34 Lab Experiment 9 Decomposition KCl O3 PDF

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Decomposition of p potassium otassium chlor chlorate ate (KClO 3)

2KClO3(s)  2KCl 2KCl(s) (s) + 3O2(g) KClO3 potassium chlorate (s) KCl

potassium chloride (s)

O2

oxygen (g)

Lets review chemical nomenclature

-ide suffix usually means that the nonmetal element is in its anionic form (it does not contain any attached oxygen) So we have chloride, bromide, ... , sulfide,..., nitride, phosphide... And there are no oxygens attached to the "-ide" element. The anion is simply the nonmetal element that has gained electrons nothing else attached

But, what is the polyatomic anion contains oxygen as well? They are called

oxyanions

one or more nonmetal elements AND also oxygen:

then the suffix depends on how much oxygen the polyatomic anion contain

two

If there are types of anions – both have the same charge but one ion has more oxygen and the other has less oxygen

ATE the oxyanion with less oxygen gets the suffix ITE

The oxyanion with more oxygen gets the suffix

Note that the ITE and ATE ions both have the same charges the same other element

they just differ in the amount of oxygens they contain

More oxygen – ATE Fewer oxygen – ITE I ATE a lot of Oxygen I only took a little bITE of oxygen Lets summarize Anions with

polyatomic anions containing oxygen

and

Oxoanions No oxygen

lower number of oxygens

ide suffix

N3- nitride ion

oxoanions greater number of oxygen

ite suffix

NO2- nitrite ion

ate suffix

NO3- nitrate ion

note nitrate and nitrite differ in amount of oxygen

SO32- sulfite ion

S2- sulfide ion

SO42- sulfate ion

these ions differ in how much oxygen atoms they have

Cl- chloride ion

ClO2- chlorite ion PO33- phosphite ion AsO33- arsenite ion

ClO3- chlorate ion PO43- phosphate ion AsO43- arsenate ion

From what we just said above -ate suffix usually means an ion where a nonmetal is attached to the largest possible number of oxygen atoms So we have chlorate [ClO3 -], sulfate [SO42- ], phosphate [PO43- ]etc.

-ite suffix refers to an ion where the nonmetal attached to oxygen contains less than the largest number of oxygen atoms. So we have sulfite SO3 2- and chlorite ClO2 -,

Seems logical but then comes along ClO- It is a polyatomic anion of chlorine charge of

1-

also with a

What to do?

Cl- chloride ion

ClO3- chlorate ion

ClO2- chlorite ion

It has the same charge (1-) as the ion with fewer oxygen - in other words, less oxygen than an ite. How do we name it?

Hypo= less under

hypodermic

hypofunction

Hypoglycemia hypothermia

Invent a new prefix, so we get HYPO chlorite. Hypo means less than. So this polyatomic anion will be called hypo chlor ite

OK, so far so good. Then , another polyatomic ion of chlorine was found that contained more oxygen atoms than was first thought possible and also with a 1- charge

ClO4 -

Hyper= more

hyperactive hypertension hypercaffeinated

so the naming system had to be patched up again to take these ions into account. So we have PERchlorate [ClO4 -]. Per means more.

Note that all of these ions have a 1- charge We started out with ClO2- chlorite and ClO 3- chlorate Then two other ions were discovered them?? Cl-

and we had to figure out how to name

chloride

ion has no oxygen = Ide suffix

ClO- hypochlorite non metal anion has even less (hypo=below) oxygen than chlorite = hypo chlorite ClO2- chlorite If a nonmetal anion containing oxygen has the smaller number of oxygen = ite ClO 3- chlorate

If the nonmetal anion containing oxygen has the

larger number of oxygen = ate ClO4- perchlorate if the nonmetal anion containing oxygen has more (per= more oxygen than even chlorate = per chlorate Lets summarize

for a nonmetal anion

ROOT+ ide  hypo. ROOT+ ite  ROOT+ ite +ROOT+ ate increasing number of oxygens  least number of oxygens



 ROOT +ate

more oxygen

 per

 most oxygen

ClO4 -

ClO3 -

ClO2 -

ClO -

all these ions have a charge of one minus Perchlorate ion Tetrahedral

chlorate ion trigonal

chlorite ion

hypochlorite ion

bent

Pyramidal Like methane

like ammonia

like water

Decomposition of p potassium otassium chlor chlorate ate (KClO3)

2KClO3(s)  2KCl( 2KCl(s) s) + 3O2(g) Balanced chemical reaction

KClO3 potassium chlorate (s) KCl

potassium chloride (s)

O2

oxygen (g)

Note 2 KClO3  3 O2

2 moles KClO3  3 moles O2

2 moles potassium chl chlor or orate ate3 moles oxy oxygen gen Every mole of O2 = 32 gr grams ams If I know how man many y gr grams ams of O2 have been formed formed,, I can calculate how man many y moles of KClO3 ha have ve been de decomposed composed I can heat KCl - no o oxygen xygen will be gener generated ated (it has no o oxygen xygen to lose) No mass loss **KCl is a tthermally hermally stable (i (i.e. .e. .e.,, heat heat-insensitiv insensitive) e) solid rre esidue **KCl does not yield O2 upon heating **KCl does not mel meltt until 770C **The flame on a Bunsen burner canno cannott melt KCl

2KClO3(s)  2KCl(s) + 3O2(g) Realize - The only source of Oxygen in this experiment is from KClO 3 There are no other O2 producing ingredients

Even though MnO2 has O2 chemically bound to it, an ordinary Bunsen Burner is not hot enough to decompose MnO2

2KClO3(s)  2KCl( 2KCl(s) s) + 3O2(g)

2 moles KClO3  3 x 32 gr grams ams O2 Every 2 moles KClO3  96 grams O2

1 mole KClO3 = 122.6 grams KClO3 One mole of KClO3 One mole

will have a mass of

K atoms +

one mole Cl atoms + 3 moles O atoms Now

One mole K atoms that has a mass of 39.098grams + One mole Cl atoms that has a mass of 35.453grams + Three moles of O atoms that has a mass of 3 X 15.999grams = 122.6g/mole

1 mole O2 = 32 grams

3 moles O2 = 96 grams

2X 122.6 grams of KClO3 that decomposes 245.2 grams KClO3

Therefore 2 moles KClO3 = 245.2 grams Will give me 96 grams of O2 245.2g

149.2g

96g

2KClO3(s)  2KCl 2KCl(s) (s) + 3O2(g) 2 x 122.6

245.2 g

2 x 74.6

=

149.2g

3 x 32

+

96g

Purpose: This lab will demonstrate the decomposition (breakdown) of KClO3 (solid) into KCl (solid) and O2 (gas). It’s a standard chemistry lab method for making oxygen.

On submarines, oxygen for breathing is normally produced through electrolysis of water. However, for emergency use electrolysis is not rapid enough

In this case, decomposition of a chlorate is used

2KClO3(s)  2KCl(s) + 3O2(g) This reaction is also used for emergency oxygen generators on spacecraft and aircraft. You are probably familiar with the oxygen masks that drop from the ceiling of an airplane during an emergency from losing air pressure. The oxygen that feeds the masks comes from metal canisters that are about 10 inches long and 3 inches in diameter. Each canister provides oxygen to 2 or 3 seats. These canisters are filled with a mixture of sodium chlorate (NaClO3), barium peroxide (BaO2), and potassium perchlorate (KClO4).

The decomposition of these 3 oxidizers is initiated by a blasting cap that is fired when you pull down on the oxygen mask.

At that point, NaClO3 decomposes to form sodium chloride (NaCl) and O2.

2KClO3(s)  2KCl(s) + 3O2(g) Chlorate (ClO3-) and perchlorate (ClO4-) can easily decompose to create pure oxygen. The reason potassium chlorate is so reactive is that its decomposition is exothermic. Once it begins, it supplies its own heat energy to keep the reaction going. Potassium chlorate (KClO3) is also used as an oxygen source in rockets and in fireworks. This concept is used by a Commercial vendor to provide emergency oxygen OC Lugo Inc. 15 Third Street | New City, NY 10956 | (845) 480-5121 Oxygen produced per 26 pound unit > 3341 Liters

A chlorate candle, or an oxygen candle, is a cylindrical chemical oxygen generator that contains a mix of sodium chlorate (not potassium chlorate) and iron powder, which when ignited smolders at about 600 °C (1,112 °F), producing sodium chloride, iron oxide, and at a fixed rate about 6.5 man-hours of oxygen per kilogram of the mixture.

The mixture has an indefinite shelf life if stored properly: candles have been stored for 20 years without decreased oxygen output. Thermal decomposition releases the oxygen gas. The burning iron supplies the heat. The candle must be wrapped in thermal insulation to maintain the reaction temperature and to protect surrounding equipment. There are various patents that teach techniques for lowering the temperature of the generated Oxygen gas so that it may be used in breathing. Why use sodium chlorate and not potassium chlorate? Sodium has a lower atomic mass (23g/mole) than potassium (39 grams/mole). On a weight basis, there is more oxygen in NaClO3 than in KClO3 Since a human being uses about 550 liters of pure oxygen per day, one candle (~26 pounds) can provide enough oxygen for a person to survive for ~ 6 days

We will use this reaction to generate Oxygen. The generated oxygen will be released into the atmosphere. In some experiments, the generated Oxygen is actually collected and the volume measured.

Sodium chlor chlorate ate – NaClO3

the facts • Sodium chlorate requires no maintenance may be stored indefinitely •It is eight times more efficient than compressed air and twice as efficient as compressed oxygen • The chemical reaction behind sodium chlorate chemistry is relatively simple; iron reacts with a small oxygen source to produce iron oxide (rust) • The heat from this reaction enables the sodium chlorate to decompose into oxygen and sodium chloride (table salt)

• Sodium chlorate is extremely stable in storage • It can be safely operated between -20°C and + 70°C • The technology has been shock, environmental and vibration tested and meets the extreme requirements of submarine usage

Lab test for O Light a wooden splint; blow flame out (leave end glowing) Insert the glowing splint into the end of the test tube The glowing splint will burst back into flames

Conservation of mass

We are playing with atomic sized lego What atoms you start with = same as with what you end up Conservation of mass

=sum of masses of reactants =sum of masses of products for ordinary chemical reactions, There is no detectable change in mass during an ordinary chemical reaction The number of individual atoms remains the same throughout the chemical reaction Since the number of atoms does not change, the mass must remain constant as well Mass (matter) is not created or destroyed –elements are merely rearranged in new combinations Immortality of matter

Example: A chemical in a sealed container is sitting on a balance. It is ignited by focusing sunlight on the sample. After the reaction, does the container weigh more, less, or the same? Answer: the Same

In an ordinary chemical reaction, mass is conserved

Think of taking wheels doors and engines and combining them to make a 4 door sedan Now I disassemble them and make a sports car I still have the same number of engines, wheels, and doors If I start with four sedans Four EW4D4 How many engines do I have available? Each sedan has one engine four sedans = 4X1=4 How many wheels do I have available? Each sedan has 4 wheels four sedans = 4 X 4= 16 wheels How many doors do I have available? Each sedan has four doors four sedans = 4X4=16 doors

We have the same number of atoms before and after an ordinary chemical reaction

Realize: total of all the Masses before the reaction and after the reaction is the same

Combination of hydrogen and oxygen molecules to form water

2H2(g) + O2(g)  2H2O(g) Started with 4 H atoms Started with 2 O atoms

ended with 4 H atoms 2 O atoms

CONSERVATION OF MASS

2KClO3(s)  2KCl(s) + 3O2(g) Can we calculate the theoretical % of oxygen in KClO 3 using mass numbers:

245.2g

149.2g

96g

2KClO3(s)  2KCl 2KCl(s) (s) + 3O2(g) 245.2 grams KClO3  96 grams O2 % oxygen in KClO3

96 grams Oxygen X 100% 245.2 grams KClO3

=

39%

The theoretical % of oxygen in KClO3 =

39% What if I had a mixture of KClO3 + KCl Sample composition 100% KClO3 100%

KCl

percent oxygen 39% 0%

Would there be any mass loss due to oxygen generation?

Realize that KCl will not lose mass upon heating!! 100% KClO3/0% KCl

100% of 39% =39%

50% KClO3/50% KCl

50% of 39% =19.5%

75% KClO3/25% KCl

75% of 39% =29.3%

25% KClO3/75% KCl

25% of 39% =9.8%

10% KClO3/90% KCl

10% of 39% =3.9%

In today’s experiment

You only need to make 3 mass measurements!!

2KClO3(s)  2KCl(s) + 3O2(g) A B C

crucible empty crucible + KClO3 mixture crucible + KClO3 mixture after heating

B-A mass of KClO3 mixture B-C mass loss due to oxygen generated C-A mass of KCl remaining Percent oxygen = B-C X 100% B-A

2KClO3(s)  2KCl(s) + 3O2(g) The theoretical % Oxygen in KClO3 = 39%

This experiment is based on the Law of Conservation of Mass Conservation of mass means that matter cannot be created or destroyed. Using the Law of Conservation of Mass we can determine the amount of oxygen released and calculate the percent of oxygen present in the original sample.

Let’s look at some sample data Data: A Mass of empty dish or test tube

40 g

B Mass of dish or test tube + KClO 3

50 g

Here is the chemical reaction

Calculations:

Mass of KClO3 I start with:

Line B – Line A

50 g - 40 g = 10 g

This is the amount of KClO 3 that will be heated

10 grams 2KClO3(s)  2KCl(s) + 3O2(g) C Mass of dish or test tube + KCl (after heating)

46.5 g

Notice it is less than the 50 grams of crucible + starting material I started with crucible + material was 50g

Mass of KCl remaining after heating: 46.5 - 40 g = 6.5 g This difference = 3.5grams is the mass of O 2 lost:

10 grams

3.5 grams

2KClO3(s)  2KCl(s) + 3O2(g)

10 grams 6.5 grams +3.5 grams 2KClO3(s)  2KCl(s) +

3O2(g)

Where did the oxygen go? It went up the fume hood and into the atmosphere

It is the difference between mass of material I started with and mass of material with which I ended up with after heating Line B minus line C

10g- 6.5g

=3.5 grams

This is how much O2 was lost due to heating 3.5 grams

10 grams 6.5 grams +3.5 grams 2KClO3(s)  2KCl(s) +

3O2(g)

Experimental % of oxygen using data: Grams O2 lost after heating divided by mass of KClO3 I started with

g of oxygen produced mass of starting material

3.5 g

X 100%

10 grams

= 35%

We determined experimentally that O2 was only 35% by weight in KClO3

However, the theoretical percent oxygen is 39% What is our error?

Error = mathematica mathematicall term

not a vvalue alue judgement

=(“Real” or “true” or “accepted” value minus your experimental value)

Error= how far away you are from the real or theoretical value

|My value – theoretical| = Error Note = error is an Absolute value

On a percent basis Error/real value x100% = percent error

I obtained 35% My error = 4

the true or theoretical value =39%

What is the percentage error? Real value = 39%

my value = 35%

(39% - 35%)/39% * 100 =

10% error

In this sample experiment, the % oxygen was slightly lower than the actual % oxygen in the formula. This means that we probably did not heat the dish long enough for all of the KClO 3 to fully decompose and that there was some undecomposed KClO3 left in the dish even after we heated it. Next time we will makes sure to heat the sample for the full amount of time at the hottest part of the flame.

Now try this set of data yourself Title: Decomposition of potassium chlorate (KClO 3)

Purpose: This lab will demonstrate the decomposition (breakdown) of KClO3 into KCl solid and O2 gas when heated. The oxygen will be released into the atmosphere and the remaining KCl can be measured to determine the change in mass. Using the Law of Conservation of Mass

2KClO3(s)  2KCl(s) + 3O2(g) 1 determine the amount of oxygen released 2 calculate the % oxygen present in the original sample 3 if the theoretical value of % oxygen is 39% what is the % error Experimental Data: A Mass of empty dish 42.56 grams B Mass of dish + KClO 3 43.66 grams C Mass of dish + KCl (after heating) 43.21 grams *************************************************** Try this yourself first Think about thisWhat is B-A? the mass of KClO3 What is B-C? What is C-A? The sum of the masses of reactant and products have to be the same CONSERVATION OF MASS

2KClO3(s)  2KCl(s) + 3O2(g) Calculations: 1. Mass of KClO3: 43.66 g -42.56 g = 1.10 g

2. Mass of KCl: 43.21 g -42.56 g = 0.69 g

3. Mass of O2: 1.10 g – 0.69 g = 0.41 g

Starting material  material

oxygen gas generated

After oxygen Has been removed

2KClO3(s)  2KCl(s) + 3O2(g) 1.10 grams

=

0.65 grams

+

0.45 grams

The total of the reactant mass =product masses CONSERVATION OF MASS 4. Experimental % of oxygen using above data: = Amount of oxygen generated/ starting mass 0.45g/1.10g X 100% =

4.9 %

% error using calculations

(39 –40.9 –40.9)/39 )/39 X 100 100% % = 4.9% error

Analysis Questions: 1. How do we use the Law of Conservation of Mass to determine the amount of oxygen? Answer: Since oxygen was part of the original compound, it had mass. The change in mass (it became less) after heating had to come from the oxygen being released. The difference in the masses of the sample before and after heating shows the amount of oxygen released. 2. What are possible sources of error in this lab? Answer; If the KClO3 was not heated long enough, not all of the oxygen would have been removed. This would give you a lower percentage for O2 If there were other compounds in the dish, such as water, the loss of mass would have been greater than if just oxygen were removed.

What if some material splattered out of the crucible during the heating? I would get a larger mass loss after heating – but this is not due to oxygen generated. I would calculate a greater %Oxygen than the true value

3. Why does diatomic oxygen O2 form when KClO3 is heated? W...