00:01
So, here we have been given with 25 gram sample of a monoprotic weak acid.
00:14
So, it is said that the combustion of 10 gram of this acid here gives 14 .99 gram of co2 and 4 .09 gram of h2o.
00:34
So, we have to find out the empirical formula of the compound.
00:44
So, if we see then the mass of co2 is equal to 14 .99 gram which means the moles of co2 is equal to 14 .99 divided by molar mass which is 44 gram mole inverse.
01:06
So, it is equal to 0 .3406 mole.
01:11
So, from here we see then the moles of carbon would be equal to 0 .3406 and then the mass of carbon would be equal to number of moles 0 .3406 multiplied by molar mass which is 12 gram mole inverse.
01:28
So, the mass comes out as 4 .0881 grams for carbon.
01:36
Now if we see the mass of h2o so it is equal to 4 .09 gram.
01:43
Now from here if we calculate moles of h2o so it is equal to 4 .09 gram divided by molar mass which is 18 gram mole inverse for water...