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The Law of Definite Proportions
in a specific chemical compound always combine in
exact, unchanging ratios.
CO2 C:O = 1:2
This was demonstrated by Joseph Proust’s
electrolysis of water (H
and oxygen gas (O
formula:
2H2O(l) →
17.1.1 Empirical / Molecular Formulas Rev.
~ 1 ~
The Law of Definite Proportions states that atoms
in a specific chemical compound always combine in
exact, unchanging ratios.
C:O = 1:2 NH3 N:H = 1:3
This was demonstrated by Joseph Proust’s
electrolysis of water (H2O) into hydrogen gas (H
and oxygen gas (O2) according to the following
→ 2H2(g) + O2(g)
+
Empirical / Molecular Formulas Rev.
states that atoms
in a specific chemical compound always combine in
O) into hydrogen gas (H2)
) according to the following
Empirical / Molecular Formulas Rev.
H2 and O2 were formed in
Proust conlcuded that in specific compounds,
elements exist in the same proportion (by atoms
and by mass).
John Dalton, continuing in this line, found that 100 g
of carbon (8.3 mol) would react completely with
either 133 g of oxygen gas (8.3 mol) or 266 g of
oxygen gas (16.6 mol).
elements could combine in mutiple different whole
number ratios. We call this the
Proportions.
Carbon Monoxide, CO
1:1
Note: There are some non
compounds, like Fe
which break these rules, but don’t worry about ‘em.
~ 2 ~
were formed in a 2:1 ratio every time
Proust conlcuded that in specific compounds,
elements exist in the same proportion (by atoms
John Dalton, continuing in this line, found that 100 g
of carbon (8.3 mol) would react completely with
either 133 g of oxygen gas (8.3 mol) or 266 g of
oxygen gas (16.6 mol). He concluded that some
elements could combine in mutiple different whole
number ratios. We call this the Law of Multiple
Carbon Monoxide, CO Carbon Dioxide, CO
1:2
There are some non-stoichiometric
compounds, like Fe3O4 or crystalline compounds
which break these rules, but don’t worry about ‘em.
a 2:1 ratio every time;
Proust conlcuded that in specific compounds,
elements exist in the same proportion (by atoms
John Dalton, continuing in this line, found that 100 g
of carbon (8.3 mol) would react completely with
either 133 g of oxygen gas (8.3 mol) or 266 g of
He concluded that some
elements could combine in mutiple different whole
Law of Multiple
Carbon Dioxide, CO2
or crystalline compounds
which break these rules, but don’t worry about ‘em.
~ 3 ~
Compounds with the same percent composition
have the same empirical formula – this shows a
compound reduced to its simplest ratios of atoms.
A molecular formula shows the “full” compound.
Given the percent composition and molar mass of a
compound, you can find both formulas. For ex:
An unknown gas is 75% C and 25% H with Mr = 32.
Determine the empirical/molecular formulas:
75% C → 75 g C 25% H → 25 g H % → g
12.011 g 1.008 g g → mol
= 6.42 mol C = 24.8 mol H
6.42 6.42 / lowest
= 1 mol C = 3.97 mol H
" CH4 "
(Scale up) CH4 (16 g mol-1
) x 2 = C2H8 (32 g mol-1
)
Empirical: CH4
Molecular: C2H8
~ 4 ~
Ex.1) A compound consists of 42.88% carbon and 57.12%
oxygen. What is its empirical formula?
Ex.2) Hexanol is 70.5% carbon, 13.8% hydrogen, and the
rest oxygen by mass. What is its empirical formula?
~ 5 ~
If you're given the masses, just go straight to moles:
Ex.3) A compound consists of 1.121 g N, 0.161 g H, 0.480
g C, and 0.640 g O. Give the empirical formula for
this compound:
Ex.4) A 2.500 g sample of an oxide of chlorine contains
1.315 g of chlorine. Determine the empirical
formula for this compound:
~ 6 ~
After dividing by the smallest number of moles, you
may occasionally end up with numbers ending
with/near 0.33, 0.67, 0.5, or 0.25. When this
happens, "multiply 'til whole": x2, x3, or x4:
Ex.1) Convert the following C:H:O ratios into formulas:
1 : 1.333 : 2 =
2.25 : 4 : 1 =
1 : 2.67 : 2 =
Ex.2) Convert the following C:H:O ratios into formulas:
2 : 1.5 : 1 =
1.75 : 3 : 2 =
4 : 5.25 : 2 =
17.1.2 Working with Fractional Ratios
~ 7 ~
You can use this knowledge to solve more complex
empirical formulas. Do so for each of the following:
Ex.3) A hydrocarbon is found to contain 71.98% C, 6.71%
H, and 21.3% O by mass.
Ex.4) A 10.50 g sample of phosphorus oxide contains
4.58 g of phosphorus.
~ 8 ~
Ex.5) A compound consists of 48.83% C, 8.12% H, and the
rest oxygen. What is the empirical formula?
Ex.6) A 14.09 g sample of iron oxide contains 3.89 g of
oxygen. Determine its empirical formula:
~ 9 ~
Empirical formulas only show the simplest ratios
between elements in a compound. Molecular
formulas reflect the actual "size" of a compound.
Knowing the molar mass of a compound, you can
"scale" up the empirical formula by a whole number
ratio to determine the molecular formula.
Use the empirical formulas and the molar masses
(Mr) given to find the molecular formulas:
Ex.1) EF = CH3 Mr = ~30 g mol-1
Molecular Formula:
Ex.2) EF = CH2O Mr = ~180 g mol-1
Molecular Formula:
Ex.3) EF = CH2 Mr = ~84 g mol-1
Molecular Formula:
17.1.3 "Scaling Up" to Molecular Formulas
~ 10 ~
Use the following data to determine the molecular
formula for each compound described:
Ex.4) A phosphorus oxide is found to be 56.36% oxygen
and has an approximate molar mass of 284 g mol-1
.
Ex.5) A 0.1000 g sample of a CHO compound contains
0.0643 g of carbon and 0.0285 g of oxygen and has
Mr = ~112 g mol-1
:
~ 11 ~
Lab measurements are often used to determine
empirical formulas. Use the differences in mass to
infer the mass of specific substances being added to
(or removed from) a substance.
Common Scenarios:
Substances burning in air: X + O2(g) → XO(?)
Hydrates being heated: X•H2O → X(s) + H2O(l)
Ex.1) In the first scenario, how would the mass of the
sample change? What would cause this?
Ex.2) For hydrates being heated, how would the mass of
the sample change? What would cause this?
17.1.4 Using Lab Data to Det. Emp/Molec Formulas
~ 12 ~
Ex.3) A 2.4101 g sample of magnesium is heated in a
crucible, producing a whitish powder with a mass of
4.0015 g. Give the name and empirical formula for
the compound being produced in this experiment:
~ 13 ~
Ex.4) 12.00 g of sulfur powder is heated in a crucible
under a pure oxygen atmosphere. When the
reaction is complete, the product of this reaction is
found to have a new mass of 24.08 g. Give the
empirical formula and name for the sulfur oxide
being formed:
~ 14 ~
Hydrates, or chemical formulas containing bonded
water molecules, can also be used in these
problems. In these cases, we work with the parent
molecule and water molecules as a whole, instead
of individual atoms. Refer to the end of 6.6 if you do
not recall how to do this.
Ex.5) Calcium chloride naturally exists as a hydrate with
the formula CaCl2•XH2O. If a 0.300 g sample of the
hydrate is found to have a mass of 0.226 g after
heating, determine the value of "X":
~ 15 ~
A 2.80 g sample of an oxide of bromine is converted
via precipitation into 4.698 g AgBr. If AgBr has a
molar mass of 187.78 g mol-1
, determine the
empirical formula of the oxide:
Ex.1) What mass of bromine is present in the sample?
Ex.2) What mass of oxygen is present in the sample?
Ex.3) Calculate the empirical formula for this bromine
oxide:
17.1.5 Empirical and Molecular FR #1
~ 16 ~
We can use the masses of a reaction's products to
calculate the masses of elements within the
reactants. By tracking these masses backwards
from the products, we can determine the masses
and empirical formulas for the reactants.
CxHy(g) + ? O2(g) → ? CO2(g) + ? H2O(l)
Ex.1) In the generic combustion reaction above, all of the
carbon contained by the hydrocarbon becomes part
of which substance as a product? What assumption
can we make between this reactant and product?
Ex.2) Assuming the mass of the carbon dioxide collected
from this reaction is 2.20 g, what mass of carbon
was originally contained by the hydrocarbon?
17.2.1 Deriving Masses from Rxn Prod. w/ DA
~ 17 ~
CxHy(g) + ? O2(g) → ? CO2(g) + ? H2O(l)
Ex.3) Which product will contain all of the hydrogen
present in the hydrocarbon when the reaction is
complete?
Ex.4) If 36.00 g of water is collected as a condensate from
the reaction, what mass of hydrogen was present in
the original hydrocarbon?
Ex.5) A hydrocarbon undergoes complete combustion in
lab, producing 7.8 g of carbon dioxide and 4.2 g of
water. What masses of carbon and hydrogen were
present in the hydrocarbon?
~ 18 ~
The easiest way to derive a mass from a reaction
product is to multiply the compound's mass by the
percentage of a particular element. For example:
(%C in CO2 as decimal) x (mass CO2 collected) = g C
(%H in H2O as decimal) x (mass H2O collected) = g H
Ex.1) A 1.50 g hydrocarbon sample undergoes complete
combustion to produce 4.40 g of CO2 and 2.70 g of
H2O. Determine this compound's empirical formula:
Determine the mass C:
Determine the mass H:
17.2.2 Using % Composition To Derive Masses
~ 19 ~
Some compounds will contain more than just carbon
and hydrogen. You can usually subtract the mass of
C and H from the total to determine the mass of the
other element present in the compound.
Ex.2) 1.80 g of water and 4.40 g of carbon dioxide is
collected from the combustion of a CHO compound
with a total mass of 2.31 g.
Determine the mass C:
Determine the mass H:
Determine the mass O:
~ 20 ~
Combustion analysis is a more advanced method of
determining an empirical formula. In this method,
we use the products of the reaction to estimate the
masses of each element within the substance being
burned (the fuel!) These masses are then used to
determine the empirical formula.
Suppose you were to burn a CHO compound:
CxHyOz + ? O2 → ? CO2 + ? H2O
You could collect the CO2 and H2O being produced
and convert them into masses of C and H,
respectively. These two masses could then be
subtracted from the total mass of the CHO sample
to find the mass of O. Knowing all 3, you could
solve for the empirical formula of the CHO
compound using what you've learned in 14.7.
17.2.3 Combustion Analysis
~ 21 ~
Ex.1) A 1.50 g hydrocarbon sample undergoes complete
combustion to produce 4.40 g of CO2 and 2.70 g of
H2O. Determine this compound's empirical formula:
Determine the mass C:
Determine the mass H:
Use the masses above to find the empirical formula
of this unknown hydrocarbon:
~ 22 ~
Ex.3) A 0.250 g hydrocarbon sample produces 0.845 g CO2
and 0.173 g H2O. Determine the empirical and
molecular formulae of the compound if the molar
mass is approximately 104 g mol-1
:
~ 23 ~
Some problems will involve other elements, such as
oxygen. In these instances, use the masses of the
elements you DO know (typically C and H) to
calculate the missing mass (typically O):
Ex.4) A carbohydrate is a compound composed solely of
C, H and O. When 10.7695 g of this carbohydrate
(MM = 128.2080 g/mol) was subjected to
combustion analysis with excess oxygen, it
produced 29.5747 g CO2 and 12.1068 g H2O. What is
its molecular formula?
~ 24 ~
Menthol is commonly found in peppermint, cough
drops, and cigarettes where it is used to directly
stimulate your "cold" receptors.
Ex.1) 0.1005 g of menthol (CHO compound) is burned,
forming 0.2829 g of CO2 and 0.1159 g of H2O. What
is menthol's empirical formula?
17.2.4 Combustion Analysis FR #1
~ 25 ~
More complex problems will require you to pull
together data from multiple problems. In these
instances, it is useful to remember the law of
definite proportions.
The combustion of 40.10 g of a compound which
contains only C, H, Cl and O yields 58.57 g of CO2
and 14.98 g of H2O. Another sample of this
compound with a mass of 75.00 g is found to
contain 22.06 g of Cl via precipitation. What is the
empirical formula of the compound?
Ex.1) Two experiments are being presented in this
scenario. What data cannot be compared between
the experiments and why?
Ex.2) What piece of data could be calculated and shared
between the two experiments?
17.2.5 Combustion Analysis FR #2
~ 26 ~
Ex.3) Calculate the mass % C:
Ex.4) Calculate the mass % H:
Ex.5) Calculate the mass % Cl:
Ex.6) Calculate the mass % O:
Ex.7) Determine the empirical formula of this compound:
~ 27 ~
Gravimetric analysis occurs when you create and
collect a precipitate in lab. By studying the mass of
the precipitate, we can predict the composition of
the mixture which produces it. For example:
An impure mixture of sugar and silver nitrate with
a total mass of 1.500 g is dissolved in pure DI water,
producing a solution. An excess of sodium chloride
solution is gradually added to this mixture,
producing the reaction below:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
A white solid is produced and then collected via
filtration. After repeated washing and drying, the
solid is found to have a mass of 0.913 g.
Ex.1) Identify the precipitate in the equation above and
explain how you know:
17.3.1 Gravimetric Analysis
~ 28 ~
In order to set up a DA, you need a pure substance
to act as your starting point.
Ex.2) Which mass in the problem should be used? Why?
Once you have a mass which relates to the
substances in the equation, you can use it to solve
for unknowns.
Ex.3) What mass of silver nitrate was actually present in
the original sample?
% purity = (pure mass) / (impure mass) x 100
Ex.4) What percent of the mixture was silver nitrate?
~ 29 ~
A 5.078-g vitamin tablet contains a mixture of
calcium carbonate, sugar, and other additives. The
tablet is powdered with a mortar/pestle and
thoroughly-dissolved in acid to release the Ca2+
ions.
A solution of sodium sulfate is added, producing a
white precipitate which is filtered, washed, and
dried to a final mass of 2.158-g according to the
equation below:
Ca2+
(aq) + Na2SO4 → CaSO4(s) + 2 Na+(aq)
Ex.1) The manufacturer claims that each tablet provides a
~600.0-mg of calcium. Assess this claim:
Ex.2) By mass, what percent of the tablet is calcium?
17.3.2 Gravimetric Free-Response #1
~ 30 ~
Iron ore contains a comparatively large percentage
of iron (III) oxide. A sample of ore is dissolved in
acid and reacted with an excess of sodium sulfide to
form a black precipitate, iron (II) sulfide. A 1.500 kg
sample of ore is found to produce 39.240 g of iron
(II) sulfide.
Ex.1) Give the formulas for both iron compounds being
described:
Ex.2) By mass, what percent of this ore sample is iron (III)
oxide? Hint: both substances contain iron.
17.3.3 Gravimetric Free-Response #2
~ 31 ~
The metric system and its prefixes form a sort of
"universal language" for scientists worldwide. You
will want to memorize the base units and prefixes!
I HIGHLY recommend making flashcards.
nano (n) a billionth 1e-9 or 0.000000001
micro (µ) a millionth 1e-6 or 0.000001
milli (m) a thousandth 1e-3 or 0.001
centi (c) a hundredth 1e-2 or 0.01
deci (d) a tenth 1e-1 or 0.1
Base Unit: length = meters (m)
time = seconds (s)
volume = liters (L)
mass = grams (g)
deca (da) ten 1e1 or 10
hecto (h) a hundred 1e2 or 100
kilo (k) a thousand 1e3 or 1,000
mega (M) a million 1e6 or 1,000,000
giga (G) a billion 1e9 or 1,000,000,000
17.4.1 Creating Metric Conversion Factors
~ 32 ~
The prefix always comes first, followed by the base
unit. Knowing this, interpret the following labels:
Ex. mg = milligrams, thousandths of a gram
Ex.1) kg =
Ex.2) ms =
Ex.3) mm =
Ex.4) dL =
Ex.5) nm =
We can also use our knowledge of prefixes to "build"
metric conversion factors relating our prefixes back
to their base unit. For larger prefixes:
Ex. A kg (kilogram) literally means "1000 grams", so
1 kg = 1000 g OR 0.001 kg = 1 g
~ 33 ~
Ex. cm (centimeter) means a hundredth of a meter:
1 cm = 0.01 m OR 100 cm = 1 m
Ex.6) Create conversion factors for each of the following:
mL / L ________ mL = _________ L
________ mL = _________ L
dg / g ________ g = _________ dg
________ g = _________ dg
Ex.7) Create conversion factors for each of the following:
dL / L ________ L = _________ dL
________ L = _________ dL
Mm / m ________ Mm = _________ m
________ Mm = _________ m
mg / g ________ g = _________ mg
________ g = _________ mg
~ 34 ~
Ex.8) Create conversion factors for each of the following:
nm / m ________ nm = _________ m
________ nm = _________ m
km / m ________ km = _________ m
________ km = _________ m
μg / g ________ g = _________ μg
________ g = _________ μg
Ex.9) Create conversion factors for each of the following:
Gs / s ________ s = _________ Gs
________ s = _________ Gs
HL / L ________ hL = _________ L
________ hL = _________ L
dam ________ dam = _________ m
________ dam = _________ m
~ 35 ~
Once you've mastered the ability to create metric
conversion factors, you can use them in your DAs:
Ex.1) 4.3 mg = ? g
Ex.2) 809 daL = ? L
Ex.3) 5.70 kg = ? g
Ex.4) 147 nm = ? m
17.4.2 Metric Conversion Review
~ 36 ~
To convert between units, when both of them have
prefixes, just "bounce" off the base unit and make it
a two-step conversion.
Ex.5) 1.8e9 nm → mm
Ex.6) 14 kg → mg
Ex.7) 2.09e3 dL → daL
Ex.8) 120 km → Mm
~ 37 ~
Occasionally, you'll have to make conversions
between values which are expressed with powers
greater than one, esp. with volume and area. In
these instances, you will typically need to square or
cube the conversion factor as well.
Ex.1) 120 m3 → cm
3
Think labels: m3 = (m)(m)(m). We just have to carry
out the "m → cm" conversion three times instead
of just once.
Long way:
Short way:
17.4.3 Squared2 and Cubed
3 Conversions
~ 38 ~
Ex.2) 450 dm2 → m
2
Ex.3) 1.2 dm3 → cm
3
Remember: 1 cm3 = 1 mL and 1 dm
3 = 1 L
Ex.4) 39.0 L → m3 (Hint: 1 dm
3 = 1 L)
Ex.5) 0.0014 dm3 → mL (Hint: 1 dm
3 = 1 L)
~ 39 ~
Some problems will give you labels which may
include two or more units, typically one on "top"
with a positive power and one on bottom with a
negative positive power. For example, all densities
are expressed as masses over volume:
grams per milliliter = g/mL = g mL-1
= g
mL
Just convert the "top" unit(s) first and then work on
the "bottom" units afterwards.
Ex.1) 240 mph → km s-1
( Hint: 1 mile = 1.609 km)
17.4.4 Double-Sided DA (DSDA)
~ 40 ~
Ex.2) 9.806 m/s2 → mm/min
2
Ex.3) 65 mg gal-1
→ g mL-1
(Hint: 3.78 L = 1 gal)
Ex.4) The density of steel is 8.05 g cm-3
. Express this in
terms of kg per cubic kilometer:
~ 41 ~
Truly difficult problems will incorporate elements of
metric conversions, cubic or squared units, and/or
DSDA. The rules never change, just keep canceling
labels!
Ex.1) Liquid mercury, Hg(l), has a density of 13.6 g/mL.
What is this density expressed in lbs m-3
?
Ex.2) Uranium naturally exists in the Earth's crust at an
average concentration of 4.0 g U / metric ton of
crust. What mass of uranium, in kg could be
extracted from 1500 lbs of the Earth's crust?
17.4.5 Advanced Dimensional Analysis
~ 42 ~
Ex.3) A physician fills out a prescription with an adult
dosage of 4.5 mg/kg of body mass. What is the
dose for a 220 lb adult male?
Ex.4) Mercury is a highly toxic substance often present in
small quantities in lake and river water. A sample of
water is taken from Lake Waco and found to contain
0.9 μg Hg/cm3. Lake Waco contains 9.7e7 m
3 of
water as of 2017. What is the total mass of
mercury, in kg, present in the lake?
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