What is the empirical forumla (lowest whole ratio) of the compounds below?
A.
13.0g Mg 1 mole/24.312 = .53 mole
87.0g Br 1 mole/79.909g = 1.09 mole
Mg .53/.53 Br 1.09/.53
Mg 1.0 Br 2.1
What do i do now?
I didn't check your math but if it's correct, round the 2.1 to 2.0 and the formula is Mg1Br2.0 or MgBr2.
you just write it as MgBr2
since when is 2.054 rounded up .
Round it to the nearest whole number
MgBr2
My teacher told me what I needed to get both numbers to whole numbers by mutiplying 2-6. But i do not know when to do it. Can you please give me an example
http://books.google.com/books?id=EKOBuCwHh9wC&pg=PA29&lpg=PA29&dq=empirical+formula+C3H8&source=web&ots=2XWO2gHpDo&sig=CQMIhjMo-rkgb0Quz5KGGpXXQMs&hl=en&sa=X&oi=book_result&resnum=3&ct=result#PPA30,M1
See prob 3-2, where they had to multipy by 2 to get a whole number ratio
Now that you have determined the mole ratio between magnesium (Mg) and bromine (Br) to be Mg 1.0 : Br 2.1, you can use this information to find the empirical formula of the compound.
To do this, you need to convert the ratio of moles into the simplest whole number ratio. In this case, you can round the number to the nearest whole number.
So, the empirical formula of the compound is MgBr2.