00:03
Let's consider the empirical formula of the compound that is the organic compound that will be cxhyoz that will be the oxygen o2 combustion reaction and from it co2 plus yyz h2o.
00:30
Now moles of co2 that is the mass will be given 10 .04 gram per mole.
00:43
From gram will be taken out that will be 0 .228 mole.
00:48
Now moles of water that will be from 8 .21 gram that will be produced divided by 15 gram per mole.
01:04
Now from 1 mole carbon dioxide that will be contained 1 mole of carbon that is 0 .228 mole of carbon dioxide contained 0 .228 multiplied by 1 that is 0 .228 mole of carbon.
01:27
Now 1 moles of water that will be contained 1 mole of hydrogen.
01:39
So 0 .456 mole of water contained multiplied by 0 .456 that will be equal 0 .912 mole of hydrogen.
01:56
Now 0 .28 mole of carbon mass that will be number of 0 .28 mole multiplied by mass of the water that will be 12 .01 gram per mole.
02:16
So 1 mole carbon will be considered as 2 .7382 gram.
02:23
This is the mass of the carbon.
02:25
Now 0 .912 mole of hydrogen mass will be number of moles 0 .912 mole multiplied by atomic charge that will be 1 .0181 per mole that is the mass of hydrogen.
02:45
So per mole mole will be equal 0 .912 into 8 .2.
02:53
Now given mass of ch, hy, oz that will be 7 .3 gram.
03:05
So mass of oxygen that will be 7 .3 minus mass of hydrogen and carbon that is mass of hydrogen will be 0 .91 minus mass of carbon that will be 0 .782 gram.
03:24
So that will be equal 3 .69 gram.
03:28
This is the mass of oxygen.
03:31
So mass of o2 that will be 3 .69 gram and molar mass that will be 32 gram per mole.
03:42
So number of moles that will be the real mass 3 .69 gram per mole.
03:50
So gram will be cancelled out that will be equal to 0 .113 mole...