic
compounds do dissolve in water, even
if it is to 1 x 10-5 grams/ 100 grams of
water (or to a lesser extent). Hence,
we do use the terms poorly soluble or
nearly insoluble to describe
precipitates formed in aqueous
solution.
When an ionic compound is soluble
in water (within reason), hydrated
ions or electrolytes in water are
produced.
Conductivity of
an electrical
current when
dissolved in water
Conductivity of
an electrical
current when
fused
Conductivity of
an electrical
current as a solid
Yes: The presence of mobile or free
moving electrolytes (+ and -) allow
for the conductance of an electrical
current
Yes: The fused compound (melted or
liquefied phase) has had the ionic
bond(s) broken and thus electrolytes
have been produced.
No: There are no free moving
electrolytes. Really, only metals
conduct electricity as a solid due to
delocalized electrons
Molecular (Covalent Compound)
Variable. Samples may be solids, liquids or gases.
The concept of “molecule” is really best applied to
the species of a molecular substance in the gaseous
phase.
Relatively low (below 150°C)
Variable: Normal Boiling Point is that temperature
at which the vapor pressure (see below) equals 1
atm. This will vary from compound to molecular
compound, due to the strength (or weakness) of the
intermolecular forces.
Variable to High: Many molecular compounds
(even when solid or liquid) exert a vapor pressure.
Sublimation is often seen as a solid molecular
substance with weak intermolecular forces and
high vapor pressure. And, of course, you will
recall my statement that “Wherever there is a
liquid of a substance, there too is its gas).
Compounds with weak intermolecular forces turn
to gas more easily & thus exert a vapor pressure
than those with stronger IMF)
Variable. Only a very few molecular compounds
are considered to be absolutely insoluble in water,
even at the individualized molecular level.
As with the ionic compounds, temperature affects
solubility.
However, when dissolved in water (as opposed to
reacted with water), electrolytes in water are NOT
produced. Water does not break the covalent
bond, but exerts a stronger columbic attraction for
molecules, which are stronger than the columbic
attractions referred to as the intermolecular forces
of attraction. The one group which breaks this rule,
are the family, of acids. Molecular acids may
dissolve and dissociate into electrolytes in water.
No …As a general rule because there are no
electrolytes. The acids are exceptions, for these
molecular compounds react with the water, and
dissociate into electrolytes.
No …melting disrupts the intermolecular forces of
attraction (Melting does NOT break the covalent
bond)
No: There are no free moving electrolytes.
Really, only metals conduct electricity as a solid
due to delocalized electrons.
4
Questions:
1) Which of the following is made of only polar covalent bonds but is a nonpolar molecule?
1) HCl
2) N2
3) H2O
4) CCl4
2) Which of the following can conduct and electrical current?
1) KCl(ℓ)
2) KCl(aq)
3) KCl(s)
4) Both 1 and 2
3) Which of the following can conduct and electrical current?
1) NaOH(aq)
2) C12H22O11(aq)
3) NaOH(s)
4) C12H22O11(ℓ)
4) Which of the following does NOT exhibit polar covalent bonding?
1)
2)
3)
4)
HCl
NO2
RbCl
H2O
For questions 5 and 6 use the following table of vapor pressures of various liquids at 50°C.
Compound
Vapor
Pressure (kPa)
ethanoic acid
9
ethanol
30
propanone
81
water
12
5) Which of the compounds has the weakest intermolecular forces of attraction? ______________
6) Which of the compounds would have the highest normal boiling point? ___________________
7) Which of the following is best classified as an organic compound?
1) CH2O
2) CaCO3
3) CO2
4) NaHCO3
Answers: 1) 4 2) 4 3) 1 4) 3 5) propanone …vapor pressure increases with weaker IMF …The molecules can move from the liquid to
gaseous phase more easily with weak IMF (and thus increase the pressure due to the vapor phase) 6) ethanoic acid: The normal boiling point is that
temperature at which the vapor pressure = 1 atmmosphere. Ethanoic acid has the strongest IMF thus it will require a greater temperature to generate
sufficient vapor pressure to equal 1 atm. 7) 1 carbonates, oxides and hydrogen carbonates are NOT considered to be organic compounds.
5
C) A Review of Classifications of Matter
1) A substance is any (pure) element or (pure) compound. Thus, by definition, a substance must
Be homogeneous … but beware … not all examples of homogeneous matter are substances.
2) A compound is any substance, which is neutral in overall charge, and has a fixed ratio
(proportion) of 2 or more different component elements. The properties of the individual
elements are lost and/or changed, relative to those of the new compound.
MATTER
Nope! Then it is a
Is it uniform
throughout?
Heterogeneous
Mixture
Yep! So, it must be
Homogeneous
Nope, so it must be a
Is the composition
variable?
Substance
Nope, it must be an
Element
Does it contain
more than one
type of species
(ion or atom)?
Yep! Then it is definitely a
Homogeneous mixture
(some type of solution)
Yep! So, it must be a
a
Compound
1) White gold contains gold and palladium (in Europe …or nickel in the USA). Two samples of white
gold differ in the relative amounts of gold and the second metal. Both samples are uniform in
composition throughout. How should white gold be classified? *Homogeneous mixture
Explain your reasoning: ____________________________________________________________
______________________________________________________________________________
2) The active ingredient in an aspirin tablet is acetylsalicylic acid is composed of 60.0% carbon,
4.5% hydrogen and 35.5% oxygen by mass, regardless of its source. How should acetylsalicylic
acid be classified? * a substance and/or more specifically a compound
3) A 500 mg aspirin tablet has a maximum of 0.325 grams of acetylsalicylic acid. The remaining mass
is due to “fillers” of various sorts, depending upon the manufacturer. How might you classify the
tablet? * could be heterogeneous or homogeneous mixture
6
D) Intensive and Extensive Properties
Table of Extensive and Intensive Properties
An Extensive Property is
An Intensive Property is
dependent upon “how much” of the
substance there is Or, it is one that changes
as the size of the sample changes
specific to the substance & is INDEPENDENT
of the system's size. An intensive property scales
with changes in,or to the size of the system .
*mass
*density
*volume
pressure
length
enthalpy (energy)
entropy (disorder)
number of molecules
weight
*temperature (for a system in thermal equilibrium)
concentration (for a homogeneous solution)
malleability
*melting point
color or odor
work
any constant pegged to “per gram” or any molar
values e.g. Molar enthalpy, Molar entropy: These are
pegged to a non-variable / defined or constant mass
… such as a mole mass
Think of two identical blocks of ice. Each block is its own system.
Each block is 50.0 grams and at −10 ºC.
Add the two blocks together
system A
system B
into a single system, (C). Now, Compare C to A or B...
1) Upon combination, what has happened to the temperature? Did it increase / decrease / remain the
same?
a. Thus, temperature is an * intensive
property
2) What has happened to the number of molecules of ice-water?
a. Thus, the number of molecules is an *extensive
Did it increase / decrease / rts?
property
3) What happens to the amount of energy (q) required to melt C into a liquid, compared to A?
a. Thus, the q (as in q = mcΔT) is an *extensive
property
4) What happens to the melting point of the ice systems as you compare C to A? _______________
a. Thus, melting point is an *intensive
property
5) What has happened to the mass of C relative to A? __________________
a. Thus, mass is an *extensive
property.
6) What has happened to the volume when comparing C to A? _________________
a. Thus, volume is an *extensive
property.
7) What has happened to the density of C relative to A? _________________
a. Thus, density is an *intensive
property
7
And, that last question raises a really interesting point ….Since the volume increased (or scaled) with the
new, larger mass, the ratio for density (M/V) is constant. Hence, volume changed in proportion with mass
(scales with mass) making the density an intensive property, or independent of the system's size.
Thus, (and here’s the interesting point….) when two extensive properties are divided by each other the
result is an intensive property
e.g ... Mass (extensive) = Density (intensive) I think that is sort of COOL! Ideas intermeshing…
Volume (extensive)
Now, go back to the table of extensive and intensive properties … and fill in the terms,
Temperature, mass, volume, density, and melting point….
E) Physical properties vs. Chemical properties
1) Physical properties can be observed without changing the identity and composition of the
substance
e.gphyscial : odor, density, melting point, hardness, color, boiling point
a) Thus, a physical change is a change in appearance, but not in the composition of a
substance. This includes all of the changes of state.
2) Chemical properties describe the way a substance may change or REACT (make new bonds)
to form other substances.
e.g.chemical: flammability, oxidation/reduction … almost any activity that will result in a
change in the electron cloud, in terms of new bonds being made.
a) Thus, a chemical reaction (change), has a substance transformed into a chemically
different substance.
F) Mixtures: Mixtures are physical combinations of various substances.

A mixture is not a substance, but rather a combination of multiple substances

A mixture may be: heterogeneous (eg. a suspension or a rock, such as granite) OR it may be
homogeneous (eg. a solution [aqueous, the gaseous stratosphere, tincture, sol, gel, some alloys] )

A mixture’s composition is not fixed. Terms such as dilute or concentrated can apply to
different mixtures. Think of Kool-Aid® or of a tossed salad per their varied composition.

The components of a mixture, maintain their essential physical properties.

Mixtures can be separated into its components by taking advantage of differences in
the component properties.
8
PRACTICE 1: For questions 1-9 identify the method you could use to separate the given sample into its
components. Some of the samples are mixtures (hence you want to select physical separation methods). Some
of the samples are compounds (hence you want to select a chemical means of decomposing a compound).
Assume all equipment necessary is available to you.
You may need to combine methods to achieve a final separation (e.g. dissolving and then filtering out insoluble
solids). Your choices are:
a) Distillation, because it takes advantage of different boiling points of the liquids
in a mixture.
b) Electrolysis, because this sample, is not a mixture, but a compound. And, electrolysis
can chemically decompose a compound into simpler substances
c) Filtration, because it takes advantage of some materials being undissolved solids
and others materials, soluble in the solvent, & thus passing through the filter paper.
d) Evaporation of a solvent, because you can take advantage of removing a liquid
solvent, and letting any (non-volatile) solute crystallize out
e) Chromatography, because the dissolved materials have different solubilities in
a particular solvent, and move at different rates along an absorbent surface.
f) Dissolving, because you can take advantage of the different solubilities of
solids in a solvent
1) ____ I would use this to separate insoluble cellulose of dead algae cells from the water of a pool
2) ____ I would use this method for separating the alcohol from the water in mouthwash.
3) ____ I would use this to separate H2O(ℓ) into (di)hydrogen gas and (di)oxgyen gas.
4) ____ I would use this method to separate the various pigmented compounds that make up black ink
5) ____ and _____I would at least 2 methods to separate a mixture of water, table salt (NaCl(s) and
water-insoluble benzoic acid crystals (C6H5COOH)
6) ____ I would use this to separate melted table salt (NaCl(ℓ)) into sodium metal and chlorine gas.
7) ____ I would use this method to separate the solvent of water from the dissolved compound, in KBr(aq)
8) ____ I would use this to separate water from any gasoline, that has contaminated a sample of water
9) ____ and ____ I would use these methods to separate a sand, water and salt mixture.
Answers: 1) c 2) a
3 ) b 4) e 5) c and then d
6) b
7) d
8) a
9) c and then d
10) State all the issues about a mixture that you can recall: ________________________________________
_________________________________________________________________________________________
_________________________________________________________________________________________
9
PRACTICE 2 Vocabulary Development
Terms that apply to Substances
Compounds
Binary Compound
Ternary Compound
Inorganic Compound
Organic Compound
Ionic Compound
Molecular Compound
Homogeneous
Substance
Elements
Metallic Element
Nonmetallic Element
Molecular Element
Homogeneous
Substance
Terms that apply to Mixtures
Aqueous Solution
Acid / Acidic Solution
Base /or Alkaline Solution
Homogeneous
Heterogeneous
Mixture
Record each of the terms which apply to the following. The number of correct terms is given in ( ). I gave you part of the answer for
two of the problems
1) CH4(g) (5) _________________________________________________________________________
2) NaIO3 (5) __________________________________________________________________________
3) O2(g) (4) ____________________________________________________________________________
4) NaCl(aq) (3) _________________________________________________________________________
5) HBr(aq) (4) Acidic solution (3 more) ______________________________________________________
6) NaOH(s) (5) _________________________________________________________________________
7) NaOH(aq) (4) Basic solution (3 more) ____________________________________________________
8) CH3Cl(l) (5) ________________________________________________________________________
9) Mg(NO3)2(s) (5) ______________________________________________________________________
10) Mg(NO3)2(aq) (4) ____________________________________________________________________
11 Mg(s) (3) ________________________________________________________________________
12) salt and pepper (2) _________________________________________________________________
13) T/F When a sample of matter is classified as homogeneous, it must be a substance.
14) T/F When a sample of matter is classified as a mixture, it can still be considered to be a substance.
Selected Answers: 1) Binary Compound, Organic Compound, Molecular Compound Homogeneous, Substance
2) Ternary Compound, Inorganic Compound, Ionic Compound, Homogeneous, Substance
3) Nonmetallic Element, Molecular Element, Homogeneous, Substance
4) Aqueous Solution, Homogeneous, Mixture
5) Aqueous Solution, Acidic, Homogeneous, Mixture
7) Aqueous Solution, Alkaline, Homogeneous, Mixture
11) Metal, Homogeneous, Substance
12) Heterogeneous Mixture
13) F 14) F
10
PRACTICE 3
___1) Which of the following is NOT a physical property?
1) liquid nitrogen (N2(l)) boils at -196°C
2) sugar dissolves in water
3) gold melts at 1064°C
4) gasoline burns in air
5) a compound of copper is blue
In order to determine the answer, what is the critical question you wish to ask / or what is the critical
piece of information you wish to know?
_____________________________________________________________________
___2) A piece of polypropylene rope (used by water skiers) floats on water, whereas a terephthalate polymer
from a soda bottle sinks in water. Using the polymers and water, order the three materials, from least
dense to most dense.
____________________________________________________________________________
least dense
most dense
What knowledge/information do you possess that allows you to answer the question, successfully?
________________________________________________________________________________
3) According to Mark Winter, at http://www.webelements.com/mercury/
Mercury is the only common metal liquid at ordinary temperatures. Mercury is sometimes
called quicksilver. It rarely occurs free in nature and is found mainly in cinnabar ore (HgS)
in Spain and Italy. It is a heavy, silvery-white liquid metal. It is a rather poor conductor of
thermal energy as compared with other metals but is a fair conductor of electricity. It alloys
easily with many metals, such as gold, silver, and tin. These alloys are called amalgams.
Which of the following is an extensive property of a sample of mercury?
1)
2)
3)
4)
shiny surface
melts at 234.22 K
has a density of 13.6 g/cm3
has a volume of 27.1 cm3
In order to determine the answer, what is the critical question you wish to ask / or what is the critical
piece of information you wish to know?
_____________________________________________________________________
11
For questions 4- 6 use the following passage, and your grasp of chemistry.
Turquoise is a semi-precious gemstone (and one of the birthstones for December!) A particular piece of
turquoise is a 2.5 grams blue-green solid, with a density of 2.65 g/cm3.
4) Calculate the volume of the piece of turquoise. Record the volume, with an appropriate unit. Show your
work. Watch your sig figs.
_______________
5) Identify the information for the piece of turquoise which are qualitative and that which is quantitative.
qualitative: __________________________________________________________
quanitative: __________________________________________________________
6) Identify the information for the piece of turquoise which is extensive and that which is intensive.
extensive: ___________________________________________________________
intensive: ____________________________________________________________
******************************
7) Identify the following as either physical changes or chemical reactions
a) ____________________ the desalination of sea water into pure water and salt
b) ____________________ the production of SO2(g) from the combustion of coal
c) ___________________ the tarnishing of silver metal, with the formation of AgS(s)
d) ___________________ dry ice (CO2(s)) subliming to CO2(g)
e) ___________________ the decrease in the density of a sample of mercury as it is heated from
-100K to 288 K
f) ____________________ KNO3(s) dissolving in water
In order to determine the answer, what is the critical question you wish to ask / or what is the critical
piece of information you wish to know?
_____________________________________________________________________
12
Answers:
1) 4 The critical question I want to ask is which of these situations, is a chemical reaction or which verb suggests that there the situation results in
new bonds (new compounds) being made?
2) polypropylene rope, water, terephthalate polymer … the floating and/or sinking in water as an indication of relative density compared with that of
water.
3) 4
I want to ask which of the choices represents a dimension or calculation which changes as the amount of matter changes...because that would
be an extensive property.
4) 0.94 cm3
5) semi-precious gemstone (maybe) … blue-green color, solid
mass of 2.5 grams, density of 2.65 g/cm3
6) mass of 2.5 grams, (you could include, as well, the volume calculated in question 4)
density of 2.65 g/cm3 , blue-green color
7) a) physical change b) chemical reaction c) chemical reaction
change in the bonding? / Are new bonds being made ?
d) physical change e) physical change
f) physical change / Is there a
II) A) Nomenclature of Ionic Compounds: Nomenclature of the ionic compounds is a relatively well defined
procedure, but you should not be surprised, as second year
students to find that there are a few issues.
I have worked to interpret the AP’s approach as well as IUPAC’s approach. They are in concert with each
other for the most part.
There may be some “hiccups” with the Main Group metals of gallium, indium, thallium, tin, lead and
bismuth. Just remain flexible. For instance, understand that the naming of GaCl3 as gallium(III) chloride
or gallium trichloride will not cost you your test grade. The nomenclature process for these main group
elements is a bit up in the air, as of now.
Use the algorithm on the next page.
Recall that when assigning oxidation states (generally) the more electro-positive species is written first.
Two exceptions which come to mind are ammonia (NH3) and phosphine (PH3), of the pnictogen family
The sum of all oxidation states must add up to be equal to 0 for a compound. The sum of all oxidation
states, for a polyatomic ion must add up to the overall charge of the polyatomic ion.
The next page has the inorganic nomenclature algorithm. The page after that has a table of common
polyatomic ions, and some help as to their names. Please note that no such table exists on the AP reference
tables. While “memory” is not the focus of the AP course, I urge you, I urge you, I urge you to know the
most common polyatomic ions by name, formula and charge. I will expect you to know them for tests. I
have marked the ones you need to know with a & next to its name on the table of selected polyatomic ions.
13
Binary Molecular Compounds
1) use prefixes to indicate the molar ratio of the
subscripts:
1 = mono (used generally for when the
second nonmetal, is an oxide and
[rarely] a halide )
2 = bi, 3 = tri, 4 = tetra, 5 = penta, 6 = hexa
7 = hepta, 8 = octa, 9 = nona, 10 = deca
Binary Ionic Compounds
1) Process:
Name the first species (a metal cation)
When the metal species is a transition metal, Pb, or
Bi, include the oxidation state as a roman numeral.
This applies to all transition metals cations.
prefix to represent the subscript
Name the second element, using an
appropriate prefix AND change the ending
of the second element to –IDE
Exemplars:
N2O5 = dinitrogen pentoxide (note the a of
penta was dropped… however, some
sources, such as the IUPAC Red book
allow the “a” naming the compound as
dinitrogen pentaoxide)
http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf
ICl = iodine monochloride or just
iodine chloride (either is acceptable)
2
Name the second species but change the ending to
–IDE
Exemplars:
3
NaCl = sodium chloride
FeO = iron(II) oxide
note there is NO Space between the cation and ( )
AgBr = silver(I) bromide
ZnCl2 = zinc chloride
When the metal species is a transition metal, Ga, In,
Tl, Sn, Pb, or Bi, include the oxidation state as a
roman numeral. Technically, the roman numeral is
written in ( ) with no space between the name of the
metal cation and the ( ). This applies to all
transition metal cations and those mentioned
above
Compounds of most Main Group metal ions are
considered to have only a single stable charge and
a roman numeral is not generally be used, unless it
is a compound of gallium indium thallium, tin,
lead or bismuth.
3
CO = carbon monoxide (note that the monowas used for the oxygen only. Note that
the –o- of mono was dropped (thus the elision
was deleted, but some sources allow its use.
Carbon monooxide can be used (but not often)
But, diiodine is more common (eg in Europe)
Reference: IUPAC Red Book p. 69
Name the first species (a metal cation or polyatomic
ion such as ammonium or dimercruy (I))
Compounds of Main Group metal ions are
considered to have only a single stable charge and
a roman numeral is not to be used, unless it is
a compound of gallium indium thallium, tin,
lead or bismuth.
2
SiF6 = silicon hexafluoride (note the a of
hexa was kept
1
I = 1, II =2, III= 3, IV = 4, V = 5, VI= 6, VII = 7
2) Process: 1Name the first element using an appropriate
CO2 = carbon dioxide
1) Process:
1
2
Ternary Ionic Compounds
Name the second species:
When the second species is a polyatomic ion,
keep the name in its entirety. Do not change
the ending of the polyatomic ion.
When the second species is a nonmetal monatomic
anion be sure to change the ending to –IDE
Exemplars:
Na2CO3 = sodium carbonate
CrO3 = chromium(VI) oxide
NH4NO3 = ammonium nitrate
PbO = lead(II) oxide
NH4Cl =
ammonium chloride
Fe(NO3)3 = iron(III) nitrate
FeS2O3 = iron(II) thiosulfate
Ca3(PO4)2 = calcium phosphate
H2S = dihydrogen mono sulfide or more
commonly hydrogen sulfide (either is
acceptable)
14
Here is a table of Polyatomic Ions. Note that I have taken some time to explain the source of the prefixes of the
oxyanions. And, I have &noted, which I want to have memorized. No such table is on your AP reference
charts.
Name
&
Ammonium
Table of Selected Polyatomic Ions
Formula
Name
Sulfite
NH4+
Formula
SO32-
(not ammonia)
Mercury (I)
&
Acetate
Cyanide
&
Carbonate
&
Hydrogen Carbonate
Oxalate
Thiocyanate
Nitrite
&
Nitrate
&
Hydroxide
Peroxide
&
Phosphate
Hg22+
C2H3O2CH3COOCNCO32HCO3C2O42SCNNO2NO3OHO22PO43-
&
Sulfate
Thiosulfate
SO42S2O32-
Hydrogen Sulfate
Hypochlorite
Chlorite
&
Chlorate
&
Perchlorate
Chromate
Dichromate
Permanganate
Arsenate
Iodate
HSO4ClOClO2ClO3ClO4CrO42Cr2O72MnO4AsO43IO3-
•For any ion, when no numeric charge is given, as in + or -, the charge value to be
assumed is “1”
•Many (not all) of the polyatomic ions can be classified as oxyanions [Negative
polyatomic ions containing oxygen].
•When an oxyanion series (or an oxoanion series) exists, such as sulfate vs sulfite, you
can determine the formulae & names because of a specific pattern for the oxyanions.
Note, that some fail to follow the general pattern (Why? Because this is chemistry!!!),
But most do (so don’t lose hope)
First, the oxyanions of a particular series have the same charge (Yes!). This
certainly helps with memorization. So this means that every polyatomic ion in the
chlorate set has an overall charge of -1. Each oxyanion in the sulfate set has an
overall charge of -2 (Phew!!!!!) …
Secondly, treat the oxyanion species ending in –ATE, as the most common or
“standard” version.
A suffix of –ITE means there is one fewer oxygen than the “standard”
Any ion with a name of HYPO__ITE has two fewer oxygen than the “standard”
Any ion named as PER _____ATE has one more oxygen, than that “standard”
A second little pattern for names, is the use of the prefix THIO-. Thio- indicates the
presence of an extra sulfur …although the transfer is not as precise as with the
oxyanions …as seen in cyanide vs. thiocyanate
&You
really want to know these 10 polyatomic ions at the very least, cold.
15
II B) Writing an Inorganic Compound Formula
1) Take a walk over to cyberspace at: http://www.kentchemistry.com/links/naming/formulawriting.htm
or (http://tinyurl.com/formula-writing )…Watch, review, practice if you need to … but only if you
need to do so. You can then scroll to the bottom for practice…
a) more practice [if needed]
i) Quite nice: http://honorsph.startlogic.com/honorsphysicalscience/quizzes/Naming_30_Compounds.htm
(Hint: Be sure the ( ) of a roman numeral is connected to the name of the metal ion)
ii) Quite Clever: http://www.sciencegeek.net/Chemistry/taters/ions/page2.htm
(Hint: Be sure to drag the species to the top line of those provided you…then check)
iii) This is “Okay:…
http://chemistry.about.com/library/weekly/blcompnamequiz.htm
Questions:
1) The formula for terbium phosphate is TbPO4. The formula for terbium sulfate is
1)
2)
3)
4)
Tb2SO4
TbSO4
Tb2(SO4)3
Tb(SO4)2
2) What is the charge on the carbonate polyatomic ion?
3) The formula for cadmium perchlorate is Cd(ClO4)2. The formula for cadmium acetate is:
1)
2)
3)
4)
Cd2C2H3O2
Cd2 C2H3O2
Cd2(C2H3O2)3
Cd(C2H3O2)2
4) The formula for indium sulfate is In2(SO4)3. What is the formula for indium carbonate?
1)
2)
3)
4)
InCO3
In2(CO3)3
In3(CO3)2
In2CO3
5) Iodate and iodite are oxyanions of the same series. Make a statement regarding the charge of each ion.
6) Iodate and iodite are oxyanions of the same series. Iodate has 3 oxygen atoms. How many oxygen atoms
must iodite have?
Answers: 1) 3 2) -2 or 2- 3) 4
4) 2 5) both are -1 or 1- 6) 2
16
III) Significant Figures There are only 2 types of measurements … those with a decimal point and those
without a decimal point. Hmm… I hope that sounds quite familiar.
A) A significant figure is a digit in a measurement that adds to the measurement’s precision. This
includes all nonzero numbers, zeroes between significant digits, and zeroes indicated to be
significant. Leading and trailing zeroes are not significant because they exist only to show the scale
(the magnitude) of the number
The overriding rule is: Significance begins at the first 1 to 9 integer, & includes that integer and
every number after it, even zeros, assuming there is a decimal point. When there is no
decimal point, do NOT include tailing or final zeroes, as significant, they are only included for
magnitude. (To avoid confusion, I recommend writing such measurements using scientific notation
….Yeah, you know I like that scientific notation idea…)
1) When dealing with values in scientific notation, only significant figures are included in what
is called the significand or mantissa of the notation.
2.000 x 1015 cm has 4 significant figures
3.8900 x 101 atoms has 5 significant figures.
When needing to write the measurement 409,022 meters with 4 significant figures, using
scientific notation, it should be written as: 4.090 x 105 m
B) There are 3 big rules when calculating with measurements and sig figs.
1) When multiplying or dividing measurements, the answer must have the same number of sig
figs as the term with the fewer sig. figs.
2) When adding or subtracting measurements, the answer must have the same number of
decimal places as the term with the fewer number of decimal places.
3) There’s one new rule (for my students) rule re: logs. The number of sig figs for a log
calculation is determined only by the # of sig figs in the decimal portion of the original
mantissa, the integer of the answer does not matter. Thus, the log of 0.0100, should be
written as -2.000 The original mantissa has 3 sig figs, thus the log must have 3 decimal
places (for we ignore that integer before the decimal point…. More on that later in the year.
C) Additionally, when determining the number of sig figs for a final answer:

Ignore “exact” or absolute numbers when determining the number of significant figures in a
calculation’s answer. Exact numbers are defined numbers or the result from a count.
For instance, 1 mole of O2 (g) equals 31.9988 grams. That 1 mole is an exact value and should
not be used to determine the number of sig figs in an answer.

When the problem is pretty straight forward (eg. completely dimensional analysis) then
adjust for sig figs at the END, using the Given value for the limit of sig figs. It is probably the
only true measurement in the problem. All other conversion factors would have a greater number
of sig figs, or be exact numbers. There may be one or two exceptions – but you’ll be fine…
17

When given a mixed operations problem (e.g. one which has a multiplication component and
an addition component), adjust for sig figs at the end of each operation, separately.

When given a multi-step problem requiring “like” operations, consider keeping a few
guard digits before adjusting for the sig figs of a final answer.
For instance: Find the radius of an aluminum cylinder, with a density of 2.71 g/cm3, a length of
2.00 cm and a mass of 15.4 grams.
You know from all of the values given, that your answer must have 3 sig figs … however, I shall
show you that I will carry some guard digits along, to “guard against” issues arising from
problems due to rounding. Every operation in this problem is a multiplication or division, so
they are “like” operations, in that these operations follow the same rules for sig figs.
step 1The
solution involves the use of D= M/V
2.71 g/cm3 = 15.4 g
V
V = 5.6826 cm3
Notice that I am carrying a few guard digits as I move into the next “like” operation. I have not
limited the volume to just 3 sig figs, immediately. You need only one or two guard digits. I let
the need for rounding be my guide.
step 2Now
we use Volume = π r2h or 5.6826 cm3 = (3.14159)(r)2(2.00 cm)
5.6826 cm3 = r2
6.28318 cm
r = 0.904414 or r = 0.904 cm
(the answer is limited to 3 sig figs, per the original values)
The guard digits just ensure that the most accurate answer is calculated, without having to carry
around an inordinate (no pun intended) number of digits.

Be sure you are watching your units … remember whatever you can do to a number,
arithmetically, you can do to a unit.
18
IV) Dimensional Analysis and Stoichiometry
Desired = Givenunit’|
|
chant the song … “desired = given, goalposts, drop the units……”
unit’
Yes, we will continue to use DA. I really no longer know of a college that does not use it at some level.
Work at recognizing that DA is a series of ratio proportions linked together, in a rational relationship.
So, Yes, there are times when equations are better … but for mole issues … golly(!) DA has is all over
pure ratio proportion.
1) The density of mercury is 13.6 g/cm3. What is the mass in kilograms of a 2.00 L commercial flask of
the metal?
kg = 13.6 g | 1,000 cm3 | 2.00 L| 1 kg | 27.2 kg
cm3
1L
1
1,000g
notice the desired unit is the only one left in the numerator
2) Convert a speed of 35.80 miles/hour to its equivalent in m/s.
m = 35.80 miles | 1.61 km | 1,000 m | 1 min | 1 hr
| = 16.01 meters/second
s
hour 1 mile
1 km
60 s
60 min
********************************************************************
Answers are on my website … Highlight wherever there is an * and turn the font to black!
3) Calculate the number of hydrogen atoms contained in 25.0 mL of H2 gas at STP.
(recall: 1 mol = 6.02 x 1023 molecules, 1 molgas @STP = 22.4 L)
* atoms = 25.0 mL| 1 L
|1 mol | 6.02 x 1023 molecules | 2 atoms | = 1.34 x1021atoms
1,000 mL 22.4 L
1 mol
1 molecule
Ans: 1.34 x1021atoms
4) What volume, in liters, of hydrogen gas at STP would contain 4.5 x 1018 hydrogen molecules?
* Liters = 4.5 x 1018 H2 molecules| 22.4 L
| = 1.7 x 10-4 L
6.02 x1023 molecules
Ans: 1.7 x 10-4 L
5) A sample of air contains 2.33 x 10-4 mg of lead per mL of gas. This air passes through an office, the
volume of which is 3.25 x 104 L. Seven people normally work in this office. How many μg of lead
will each person in the office receive from this sample of air? (1 mg = 1,000μg)
* μg = 2.33 x 10-4 mg| 1,000 μg | 1,000 mL| 3.25 x 104 L | 1
| = 1.08 x 106 μg Pb/person
person
mL 1 mg
1L
1
7 persons
Ans: 1.08 x 106 μg Pb/person
19
6) A molecule of hydrogen moves at a speed of 115 cm/s. How many seconds will it take to travel
the length of a 100 yard football field?
* seconds = 100 yds | 3 ft | 12 inches | 2.54 cm | 1 second| = 80 seconds (only 1 sig fig)
1 yd 1 ft
1 inch
115 cm
Ans: 80 seconds (only 1 sig fig)
7) A doctor orders that a patient receive 1.50 x 10-3 mole of sodium chloride. The only solution
available contains 1.00 g per 100. mL of solution. Calculate the volume in mL of solution which
should be given to the patient. 1 mol of sodium chloride is 58.44277 g/mol
*mL = 100. mL | 58.44277 g | 1.5 x 10-3 mol | = 8.77 mL
1.00 grams
1 mol
Ans: 8.77 mL
8) What is the density of mercury (13.6 g/cm3) in units of kg/m3?
Note: 1,000 cm3 = 1 dm3 and 1,000 dm3 = 1m3
*kg = 13.6 g | 1 kg
m3
cm3 1,000 grams
| 1,000 cm3 | 1,000 dm3 | = 13,600 kg/m3
1 dm3
1 m3
Ans: 13,600 kg/m3
9) How many atoms of hydrogen can be found in 45 g of ammonia, NH3?
atoms H =* 45 grams | 1 mole | 3 atoms
17 grams
1 molecule
| 6.02 x 1023 molecules |
1 mol
Ans: 4.8 x 1024 atoms of H
20
10) An unknown mass of calcium nitrate, Ca(NO3)2, has 5.00 x 1027 atoms of oxygen. How many
kilograms of Ca(NO3)2 are in the sample? (recall, an ionic compound is referred to as a formula unit)
*kg = 5.00 x 1027 atoms O | 1 f.u. | 1mole
| 164 grams | 1 kg
| = 227 kg
6 atoms O 6.02 x 1023 f.u. 1 mole
1,000 grams
Ans: 227 g Ca(NO3)
11) When you are going 50.0 miles per hour, how many feet per second are you traveling?
(1 mile = 5,280 feet)
*ft = 50.0 miles | 1 hr
s
hr
60 minutes
| 1 minutes
60 s
| 5,280 feet | = 73.3 ft/s
1 mile
Ans: 73.3 ft/s
12) Calculate the number of mL of bleach required to make 1.0 quart of 5.0 % bleach solution?
(1 liter = 1.06 quarts)
mL =* 1.0 quart | 1 Liter
1.06 qt
| 5 mL
100 mL
| 1,000 mL|
= 47 mL
1L
Ans: 47 mL
13) A gas sample containing 1.32 grams of helium at a pressure of 900. torr is cooled to 288 Kelvin.
What is the volume, in Liters, occupied by the gas, at these conditions?
Review: PV = nRT where R = 0.08206 L∙atm ∙ mol-1 ∙ K-1 and 760 torr = 1 atm
Ans: 6.61 L
21
14) The complete combustion of ethanol (C2H5OH) is represented by the following reaction.
C2H5OH(l) + 3 O2(l)  2 CO2(g) + 3 H2O(g)+ 1,236 kJ
a) Calculate the maximum number of grams of carbon dioxide (mol mass = 44.0095g) produced when
77.0 grams of ethanol (mol mass = 46.0684g) and 19.5 grams of O2 (mol mass = 31.9988 g) are
allowed to react.
*grams CO2 = 19.5 gO2 | 1mole O2| 2 mol CO2 | 44.0095 grams CO2| = 17.9 grams
31.9988 g 3 mol O2
1 mole CO2
*grams CO2 = 77.0 g eth | 1mole eth
| 2 mol CO2 | 44.0095 grams CO2| = 147 grams
46.0684 g eth 1 mol eth
1 mole CO2
Ans: 17.9g CO2
b) How many grams of the excess reagent should be left in the reaction vessel, once the limiting reagent
is consumed and the reaction has ended?
*g eth used = 19.5 g O2 | 1 mol O2
| 1mol eth | 46.0684 g eth | = 9.36 g eth used
31.9988 grams 3 mol O2 1 mol eth
*Since the vessel contained 77.0 grams of ethanol, but ONLY 9.34 grams were consumed then:
77.0 - 9.34 = 67.6 grams are left over unreacted...
Ans: 67.6 g ethanol
15) Potassium perchlorate, KClO4 (mole mass = 138.550g) may be prepared by reacting KOH with Cl2 and
the following series of chemical reactions. What mass, in grams, of KClO4 (GFM = 138 g) can be prepared
using 65.0 g of KOH (mol mass =56.1056 ) in an excess of Cl2 and all other reactants?
2 KOH + Cl2  KCl + KClO + H2O
3 KClO  2 KCl + KClO3
4 KClO3  4 KClO4 + KCl
*g KClO4 = 65.0 g KOH | 1 mol KOH | 1 mol KClO | 1 mol KClO3 | 4 mol KClO4 | 138.550 g KClO4
56.1056 grams 2 mol KOH 3 mol KClO
4 mol KClO3 1 mol KClO4
Ans: 26.8 grams KClO4
22
Combustion Analysis: An application of Stoichiometry
The products of the complete combustion of a hydrocarbon are carbon dioxide and water. Using the system of
combustion analysis, the products (CO2 and H2O) are trapped separately in compounds that can absorb carbon
dioxide or water. These absorbent compounds are then re-massed to determine a change in mass, due to the
absorbed compounds.
e.g. CaCl2,
or P4O10 or
Mg(ClO4)2
e.g. NaOH
on asbestos
http://preparatorychemistry.com/Bishop_Combustion_Analysis.htm
Before attacking the problems, you should know that you can assume:
1) all of the C of the hydrocarbon ends up in the CO2
2) all of the H of the hydrocarbon ends up in the H2O
For instance, let’s just start with interpreting what the above means …
A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO2 and 2.70 g of
H2O. What is the empirical formula of this compound?
CxHy + O2 → CO2 + H2O
So we can assume that ALL of the C in the unknown compound CxHy ends up in CO2. All of the hydrogen
in the compound CxHy ends up in the H2O. Using these assumptions, we have the chance to figure out
the number of moles of carbon and hydrogen that are in the unknown sample.
In your unit cancellation process, it is important to recognize that in 1 mole of CO2 molecules, there is 1 mole of
C atoms (there is a 1:1 ratio) …and in 1 mole of H2O molecules there are 2 moles of H atom (or a 1:2 ratio)
The data and the ratios can be used to determine the empirical formula and then ultimately the molecular
formula of the original compound.
There are a couple of ways you can go about this analysis – Each has the goal to determine the moles of C
and moles of H … I have given you, in this packet, the simpler of the two. Anyone wishing to know
another way, using percent composition and stoichiometry, is welcome to ask. I will happily show you
that method.
23
Process Using Combustion Analysis to Find a Formula
•Use the masses of CO2 and H2O produced by the combustion of the organic compound, in oxygen.
•Use stoichiometry to convert from grams of CO2 to moles of Carbon (the examples will help)
•Use stoichiometry to convert from grams of H2O to moles of H.
•Find the empirical formula by dividing each mole value by the smaller value (mole ratio …as you did
when determining an empirical formula …See the Empirical formula independent packet if you need
to do so … or check out: http://crescentok.com/staff/jaskew/ISR/chemistry/class10.htm or check out
https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/empirical-molecularformula/v/formula-mass-composition )
•If required, multiply by a factor until a whole number ratio is achieved.
Guided
Practice:
•When
asked for the molecular formula, divide the mass of the empirical formula (EFM) by the given
mole mass to find the “multiple”. Multiply the subscripts of the empirical formula by that multiple
1) A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO2 and 2.70 g of
H2O. The molecular mass of the compound equals 30 μ. Find the empirical formula AND molecular
formula for this compound.
Finding the empirical formula
Here is where we use the 1:1 of 1 C per 1 CO2. What is the ratio
of H per mol of H2O molecules?
Find moles of C in CO2 = 4.40 g CO2 | 1 mol CO2 | 1 mol of C atoms
| = 0.100 mol C
44 g
1 mol of CO2 molecules
Find moles of H in H2O = 2.70 g H2O* | 1 mol H2O | 2 mol of H atoms
| = 0.300 mol H
18 g
1 mol of H2O molecules
Find the mole ratio:
for C: *0.100 mol C = 1 C
0.100
For H: * 0.300 mol H = 3
0.100
The empirical formula is: *CH3, with an empirical formula mass of 15μ
Finding the molecular formula
The multiple = * 30 μ = 2 …. Thus the molecular formula = C2H6
15μ
Ans: CH3 and C2H6
24
2) A 0.250 g sample of hydrocarbon undergoes complete combustion to produce 0.845 g of CO2 and 0.173 g
of H2O. The molecular mass of the compound is 78 μ. Determine both the empirical formula
and the molecular formula of this compound.
Finding the empirical formula
Find moles of C in CO2 = 0.845 g CO2 | 1 mol CO2 | 1 mol of C atoms
| = 0.0192 mol C
44 g
1 mol of CO2 molecules
Notice, with the exception of the given mass of CO2 …the math is just like the first example.
Find moles of H in H2O = 0.173 g H2O* | 1 mol H2O | 2 mol of H atoms
| = 0.300 mol H
18 g
1 mol of H2O molecules
Find the mole ratio:
for C: *0.100 mol C = 1 C
0.100
For H: * 0.300 mol H = 0.0192 mol H
0.100
The empirical formula is: *CH, with an empirical formula mass of 13μ
Finding the molecular formula
The multiple = * 78 μ = 6 …. Thus the molecular formula = C6H6 …Hey! The compound is benzene!
13μ
Ans: CH and C6H6
3) A touch more complicated … but it is all about finding the number of moles and then that simplest ratio…
A hydrocarbon burns completely, producing 7.2 grams of water and 7.2 Liters of CO2 at STP. What is
the empirical formula of the hydrocarbon?
Finding the empirical formula
Find the moles of C … You are given Liters (so, you could use PV=nRT) or stoichiometry, since the
conditions are STP, you can use the 1 mol = 22.4 L conversion factor.
Using PV=nRT
nCO2 = (1 atm)(7.2 L) = 0.32 mole CO2 which means…0.32 mol of C
(0.08206)(273)
HEY! It’s the same number
no matter how I do it!
Using stoich:
moles C = 7.2 Liters CO2| 1mol CO2 | 1 mol C | = 0.32 mol C
22.4 L
1mo CO2
Find the moles of hydrogen = *7.2 grams H2O| 1 mol H2O | 2 mol H
| = 0.80 mol H
18 grams 1 mol H2O molecules
25
Find the mole ratio: For C: * 0.32 = 1
0.32
For H =* 0.80 = 2.5
0.32
The empirical formula must be * C2H5
Question: Which of the following could be a molecular formula of the compound?
1) C4H10
2) C2H5
3) C6H10
4) all of these
ans: *1
Practice: A few of these problems may push you around … but just recall it is all about finding moles … and
then the ratio…. Answers are at the end or you may go online and highlight wherever there is an * and turn the
font to a color like black or red….
1) When 1.00 g of a compound containing only carbon and hydrogen is burned completely, 3.14 g of CO2
and 1.29 g of H2O is produced. What is the empirical formula?
*find moles of C = 31.4 g of CO2| 1mole CO2| 1 mol C atom
| = 0.0714 mol C atoms
44 grams 1 mol CO2 molecule
*find moles of H = 1.29 g of H2O| 1 mole H2O | 2 mol H atoms
| = 0.143 mol H atoms
18 grams 1 mole molecules
*ratio = 1:1.9 or 1:2 ratio ….
1) C2H
2) CH3
3) CH4
4) CH2
2) Assume a sample of an unknown hydrocarbon is completely burned in the presence of oxygen, producing
3.45 grams of carbon dioxide and 1.65 grams of water. Determine the empirical formula.
1) C2H5
2) CH
3) C3H7
4) CH3
26
3) If the molecular mass of the compound in question 2 equaled 86 grams, what would be the molecular
formula?
ans: *C6H14
4) Complete combustion of a 0.523 g of unknown hydrocarbon produces 1.61 g of carbon dioxide and
0.743 g of water. In a separate experiment, it is determined that the molecular mass of the compound
is 114 μ. What is the molecular formula?
ans: *C8H18
5) Complete combustion of a 0.0150 mol sample of a hydrocarbon, CxHy, gives 2.352 L of CO2 at STP and
1.891 grams of H2O. The molecular mass of the compound is 84 μ. Determine both the empirical formula
and the molecular formula
*ans: The formulae are: CH2 and C6H12
*nCO2 = PV/RT
…. n = (1)(2.352)/ (0.08206)(273) = 0.1050 mol of CO2 …thus C
* mol H = 1.891 mol H2O| 1 mol water | 2 mol H
| = 0.2101 mol H
18 grams 1mol molecules
* mole ratio = 1:2
27
6) The combustion analysis of an unknown hydrocarbon was completed by using an apparatus, and appropriate
chemicals as described on the second page of this packet. The following data were collected. Using the
reading and the data table, determine the empirical formula of the compound.
Mass of Drying Agent Mg(ClO4)2 Before Combustion: 25.00 grams
Mass of Drying Agent Mg(ClO4)2 After Combustion:
27.70 grams
Mass of Water Due to Combustion
Mass of NaOH Before Combustion
25.00 grams
Mass of NaOH After Combustion
29.40 grams
Mass of Carbon Dioxide Due to Combustion
1) CH3
2) CH2
3) C3H4
4) CH
Answers: 1) 4 2) 3 3) C6H14 4) C8H18 5) CH2 and C6H12 6) 1
28