THE ATMOSPHERE

THE ATMOSPHERE

1. The atmosphere is made up of air. Air is a mixture of colourless, odourless gases which is felt as wind (air in motion).

All living things breathe in air for respiration. Plants use air for respiration and photosynthesis.

2. The main gases present in the atmosphere/air
3. The following experiments below shows the presence and composition of the gases in air/atmosphere

 

(a) To find the composition of air supporting combustion using a candle stick

Procedure

Measure the length of and empty gas jar M1.

Place a candle stick on a petri dish.

Float it on water in basin/trough.

Cover it with the gas jar. Mark the level of the water in the gas jar M₂.

Remove the gas jar.

Light the candle sick. Carefully cover it with the gas jar.

Observe for two minutes. Mark the new level of the water M₃.

 

Sample observations

Candle continues to burn then extinguished/goes off

Level of water in the gas jar rises after igniting the candle

Length of empty gas jar = M₁= 14cm

Length of gas jar without water before igniting candle = M₂= 10 cm

Length of gas jar with water before igniting candle = M₁ -M₂=14- 10 = 4cm

Length of gas jar with water after igniting candle = M₃= 8cm

Length of gas jar without water after igniting candle = M₁ – M₃= 10 -8 = 2 cm

 

Explanation

Candle burns in air. In a closed system (vessel), the candle continues to burn using the part of air that support burning/combustion. This is called the active part of air.

The candle goes off/extinguished when all the active part of air is used up.

The level of the water rises to occupy the space /volume occupied by the used active part of air.

The experiment is better when very dilute sodium/potassium hydroxide is used instead of water .

Dilute Potassium/ sodium hydroxide absorb Carbon(IV)oxide gas that come out from burning/combustion of candle stick.

From the experiment above the % composition of the:

 

(i) Active part of air can be calculated:

M₂– M₃x 100% => 10- 8x 100% = 20%

M₂ 10cm

(ii) Inactive part of air can be calculated:

100% -20% = 80% // M3=> 8x 100%= 80%

M₂10cm

 

(b)To find the composition of active part of air using heated copper turnings.

Procedure

Clamp a completely packed/filled open ended glass tube with copper turnings.Seal the ends with glass/cotton wool. Label two graduated syringes as “A” and “B” Push out air from syringe “A”. Pull in air into syringe “B”.

Attach both syringe “A” and “B” on opposite ends of the glass tube. Determine and record the volume of air in syringe “B” V₁. Heat the glass tube strongly for about three minutes. Push all the air slowly from syringe “B” to syringe “A” as heating continues. Push all the air slowly from syringe “A” back to syringe “B” and repeatedly back and forth. After about ten minutes, determine the new volume of air in syringe “B” V₁

Set up of apparatus

Sample observations

Colour change from brown to black

Volume of air in syringe “B” before heating V₁ = 158.0 cm₃

Volume of air in syringe “B” after heating V₂ = 127.2 cm₃

Volume of air in syringe “B” used by copper V₁ – V₂= 30.8cm₃

 

Sample questions

1. What is the purpose of:

(i) Glass/cotton wool

To prevent/stop copper turnings from being blown into the syringe/out of the glass tube

(ii) Passing air through the glass tube repeatedly

To ensure all the active part of air is used up

(iii) Passing air through the glass tube slowly

To allow enough time of contact between the active part of and the heated copper turnings.

2. State and explain the observations made in the glass tube.

Colour change from brown to black Brown copper metal reacts with the active part of air/oxygen to form black copper(II)oxide.

 

Chemical equation Copper + Oxygen -> Copper(II)oxide

2Cu(s) + O₂(g) -> 2CuO(s)

The reaction reduces the amount/volume of oxygen in syringe “B” leaving the inactive part of air. Copper only react with oxygen when heated.

3. Calculate the % of

(i) Active part of air

% active part of air = V1 – V2x 100% => 30.8cm3 x 100% =19.493%

V1 158.0cm3

 

(ii)  Inactive part of air

Method 1

% inactive part of air = V2 x 100% =>127.2cm3 x 100% = 80.506%

V1 158.0cm3

Method 2

% inactive part of air = 100% -% active part of air

=> 100 % – 19.493 % = 80.507%

 

4. The % of active part of air is theoretically higher than the above while % of inactive part of air is theoretically lower than the above. Explain.

Not all the active part of air reacted with copper

5. State the main gases that constitute:

(a) Active part of air.

Oxygen

(b) Inactive part of air

Nitrogen, carbon(IV)oxide and noble gases

 

6. If the copper turnings are replaced with magnesium shavings the % of active part of air obtained is extraordinary very high. Explain.

Magnesium is more reactive than copper. The reaction is highly exothermic.

It generates enough heat for magnesium to react with both oxygen and nitrogen in the air.

A white solid/ash mixture of Magnesium oxide and Magnesium nitride is formed.

This considerably reduces the volume of air left after the experiment.

Chemical equation Magnesium + Oxygen -> magnesium (II)oxide

2Mg(s) + O₂(g) -> 2MgO(s)

Magnesium + Nitrogen -> magnesium (II)nitride

3Mg(s) + N₂(g) -> Mg₃N₂ (s)

 

(c) To find the composition of active part of air using alkaline pyrogallol.

Procedure

Measure about 2cm₃of dilute sodium hydroxide into a graduated gas jar. Record the volume of the graduated cylinder V₁. Place about two spatula end full of pyrogallol/1,2,3-trihydroxobenzene into the gas jar.₂Immediately place a cover slip firmly on the mouth of the gas jar. Swirl thoroughly for about two minutes.

Invert the gas jar in a trough/basin containing water. Measure the volume of air in the gas jar V₂ Sample observations Colour of pyrogallol/1,2,3-trihydroxobenzene change to brown. Level of water in gas jar rises when inverted in basin/trough. Volume of gas jar/air in gas jarV1= 800cm₃ Volume of gas jar/air in gas jarafter shaking with alkaline pyrogallol/1,2,3-trihydroxobenzeneV2= 640 cm₃

 

Sample questions

1. Which gas is absorbed by alkaline pyrogallol/1,2,3-trihydroxobenzene?

Oxygen

2. Calculate the

(i) % of active part of air

V1-V2 x 100% =>(800cm3 – 640 cm₃) x 100% = 20%

V₁ 800cm₃

(ii) % of inactive part of air V₂ x 100% => 640 cm₃ x 100% = 80%

V₁ 800cm₃

 

(d) To establish the presence of carbon(IV)oxide in air using lime water Pass tap water slowly into an empty flask as in the set up below

Sample observation questions

1. What is the purpose of paper cover?

To ensure no air enters into the lime water.

2. What happens when water enters the flask?

It forces the air from the flask into the lime water.

3. What is observed when the air is bubbled in the lime water

A white precipitate is formed. The white precipitate dissolves on prolonged bubbling of air.

4. (a) Identify the compound that form:

(i) lime water

Calcium hydroxide / Ca(OH)₂

(ii)white precipitate

Calcium carbonate/ CaCO₃

(iii)when the white precipitate dissolves

Calcium hydrogen carbonate/ CaHCO₃

 

(b) Write the chemical equation for the reaction that take place when:

(i) White precipitate is formed

Calcium hydroxide + carbon(IV)oxide ->Calcium carbonate + water

Ca(OH)2(aq) +CO2(g) ->CaCO₃(s) +H2O(l)

(ii) White precipitate dissolves

Calcium carbonate+ water+ carbon(IV)oxide ->Calcium hydrogen carbonate

CaCO₃(s) +H₂O(l)+CO₂(g)->CaHCO₃(aq)

5. State the chemical test for the presence of carbon(IV)oxide gas based on 4(a) and (b)above:

Carbon(IV)oxide forms a white precipitate with lime water that dissolves in excess of the gas.

6. State the composition of carbon(IV)oxide gas by volume in the air.

About 0.03% by volume

 

See also:

PROPERTIES OF ACIDS

INTRODUCTION TO ACIDS, BASES AND INDICATORS

PHYSICAL/TEMPORARY AND CHEMICAL CHANGES

METHODS OF SEPARATING MIXTURES

CLASSIFICATION OF SUBSTANCES