00:01
Good day in this question we are asked to identify the empirical formula and molecular formula of ethylene glycol.
00:08
So we assume we have 100 grams of the compound.
00:14
Therefore, the percent compositions here will be our masses, which is 37 .8 grams of carbon, 9 .7 grams of hydrogen, and 51 .6 grams of oxygen.
00:28
Next we solve for moles of each.
00:31
So mole of carbon is equals to 38 .7 grams divided by 12 .01 grams per mole.
00:46
This is equals to 3 .22 moles.
00:51
Next, we have mole of hydrogen, which is equals to 9 .7 grams, divided by 1 .01 grams per mole.
01:01
This is equals to 9 .60 mole.
01:06
Next, we have mole of oxygen, which is equals to 51 .6 grams divided by 16 grams per mole.
01:17
This is equals to 3 .22 moles.
01:23
Next, we divide the moles that we got to the smallest number of mole...