empirical formula problem?
a compound is composed of 54.14% carbon, 6.16%hydrogen, 9.52%nitrogen and 27.18% oxygen.
what is the simplest/empirical formula?
element g mol
C 57.14 57.14 mol/12.01 = 4.758 mol
H 6.16 6.16 mol/1.01 = 6.10mol
N 9.52 9.52 mol/14.01 = 0.680mol
O 27.18 27.18 mol/16.00 = 1.699mol
C 4.758/0.680=7x2=14 N 0.680/0.680=1X2=2
H 6.10/0.680=9x2=18 O 1.699/0.680=2.50x2=5 empirical formula=C14 H18 N2 O5
i understand most of it except the part where you multiply by 2 my teacher said anything under 0.1 you round down and 0.9 you round up so why do you round down if its 2.50 shouldnt you round up instead. which would make it a 3 and multiply everything by 3 plz explain thanks for any help
Comments
If the data in the problem are good, then you should not round a result that ends in 0.50. I would round down or up if the fraction were 0.10 or less and greater than 0.90. But if you get 0.50, double everything to change the X.50 into an integer. The same thing goes for 0.33 (1/3) or 0.66 (2/3). Triple everything in those cases.
Let's take this for example.
4.758 mol C
6.10 mol H
0.680 mol N
1.699 mol O
Divide by the smallest, and...
4.768/0.68 = 7 C
6.10/0.68 = 9 H
0.68/0.68 = 1 N
1.699/0.68 = 2.5 O
You cannot have 2 and a HALF, so you multiply everything by two so they are all integers.
14C, 18H, 2N, 5O
If the copper is heated strongly, you are able to assume that one and each and each of the copper reacts. you realize the formula n = m/M or kind of moles = mass/molar mass? you are able to artwork out what number moles of copper you all started with, and subsequently what number moles of copper you end with. 10.04/sixty 3.fifty 5 = 0.158 moles. you realize the mass of oxygen you eventually end up with, so that you'll artwork out what number moles of oxygen you've. 2.fifty 3/16 = 0.158 moles. you keep in mind that you're merely going to make an oxide (ie no different species will be present day in this compound) so that you've a a million:a million ratio of moles. hence you've CuO.
When the number comes right in the middle, like 2.5, then you double it. When it would be like 2.05 or 2.95, then when one considers that elemental analysis has a variation in accuracy, you simply round that up or down.