A compound contains 57.54% of c , 3.45% of h and 39.01% f what is it's empirical formula? why we multiply it by 3 in end?? can anyone help?
Answers
Answer:
Notice below how I do the first problem with some attention to using proper atomic weights, as well as keeping close to the proper number of significant figures. Then, notice how I get away from that (as well as being real consistent with units) in the following problems.
Notice also how it really doesn't make much of a difference. The trick is to know when to do that and it comes only via experience. Generally speaking, in empirical formula problems, C = 12, H = 1, O = 16 and S = 32 are sufficient.
There are times when using 12.011 or 1.008 will be necessary. If you hit a problem that just doesn't seem to be working out, go back and re-calculate with more precise atomic weights. These problems, however, are fairly uncommon.
For what it is worth, one piece of advice on rounding: don't round off on the moles if you see something like 2.33 or 4.665. That first one can be rendered as two and one-third (or seven thirds) and the second one as four and two-thirds (or fourteen thirds). In a situation like that, you would multiply by three to reach the smallest whole-number ratio rather than dividing by the smallest.
I know it's easy to say, harder to demonstrate. Some of the problems below involve this thirds issue. Look for a problem involving citric acid. Just be aware that rounding off too early and/or too much is a common problem in this type of problem.