Maleic acid is an organic compound composed of 41.39% C, 3.47% H, and the rest oxygen. If 0.271 mol of maleic acid has a mass of 31.4 g, what are the empirical and molecular formulas of maleic acid?

So I checked the problem again, and yes this is the whole problem. I even copied and paste it from the homework page online. There's no part a or b before it either.... so do you think this problem is doable?

Excuse me, I was sleepy last night OR my brain was out of town.

I think the empirical formula you had was CHO and the formula mass for CHO is 12+1+16=29

moles = grams/molar mass and rearrange to
molar mass = grams/moles = 31.4 g/0.271 moles = about 117 or so. You can do it more accurately. So how many units of CHO are in the 117; that will be 117/29 = 4.03 which rounds to 4.0 and the molecular formula is (CHO)4 or to put it in more conventional terms it would be C4H4O4

Yes, this problem is indeed doable. To determine the empirical formula of maleic acid, we need to find the ratios of the elements present in the compound based on the given percentages.

Step 1: Convert the given percentages to masses.

Given:
Mass of C = 41.39%
Mass of H = 3.47%
Mass of O = (100% - (41.39% + 3.47%))

To find the mass of each element, assume 100g of the compound, which will allow us to convert the percentages to masses:

Mass of C = (41.39g/100g) × 100g = 41.39g
Mass of H = (3.47g/100g) × 100g = 3.47g
Mass of O = (100% - (41.39% + 3.47%)) = 55.14g

Step 2: Convert the masses of the elements to moles.

Use the molar mass of each element to convert the masses to moles:
Molar mass of C = 12.01g/mol
Molar mass of H = 1.01g/mol
Molar mass of O = 16.00g/mol

Moles of C = mass of C / molar mass of C = 41.39g / 12.01g/mol ≈ 3.447 mol
Moles of H = mass of H / molar mass of H = 3.47g / 1.01g/mol ≈ 3.436 mol
Moles of O = mass of O / molar mass of O = 55.14g / 16.00g/mol ≈ 3.446 mol

Step 3: Determine the empirical formula.

To find the empirical formula, divide the number of moles of each element by the smallest number of moles obtained (to obtain a simple whole number ratio):

Empirical formula of maleic acid: C3H3O3

Step 4: Determine the molecular formula.

To find the molecular formula, we need to know the molar mass of the compound. Given that 0.271 mol of maleic acid has a mass of 31.4g, the molar mass can be calculated as:

Molar mass of maleic acid = mass / moles = 31.4g / 0.271 mol ≈ 115.8g/mol

Next, compare the molar mass of maleic acid to the empirical formula mass:

Empirical formula mass = (3 moles of C × 12.01g/mol) + (3 moles of H × 1.01g/mol) + (3 moles of O × 16.00g/mol)
= 36.03g + 3.03g + 48.00g
= 87.06g/mol

Divide the molar mass of maleic acid by the empirical formula mass to find the whole number multiple:

Whole number multiple = molar mass of maleic acid / empirical formula mass
= 115.8g/mol / 87.06g/mol
≈ 1.33

Finally, multiply the subscripts in the empirical formula by the whole number multiple:

Molecular formula of maleic acid: C4H4O4

Yes, this problem is definitely doable. To solve this problem, we need to determine the empirical and molecular formulas of maleic acid based on the given information.

First, let's calculate the molar mass of maleic acid using the given mass and molar quantity. We know that the molar mass of maleic acid is equal to the mass (31.4 g) divided by the molar quantity (0.271 mol):

Molar mass = 31.4 g / 0.271 mol
Molar mass ≈ 115.5 g/mol

Now, let's calculate the number of moles of each element present in maleic acid. We can assume we have 100 grams of maleic acid since the percentages of each element in the compound are given.

C (Carbon):
Percentage = 41.39%
Mass of carbon = (41.39 g/100 g) × 100 g = 41.39 g
Number of moles of C = mass of carbon / molar mass of C = 41.39 g / 12.01 g/mol ≈ 3.448 mol

H (Hydrogen):
Percentage = 3.47%
Mass of hydrogen = (3.47 g/100 g) × 100 g = 3.47 g
Number of moles of H = mass of hydrogen / molar mass of H = 3.47 g / 1.01 g/mol ≈ 3.436 mol

O (Oxygen):
Percentage of oxygen = 100% - (percentage of carbon + percentage of hydrogen)
Percentage of oxygen = 100% - (41.39% + 3.47%) = 100% - 44.86% = 55.14%

Now, let's calculate the number of moles of oxygen assuming we have 100 grams of maleic acid:

Mass of oxygen = (55.14 g/100 g) × 100 g = 55.14 g
Number of moles of O = mass of oxygen / molar mass of O = 55.14 g / 16.00 g/mol ≈ 3.446 mol

Now that we have the number of moles for each element, we need to find the simplest whole-number ratio of moles to determine the empirical formula. Divide each number of moles by the smallest number of moles (which is approximately 3.436 in this case) and round to the nearest whole number:

C: 3.448 mol / 3.436 mol ≈ 1
H: 3.436 mol / 3.436 mol = 1
O: 3.446 mol / 3.436 mol ≈ 1

Therefore, the empirical formula of maleic acid is CH2O.

To find the molecular formula from the empirical formula, we need to know the molecular mass of maleic acid. The molecular mass can be calculated by adding up the molar masses of all the atoms in the empirical formula:

Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol

Molar mass of CH2O = (12.01 g/mol × 1) + (1.01 g/mol × 2) + (16.00 g/mol × 1) ≈ 30.03 g/mol

To find the ratio between the molecular mass and the empirical formula mass, divide the molar mass of maleic acid by the empirical formula mass:

Molecular formula ratio = molecular mass / empirical formula mass
Molecular formula ratio = 115.5 g/mol / 30.03 g/mol ≈ 3.85

Since the ratio is approximately 3.85, we need to multiply the empirical formula by this number to get the molecular formula:

Molecular formula = (CH2O) × 3.85 ≈ C3H3O3

Therefore, the molecular formula of maleic acid is C3H3O3.