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.
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.
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.
To determine the empirical formula of a compound, we need to calculate the simplest ratio of the elements present.
1. Start by assuming you have 100 grams of the compound. This allows us to work with percentages as grams.
2. Convert the percentage of each element into grams.
- Carbon: 61% of 100g = 61g
- Hydrogen: 6.6% of 100g = 6.6g
- Oxygen: 32.6% of 100g = 32.6g
3. Next, convert the grams of each element into moles by dividing by their respective atomic masses.
- Carbon: 61g / 12.01 g/mol = 5.08 mol
- Hydrogen: 6.6g / 1.01 g/mol = 6.53 mol
- Oxygen: 32.6g / 16.00 g/mol = 2.04 mol
4. Once we have the moles of each element, we need to find the simplest whole-number ratio. Divide each of the moles by the smallest number of moles.
- Carbon: 5.08 mol / 2.04 mol = 2.49
- Hydrogen: 6.53 mol / 2.04 mol = 3.20
- Oxygen: 2.04 mol / 2.04 mol = 1.00
The ratios we obtained are not whole numbers, so we need to multiply all the ratios by the smallest number that will give us whole numbers. In this case, the smallest ratio is 2.49, which we can round to 2.
5. Round the ratios to the nearest whole number.
- Carbon: 2.49 rounded to 2
- Hydrogen: 3.20 rounded to 3
- Oxygen: 1.00 rounded to 1
Based on the rounded ratios, the empirical formula of the compound is C2H3O.
Your initial attempt at the empirical formula (C25H32H10) is incorrect. It seems you made a mistake during the calculations. Remember to divide each mole value by the smallest mole value to obtain the simplest whole-number ratio.