A compound is found to have 61% carbon, 6.6% hydrogen, and 32.6% oxygen. What is the empirical formula of this compound?
I got C25H32H10 but the subscripts seem absurdly large.
3 answers
I think you are right about that. If you had shown your work I would have found your error.
61/12.01 = 5.079/2.0375 = 2.5
6.6/1.008 = 6.5476/2.0375= 3.21
32.6/16 = 2.0375/2.0375= 1
I know to get the empirical formula, i Have to make these all whole numbers so I multiplied by 10 to get C25H32O10.
6.6/1.008 = 6.5476/2.0375= 3.21
32.6/16 = 2.0375/2.0375= 1
I know to get the empirical formula, i Have to make these all whole numbers so I multiplied by 10 to get C25H32O10.
The error is you used H2 and not H.
6.6/1 = 6.6 so you have
C = 5
H = 6.6
O = 2
Now multiply everything by 3 to get
C 15
H 19.8
O = 6
Then round to whole numbers.
I didn't divide by the smallest number, as I'm supposed to do, I just skipped that step since I had two whole numbers already so by multiply everything by 3 I got all whole numbers. If you do it the other way, you get
C 5/2 = 2.5
H 6.6/2 - 3.3
O 2/2 = 1
Multiply everything by 6 to get
C 15
H 19.8
O 6
Hope this helps.
6.6/1 = 6.6 so you have
C = 5
H = 6.6
O = 2
Now multiply everything by 3 to get
C 15
H 19.8
O = 6
Then round to whole numbers.
I didn't divide by the smallest number, as I'm supposed to do, I just skipped that step since I had two whole numbers already so by multiply everything by 3 I got all whole numbers. If you do it the other way, you get
C 5/2 = 2.5
H 6.6/2 - 3.3
O 2/2 = 1
Multiply everything by 6 to get
C 15
H 19.8
O 6
Hope this helps.
1 answer