Assume our whole sample weighs 100 grams.
C = 12.77 % = 12.77/100 = 12.77 g C
H = 1.60 % = 1.60 g H
100 g - 12.77g - 1.60 g = 85.63g = our bromine
Now that we know the element components of the unknown compound we have to find out what they are in moles.
C : 12.77 g C ( 1 mol C / 12.01 g C) = 1.06 mol C
H: 1.60 g H ( 1 mol H / 1.008 g H) = 1.59 mol = 1.5 mol H
Br: 85.63 g Br ( 1 mol Br / 79.904 g Br) = 1.07 mol Br
C : 1.06 mol/ 1.06 = 1 * 2 = 2
H : 1.59 mol/1.06 = 1.5 * 2 = 3
Br : 1.07 mol/ 1.06 = ~1 * 2= 2
therefore we get C2H3Br2. To double check, we find the molar mass of one unit of this. We get ~186.8 which is pretty close.
(12.011 * 2) + (1.008 * 3) + (79.904 * 2) = 186.854 grams per mole of C2H3Br2
An unknown organic compound contains carbon, hydrogen, and bromine only. The mass percentage of
C is 12.81% C. Bromine from 1.88 g of this compound has been completely precipitated as 3.78 g of AgBr.
The molar mass of unknown organic compound is 188 u. Determine the molecular formula of the unknown
1 answer
1 answer