Vanillin contains 63.15% C, 5.30%H, and oxygen. What is the empirical formula of vanillin?
Take 100 g sample which will contain 63.15 g C, 5.30 g H, and 31.6 g oxygen.
Convert to mols. mols = grams/atomic mass
63.15/12 = 5.26 mols C
5.30/1 = 5.30 mols H
31.6/16 = 1.98 mols O
Now you want to find the ratio of these elements to one another with none smaller than 1. The easy way to do that is to divide all of the numbers by the smallest one; i.e., that of C, so
5.26/1.98 = 2.66 C
5.30/1.98 = 2.68 2.68 H
1.98/1.98 = 1.00 O but C and H are too far from a whole number. What we do in these situations is to multiply by whole numbers until we get close to a whole number for C and H.
by 2 gives us
C = 2.66 x 2 = 5.32
H = 2.68 x 2 = 5.36
O = 1.00 x 2 = 2.00
Doesn't work so multiply by 3
C = 2.66 x 3 = 7.98
H = 2.68 x 3 = 8.04
O = 1.00 x 3 = 3.00
So the empirical formula must be C8H8O3. Check my work.
Well, it seems vanillin just loves the elements C and H, but it's a bit shy about oxygen. Now, let's calculate its empirical formula, shall we? To do that, we need to convert those lovely percentages into moles.
So, assuming we have 100g of vanillin, we have 63.15g of carbon and 5.3g of hydrogen. To convert those into moles, we divide by their respective atomic masses.
The atomic mass of carbon is roughly 12 g/mol, so moles of carbon would be 63.15g / 12g/mol = 5.26 mol.
The atomic mass of hydrogen is roughly 1 g/mol, so moles of hydrogen would be 5.3g / 1g/mol = 5.3 mol.
As for oxygen, we don't know the exact percentage, but we can find it by subtracting the sum of carbon and hydrogen percentages from 100%. This gives us: 100% - 63.15% - 5.30% = 31.55%.
Now we need to convert this percentage into grams by multiplying by the total mass of vanillin (100g).
31.55% oxygen = 31.55g oxygen.
To find the moles of oxygen, we divide by its atomic mass of 16 g/mol:
31.55g / 16g/mol ≈ 1.97 mol.
Now let's find the simplest ratio of the elements by dividing all the moles by the lowest number of moles (in this case, carbon):
Carbon: 5.26 mol / 5.26 mol = 1
Hydrogen: 5.3 mol / 5.26 mol ≈ 1
Oxygen: 1.97 mol / 5.26 mol ≈ 0.373
Now, we just need to multiply these numbers by a whole number to get rid of the decimal points. In this case, multiplying by 3 should do the trick.
Carbon: 1 x 3 = 3
Hydrogen: 1 x 3 = 3
Oxygen: 0.373 x 3 ≈ 1
So, the empirical formula of vanillin is C3H3O. Voila!
To determine the empirical formula of vanillin, we need to find the ratio of the atoms present in the compound.
1. Start by assuming we have 100g of vanillin.
2. Convert the given percentages into grams.
- Carbon (C): 63.15g (63.15% of 100g)
- Hydrogen (H): 5.30g (5.30% of 100g)
3. Determine the grams of oxygen by subtracting the sum of the carbon and hydrogen masses from the total mass (100g).
- Oxygen (O): 100g - (63.15g + 5.30g) = 31.55g
4. Find the moles of each element by dividing their respective masses by their atomic masses:
- Carbon (C): 63.15g / 12.01 g/mol = 5.26 mol
- Hydrogen (H): 5.30g / 1.01 g/mol = 5.25 mol
- Oxygen (O): 31.55g / 16.00 g/mol = 1.97 mol
5. Divide the moles by the smallest number of moles to get the simplest ratio of atoms.
- Carbon (C): 5.26 mol / 1.97 mol = 2.67
- Hydrogen (H): 5.25 mol / 1.97 mol = 2.66
- Oxygen (O): 1.97 mol / 1.97 mol = 1.00
6. Round the ratio to the nearest whole number.
- Carbon (C): 3
- Hydrogen (H): 3
- Oxygen (O): 1
7. The empirical formula of vanillin is C3H3O.
To determine the empirical formula of vanillin, we need to find the simplest whole number ratio of the atoms present in the compound.
Given that vanillin contains 63.15% carbon and 5.30% hydrogen, we can assume the remaining percentage represents oxygen.
To calculate the empirical formula, we need to convert the percentages into moles. We will assume we have 100 grams of vanillin for convenience.
1. Calculate the grams of carbon (C) present:
Mass of C = 100 grams x 63.15% = 63.15 grams
2. Calculate the moles of carbon (C):
Moles of C = Mass of C / Molar mass of C
The molar mass of carbon is 12.01 grams/mole, so:
Moles of C = 63.15 grams / 12.01 grams/mole ≈ 5.261 moles
3. Repeat this process to find the moles of hydrogen (H) and oxygen (O):
Mass of H = 100 grams x 5.30% = 5.30 grams
Moles of H = Mass of H / Molar mass of H
Molar mass of hydrogen is 1.01 grams/mole, so:
Moles of H = 5.30 grams / 1.01 grams/mole ≈ 5.248 moles
To find the moles of oxygen (O), subtract the moles of carbon and hydrogen from the total moles:
Moles of O = 100 grams - (Moles of C + Moles of H) ≈ 100 grams - (5.261 moles + 5.248 moles) ≈ 89.491 moles
4. Divide the number of moles obtained for each element by the smallest number of moles to find the simplest whole number ratio. In this case, the smallest number of moles is approximately 5.248.
Moles of C / 5.248 ≈ 1
Moles of H / 5.248 ≈ 1
Moles of O / 5.248 ≈ 17
The empirical formula of vanillin is CH2O.
Therefore, the empirical formula of vanillin is CH2O.