17.1 .1 Empirical / Molecular Formulas Rev.

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


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


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






oxygen gas (16.6 mol). He concluded that some


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


a 2:1 ratio every time;






He concluded that some


Law of Multiple

Carbon Dioxide, CO2

or crystalline compounds


~ 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 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:



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


dL / L ________ L = _________ dL

________ L = _________ dL

Mm / m ________ Mm = _________ m

________ Mm = _________ m

mg / g ________ g = _________ mg

________ g = _________ mg

~ 34 ~


nm / m ________ nm = _________ m

________ nm = _________ m

km / m ________ km = _________ m

________ km = _________ m

μg / g ________ g = _________ μg

________ g = _________ μg


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?

17.1 .1 Empirical / Molecular Formulas Rev.

Documents

Transcript of 17.1 .1 Empirical / Molecular Formulas Rev.