Find the empirical formula of a compound found to contain 63.15 percent carbon 5.30 percent hydrogen and 31.55 percent oxygen
I got C27H27O
That is a very odd answer.
assume you had 100grams.
63.15g c
5.3g H
31.55g O
in moles, that is
C 63.15/12.011= 5.26Mol
H 5.3/1.008 =5.26 mol H
O 31.55/15.999=1.971 mol O
Now divide each by the lowest number in the mole ratio:
C 5.26/1.971=2.69
H = 2.69
O=1
Now, to covert to whole numbers, note that 2.69*3=8
empirical is C8H8O3
Which is again, weird, but check my work.
Oh I got C and H = to 2.71
To find the empirical formula, we need to determine the simplest whole-number ratio of atoms present in the compound.
First, we can assume we have 100 grams of the compound.
Then, convert the percentages into grams:
- Carbon (C): 63.15 grams
- Hydrogen (H): 5.30 grams
- Oxygen (O): 31.55 grams
Next, we need to convert the masses into moles. To do this, we divide the grams by their respective molar masses:
- Carbon (C): 63.15 g / (12.01 g/mol) = 5.257 moles
- Hydrogen (H): 5.30 g / (1.01 g/mol) = 5.247 moles
- Oxygen (O): 31.55 g / (16.00 g/mol) = 1.972 moles
Now, we need to find the simplest ratio between the number of moles of each element. We can do this by dividing the number of moles of each element by the smallest number of moles (in this case, 1.972 moles of oxygen):
- Carbon (C): 5.257 moles / 1.972 moles ≈ 2.67
- Hydrogen (H): 5.247 moles / 1.972 moles ≈ 2.66
- Oxygen (O): 1.972 moles / 1.972 moles = 1
The numbers obtained are close to 2.5, but we need whole numbers. Multiply all the numbers by 2 to get whole numbers:
- Carbon (C): 2.67 * 2 = 5.34 ≈ 5
- Hydrogen (H): 2.66 * 2 = 5.32 ≈ 5
- Oxygen (O): 1 * 2 = 2
Therefore, the empirical formula of the compound is C5H5O2.
To find the empirical formula of a compound, you need to determine the ratio of the different elements present in the compound. Here are the steps to calculate the empirical formula:
1. Assume that you have 100 grams of the compound, which makes it easier to calculate the percentages as grams.
2. Convert the percentages of each element to grams. Based on the percentages given:
- Carbon (C): 63.15 grams
- Hydrogen (H): 5.30 grams
- Oxygen (O): 31.55 grams
3. Convert the grams of each element to moles. To do this, divide the grams of each element by their respective atomic masses:
- Carbon (C): 63.15 g / 12.01 g/mol = 5.26 mol
- Hydrogen (H): 5.30 g / 1.01 g/mol = 5.25 mol
- Oxygen (O): 31.55 g / 16.00 g/mol = 1.97 mol
4. Determine the simplest whole-number ratio of the elements. To do this, divide each moles value by the smallest number of moles present (in this case, oxygen):
- Carbon: 5.26 mol / 1.97 mol = 2.67 (approximately)
- Hydrogen: 5.25 mol / 1.97 mol = 2.66 (approximately)
- Oxygen: 1.97 mol / 1.97 mol = 1
5. Round the numbers obtained in step 4 to the nearest whole number. In this case, the ratios are approximately:
- Carbon: 3
- Hydrogen: 3
- Oxygen: 1
The empirical formula of the compound is C3H3O1, which can be simplified to CHO. Therefore, the empirical formula of the compound is CHO.