A compound contains 12.8% Carbon, 2.10% Hydrogen, 85.1% Bromine (by mass) Calculate the empirical forumla of this compound. If the molar mass of tehis comopound is 188 g/mole, what is teh molecular formula
For teh empirical forumla would i just find the ratios of all those elements by doing C/H and Br/C??
For teh molecular forumla, I have no idea how to do it
2006-12-21 05:28:54 · 5 answers · asked by silentcargo 3 in Science & Mathematics ➔ Chemistry
12.8%C/12 = 1.07
2.10%H/1 = 2.10
85.1%Br/80 = 1.06
Dividing through by 1.06 gives a ratio of 1:2:1. So the empirical formula is CH2Br. The formula weight of CH2Br is 12+2+80 = 94. 188/94 = 2. So multiply the empirical formula by 2 to get the molecular formula of C2H4Br2. The compound is a dibromoethane, either Br-CH-CH-Br or Br2CH-CH3.
2006-12-21 05:51:06 · answer #1 · answered by steve_geo1 7 · 1⤊ 0⤋
Br Molecular Weight
2016-11-16 08:58:19 · answer #2 · answered by ? 4 · 0⤊ 0⤋
No, for the empirical formula you need to change each of the percentages to masses. You can recognize that 12.8% would be the same as having 12.8 g in a 100g sample. Do this for each of the elements present. Then convert the GRAMS to MOLES...THEN you divide each of the MOLES of elements by the one that has the smallest NUMBER OF MOLES. The quotients, rounded off appropriately will be your molar ratios.
By the way, with all DUE RESPECT to Raymond...his answer is completely wrong....disregard it outright.
For the molecular formula...find out first what the EMPIRICAL formula mass is. You may notice that the Empirical MW and the given MW are off by a ratio. Get this ratio by dividing the MW by the Emp. MW. Then multiply each of the molar ratios (coefficients in the empirical formula) by the obtained ratio.
Simple, if you had DONE YOUR HOMEWORK!!!!
2006-12-21 05:42:30 · answer #3 · answered by ? 4 · 0⤊ 1⤋
No, to find the empirical formula divide each percentage mass by the molar mass of the respective element.
C H Br
12.8/12 2.10/1 85.1/79.9
1.067 2.10 1.065
then divide the three no of moles by the smallest amongst them
1.067/1.065 2.10/1.065 1.065/1.065
1 1.97 1
1 2 1
therefore the empirical formular is CH2Br.
to find molecular formular
(CH2Br)*n= 188
(1*12+2*1+79.9*1)*n =93.9n
n=188/93.9
n= 2.
then multiply through using n=2.
therefore molecular formular is
C2H4Br2.
this is a halogenoakane. A bromoethane to be precise. better still 1,2-dibromoethane.
structural formular is
CH2Br-CH2Br or CH3-CHBR2 or CHBr2-CH3
i hope this will helpful to you
2006-12-21 08:51:20 · answer #4 · answered by tomzy 2 · 0⤊ 0⤋
Begin with:
mass of Carbon = 12.01
mass of hydrogen = 1.008
mass of Bromine = 79.90
a = number of carbon atoms
b = number of hydrogen atons
c = number of bromine atoms
a(12.01)+b(1.008)+c(79.90)=188
a(12.01) = 0.128*188 (0.128 = 12.8%)
2006-12-21 05:38:45 · answer #5 · answered by Raymond 7 · 0⤊ 0⤋