Combustion Analysis of Gaseous Hydrocarbons

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In this video we want to discuss how to use combustion analysis to determine the molecular formula of an unknown gaseous hydrocarbon CxHy.

The balanced equation of the combustion of hydrocarbon in excess oxygen to give carbon dioxide and water is as follows:

Notice that all the carbon in the hydrocarbon is converted to carbon dioxide CO2 which is acidic in nature.

This means we can use a base to absorb acidic CO2 to measure the amount of CO2 produced and determine the amount of carbon in the hydrocarbon.

Also, all the hydrogen is converted to water H2O which is a liquid at room temperature or standard temperature.

We can simply cool the gaseous product mixture to collect the liquid water, or pass the gaseous mixture through an anhydrous salt to absorb water to determine the amount of water produced and the amount of hydrogen in the hydrocarbon.

Therefore combustion analysis is a simple method to determine the amount of carbon and hydrogen present in the hydrocarbon, and we can deduce the molecular formula of that hydrocarbon.

Let's take a look at an example question.

To help visualise the question we can use the following diagram to determine the volumes of gases involved:

1. Volume of hydrocarbon CxHy

The volume of CxHy is given in the question which is 20 cm3

2. Volume of CO2 produced

The 130 cm3 volume of residual gases consists of volume of CO2 produced and volume of unreacted O2 in excess.

Water is a liquid at room temperature and pressure hence its volume is not considered as part of the 130 cm3 of residual gases.

After shaking with alkaline sodium hydroxide, carbon dioxide is absorbed and final volume of 90 cm3 is the volume of O2 in excess.

This means that volume of CO2 will just be the difference which is 40 cm3.

3. Volume of O2 reacted

Total volume of O2 is 150 cm3 and volume of O2 in excess is 90 cm3. Therefore the volume of O2 reacted will be the difference which is 60 cm3.

We can now fill up the following table which lists down the mole ratio and volume ratio of the gases.

The mole ratio is always fixed since it is based on the balanced equation.

The volume ratio is the one that varies and we have already determined the volumes of CxHy, O2 reacted and CO2 produced from the information given in the question.

Again since water is a liquid at rtp, we do not need to compare the mole and volume ratio of water hence its mole and volume are not required.

We can now compare mole ratio and volume ratio of these gases to solve for x and y.

1. Solve for x

We compare mole ratio and volume ratio of CO2 and CxHy to determine that x=2.

2. Solve for y

We compare mole ratio and volume ratio of O2 and CxHy to solve for y. We need to substitute x=2 to determine that y=4.

Finally we can determine the molecular formula for this hydrocarbon is C2H4.

For the detailed step-by-step discussion on how to determine the molecular formula of an unknown hydrocarbon from combustion analysis, check out this video!

Topic: Mole Concept, Physical Chemistry, A Level Chemistry, Singapore

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This helped a lot, as compared to the rather unintuitive way of finding y using decrease in combustion, the use of Avogadro's Law is extremely elegant and way easier to visualise! Thank you!

megumin
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Very good, we need teachers like you!!!!

zacharialushika
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Thanks been stuck on these type of questions for hours.

siyamrafiq
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Thank you so much sir this was so helpful compared to my teachers 😭

luciadroplet
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Wow! Thanks, you made it simple more than my teacher

leilani
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You have no idea how my chemistry teacher confused me on this one, our teachers prefer teaching us to memorize expressions rather than understand where they come from, thanks alot

musangofrancis
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Thank you for giving the best idea for solving this equation...

nimeshpalungwa
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You’re a great chemistry teacher, much love

fareedasamer
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Dude's really good... keep up the good work

ceejaysvlog
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You are indeed a God sent. Thank you so much

melisapryce
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Thanks a lot.I finally got that concept

patriciaakwero
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Thanks for the video! Is it possible to have a complete combustion reaction in which oxygen is actually the limiting reactant?

teddy
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bro you explained it way better than my teacher... arigato

jmvw
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thanks you so much!!! had been struggling since a long time

siddhanthduggal
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i was using this method but i got x=9/2 for the first two fraction and this kind of complicate things. what should i do

RomarioBryan-yi
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It’s very useful video. Thx for sharing!

葉赫那拉榮禹
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Sir, I have a question to ask. How can I find the grams of carbon and water if the given are grams sample of hydrocarbon, pressure, temperature, and volume? Thank you sir

starlight
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Why take ratio of co/co2 to check combustion in LPG stove?

vijaygupta
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Hi chemistry guru, i have a question . how will you solve the question if the question doesnt specify the volume of excess oxygen ?

Aarvindsundar
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hey, sorry for the inconvinience but how is y=4 in the last problem if 3x4 is 12 -2 is 10

raphaelfarrugiawismayer
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