- 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.
- The main gases present in the atmosphere/air
- 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
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.
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
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%
(ii) Inactive part of air can be calculated:
100% -20% = 80% // M3=> 8x 100%= 80%
(b)To find the composition of active part of air using heated copper turnings.
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
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
- 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.
- 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.
- Calculate the % of
(i) Active part of air
% active part of air = V1 – V2x 100% => 30.8cm3 x 100% =19.493%
(ii) Inactive part of air
% inactive part of air = V2 x 100% =>127.2cm3 x 100% = 80.506%
% inactive part of air = 100% -% active part of air
=> 100 % – 19.493 % = 80.507%
- 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
- State the main gases that constitute:
(a) Active part of air.
(b) Inactive part of air
Nitrogen, carbon(IV)oxide and noble gases
- 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.
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
- Which gas is absorbed by alkaline pyrogallol/1,2,3-trihydroxobenzene?
- Calculate the
(i) % of active part of air
V1-V2 x 100% =>(800cm3 – 640 cm3) x 100% = 20%
(ii) % of inactive part of air V2 x 100% => 640 cm3 x 100% = 80%
(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
- What is the purpose of paper cover?
To ensure no air enters into the lime water.
- What happens when water enters the flask?
It forces the air from the flask into the lime water.
- 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.
- (a) Identify the compound that form:
(i) lime water
Calcium hydroxide / Ca(OH)2
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
- 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.
- State the composition of carbon(IV)oxide gas by volume in the air.
About 0.03% by volume