Question

what is the empirical formula of a compound that contains 10.52 g Ni, 4.38g C , and 5.10g N

Answer 1

okay..i'll not give u direct answers...i want u to try first..

Answer 2

http://chemistry.about.com/od/workedchemistryproblems/a/How-To-Calculate-The-Empirical-And-Molecular-Formula-Of-A-Compound.htm

Answer 3

chk the above link and try to do it first urself

Answer 4

10.52*1 mole/58.69 = .179
4.38*1mole/12.01= .3646
5.10*1 mole/14.01= .3640

Answer 5

NiC2N2 If the data @kajaraaaa has posted is correct because you divide each of those 3 results by the lowest number.

Answer 6

I'm not gonna tell you the answer outright because you need to understand how to figure it out: basically the empirical formula is the ratio of the quantity of atoms of those elements that are present in the matter. For example, water is H2O (since its already in its simplest form with a 2:1 ratio). However, something like Benzene, C6H6, is simply CH (1:1).

To figure this problem out you want to know the atomic masses of those elements and from that figure out how many atoms (or should I say mol) of each of those three.

In short, this sample contains .179 mol of Ni, .365 of C, and .362 of N. Therefore it is roughly NiC(2)N(2)