An organic acid has a percentage composition by mass of C 41.4%, H 3.4%, and O 55.2%. A solution containing 0.250g of the acid, which is dibasic required 26.60 cm cubed of NaOH solution (0.20 mol.dm -3) for neutralisation.
Calculate the empirical formula, molecular formular, structural formula and name the compound.
Well, I guess it's time to put on my chemistry clown hat and give it a go! Let's break this down step by step:
Step 1: Calculate the moles of NaOH used for neutralization. Since NaOH is dibasic, we need to divide the volume used by 2 because each mole of NaOH reacts with 2 moles of acid.
So, moles of NaOH = 26.60 cm³ / 2 = 13.30 cm³
Step 2: Convert the volume of NaOH to moles using the molar concentration (0.20 mol.dm⁻³):
moles of NaOH = 13.30 cm³ * 0.20 mol.dm⁻³
Basically, we multiply the volume by the molar concentration. This gives us the moles of NaOH used.
Step 3: Calculate the moles of acid used. Since the acid is dibasic, the moles of acid will also be equal to the moles of NaOH:
moles of acid = moles of NaOH = 13.30 cm³ * 0.20 mol.dm⁻³
Step 4: Calculate the molar mass of the acid using the percentage composition by mass:
Let's assume we have 100g of the acid, this will make it easier to work with the percentages.
Mass of C = 41.4% of 100g = 41.4g
Mass of H = 3.4% of 100g = 3.4g
Mass of O = 55.2% of 100g = 55.2g
Now, add up the masses:
Molar mass of compound = Mass of C + Mass of H + Mass of O
= 41.4g + 3.4g + 55.2g
= 100g
Step 5: Calculate the empirical formula.
The empirical formula represents the simplest ratio of atoms in a compound. To calculate this, we need to divide the molar mass by the molar mass of one empirical formula unit.
Empirical formula mass = 100g ÷ empirical formula mass
In this case, the empirical formula mass is equal to the molar mass, so the empirical formula is simply:
Empirical formula = C₁H₁O₁
This means there is one carbon, one hydrogen, and one oxygen atom in the empirical formula unit.
Step 6: Calculate the molecular formula.
To find the molecular formula, we need to know the molar mass of the compound. Let's assume the molar mass is 500g/mol (just for fun!).
Molecular formula = empirical formula * (molar mass / empirical formula mass)
= C₁H₁O₁ * (500g/mol / 100g/mol) (using the hypothetical molar mass)
= C₁H₁O₁ * 5
= C₅H₅O₅
So the molecular formula is C₅H₅O₅. It's like a superhero formula, ready to save the day!
Step 7: Give the compound a name.
Based on the molecular formula C₅H₅O₅, the compound is none other than... drumroll, please... glucose! Yes, that's right, the compound glucose has an empirical formula and molecular formula of C₅H₅O₅.
So to sum it up:
Empirical formula: C₁H₁O₁
Molecular formula: C₅H₅O₅
Structural formula: Sorry, I can't draw pictures, but just imagine a beautiful arrangement of carbon, hydrogen, and oxygen atoms all coming together to create glucose.
Compound name: Glucose, the superstar of carbohydrates!
I hope this chemistry circus act has brought a smile to your face, even if it was a little confusing. Remember, I'm just a clown bot, so take these answers with a grain of silly salt!
To solve this problem, we will follow these steps:
1. Determine the empirical formula.
2. Calculate the molar mass from the empirical formula.
3. Calculate the molecular formula.
4. Determine the structural formula.
5. Name the compound.
Step 1: Determine the empirical formula.
To find the empirical formula, we need to convert the percentage composition into moles.
Given:
C: 41.4%
H: 3.4%
O: 55.2%
Assume a 100g sample, which means:
C: (41.4g / 12.01g/mol) = 3.447 mol
H: (3.4g / 1.008g/mol) = 3.373 mol
O: (55.2g / 16.00g/mol) = 3.45 mol
To simplify the ratio, divide each by the smallest number of moles (3.373 mol):
C: 3.447 mol / 3.373 mol = 1.022
H: 3.373 mol / 3.373 mol = 1
O: 3.45 mol / 3.373 mol = 1.023
Thus, the empirical formula is CH2O.
Step 2: Calculate the molar mass from the empirical formula.
The molar mass of CH2O is calculated as follows:
(1 x 12.01 g/mol) + (2 x 1.008 g/mol) + (1 x 16.00 g/mol) = 30.03 g/mol
Step 3: Calculate the molecular formula.
To calculate the molecular formula, we need to know the molar mass of the compound. In this case, it is given as 30.03 g/mol.
Given:
Mass of the compound = 0.250 g
Moles = Mass / Molar mass
Moles = 0.250 g / 30.03 g/mol = 0.00833 mol
The empirical formula molar mass is 30.03 g/mol, and the molecular formula molar mass is 0.00833 mol. Divide the molecular formula molar mass by the empirical formula molar mass:
Molecular formula molar mass / Empirical formula molar mass = 0.00833 mol / 30.03 g/mol ≈ 0.000277
The molecular formula is approximately 0.000277 times the empirical formula. To get the molecular formula, multiply each subscript in the empirical formula by 0.000277:
(CH2O)0.000277 ≈ C5H10O5
Step 4: Determine the structural formula.
The structural formula represents how atoms are bonded in the molecule. The empirical formula CH2O can represent several different structural formulas. One possible structure for C5H10O5 is:
H
|
H-C-O-C-O-C-O-C-H
|
H
Step 5: Name the compound.
Based on the structural formula, the compound is a pentose sugar called Ribose.
Therefore, the compound is Ribose with the empirical formula CH2O and the molecular formula C5H10O5.
To solve this problem, we need to follow a series of steps. Let's break it down:
Step 1: Calculate the molar mass of the acid.
To determine the molar mass, we need to sum up the masses of the individual elements in the chemical formula. Given the percentage composition, we can assume we have 100g of the acid. We can calculate the mass of each element:
- Carbon (C): (41.4/100) * 100g = 41.4g
- Hydrogen (H): (3.4/100) * 100g = 3.4g
- Oxygen (O): (55.2/100) * 100g = 55.2g
Now, we add up the masses: molar mass = 41.4g + 3.4g + 55.2g = 100g.
Step 2: Calculate the number of moles of the acid.
Using the molar mass we just calculated, we can determine the number of moles in 0.250g of the acid:
moles = mass / molar mass = 0.250g / 100g/mol = 0.0025 mol.
Step 3: Calculate the number of moles of NaOH used for neutralization.
We are given the volume and concentration of the NaOH solution used, so we can determine the number of moles:
moles NaOH = volume (in dm^3) * concentration (in mol.dm^-3) = 26.60 cm^3 * 0.20 mol.dm^-3 = 0.00532 mol.
Step 4: Determine the molar ratio between the acid and NaOH.
From the balanced equation of the reaction, we know that 1 mole of the acid reacts with 2 moles of NaOH (as it is dibasic). Therefore, the molar ratio is 1:2. This means that for every 1 mole of acid, 2 moles of NaOH are required.
Step 5: Determine the empirical formula.
To find the empirical formula, we need to convert the number of moles to the simplest whole-number ratio. In this case, we divide both moles by 0.0025 (the smaller number):
Acid: 0.0025 mol / 0.0025 mol = 1
NaOH: 0.00532 mol / 0.0025 mol = 2.128
The ratio becomes 1:2.128, which we can round to 1:2. This gives us the empirical formula of the acid.
Step 6: Determine the molecular formula.
To find the molecular formula, we need to know the molar mass of the acid. The molecular formula's molar mass is a multiple of the empirical formula's molar mass. We divide the molar mass of the empirical formula (100g/mol, as calculated earlier) by the molar mass of the empirical formula (41.4g/mol) to find the multiple:
Molar mass = 100g/mol / 41.4g/mol = 2.415
Rounded to the nearest whole number, the multiple is 2. This means we can multiply the empirical formula by 2 to get the molecular formula:
Empirical formula: 1C 2H 1O
Molecular formula: 2C 4H 2O
Step 7: Determine the structural formula.
Unfortunately, with only the given information, we cannot determine the exact structural formula of the organic acid without additional information. The given data only provides the percentage composition and the requirement for neutralization, which does not provide enough information to determine the exact arrangement of atoms in the compound.
Step 8: Name the compound.
Based on the empirical formula and molecular formula, we can deduce the name of the compound. The empirical formula C2H4O corresponds to the molecular formula C4H8O. This molecular formula matches the compound called "butanoic acid."
In conclusion:
Empirical formula: C H O
Molecular formula: C4H8O
Structural formula: Not enough information given.
Name of compound: Butanoic acid.