00:01
Here we have to find the empirical formula of an unknown compound that have carbon, hydrogen and oxygen.
00:14
So what are the given values in the question is that combustion of unknown compound, mass of unknown compound is given here is 5 .40 gm whereas mass of carbon dioxide produced is 12 .3 gm and mass of water produced is 5 .02 gm.
00:57
So the first step is we have to follow is to calculate the mole of co2 and water produced.
01:05
So to calculate moles of co2 and water produced, moles of co2 is equal to mass upon molecular mass, mass of co2 divided by its molar mass.
01:26
So mass of co2 is given already 12 .3 gm and molar mass is 44 .01 gm per mole and dividing this you will get approximate value of 0 .279 mole.
01:45
Ok second thing is you can do is calculate moles of water produced.
01:52
Similarly you can calculate mass of water and molar mass.
02:05
That you have to put.
02:06
So the mass of water given is 5 .02 gm and the molar mass is 18 .02 gm.
02:15
Dividing that you will get approximate value of 0 .278 moles.
02:24
So now we know the number of moles of water and co2 produced.
02:29
From here the first thing we can do is to write the combustion equation.
02:34
So suppose the molecule has cx, hy and oz.
02:41
When it reacts with oxygen it will produce co2 and h2o.
02:50
So you can see that on the product side the number of carbon here is fixed.
02:59
So each mole of co2 will produce each single mole of carbon.
03:03
So moles of carbon will be equal to 0 .279 mole because 0 .279 mole of carbon dioxide was produced.
03:21
Similarly moles of hydrogen can be calculated as.
03:28
So each mole of water produced consists of 2 moles of hydrogen.
03:37
So hydrogen will be 2 into 0 .278...