An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% Carbon and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20 celsius is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g of water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a non electrolyte.

C = 31.57%

H = 5.30%
Therefore, O = 100%-31.57-5.30 = about 63%.

Take a 100g sample which will give you
31.57g C atoms.
5.30g H atoms.
63g O atoms.

Convert to mols and I'll estimate.
31.57/12 = about 2.63 mols C
5.30/1 = 5.30 mols H
63/16 = about 3.94 mols O

Now find the ratio to each other with the smallest being 1.0. The easy way to do that is to divide the smallest number by itself and follow suit with the other numbers.
2.63/2.63 = 1.0 mols C
5.30/2.63 = 2.01 mols H
3.94/2.63 = 1.49 mols O

Normally one would round these off to the nearest whole number but the O is too much to round. We can see that is about 1.5 so the ratio is 2C, 4H, 3O and the empirical formula is C2H4O3 (I just multilied the 1, 2, and 1.5 by 2 to get whole numbers). That makes the mass of the empirical formula about 76.

For the molar mass we use the freezing point data.
delta T = Kf*m
5.20 = 1.86*m
Solve for m = molality

Then m = mols/kg solvent. You know m and kg solvent (0.025g), solve for mols.

Then mols = grams/molar mass. You know mols and grams, solve for molar mass
I approx about 150 for molar mass.

empirical formula mass x ? = 150
76 x ? = 150 so ? (a whole number) = 150/76 = about 1.97 which rounds to 2.0 as a whole number so the molecular formula is
(C2H4O3)2 or C4H8O6 with a molar mass of 152. The 150 determined by the freezing point method is only CLOSE to the right answer and it allows you to find that whole number of 2 which then gives you the correct molecular formula and correct molar mass.

Well, it seems like we have a scientific mystery on our hands! Let's tackle it one step at a time.

First, let's find the empirical formula of the compound. From the combustion analysis, we know that the compound contains 31.57% carbon and 5.30% hydrogen. We can assume that the remainder is oxygen. To find the empirical formula, we need to convert these percentages into moles.

Assuming we have 100g of the compound, we have 31.57g of carbon and 5.30g of hydrogen. To find the number of moles, we divide the mass by the molar mass of each element:

Molar mass of carbon (C) = 12 g/mol
Molar mass of hydrogen (H) = 1 g/mol

Moles of carbon = 31.57g / 12 g/mol ≈ 2.63 mol
Moles of hydrogen = 5.30g / 1 g/mol ≈ 5.30 mol
Moles of oxygen ≈ (100g - (31.57g + 5.30g)) / 16 g/mol ≈ 3.76 mol

Next, we need to find the simplest whole-number ratio of the elements in the compound. To do this, we divide the number of moles of each element by the smallest number of moles (which is 2.63 moles in this case):

Carbon: 2.63 mol / 2.63 mol ≈ 1 mol
Hydrogen: 5.30 mol / 2.63 mol ≈ 2 mol
Oxygen: 3.76 mol / 2.63 mol ≈ 1.43 mol

Since we want to find whole numbers, we multiply these values by 2 to get:

Carbon: 1 mol x 2 = 2
Hydrogen: 2 mol x 2 = 4
Oxygen: 1.43 mol x 2 = 2.86 ≈ 3

So, the empirical formula of the compound is C2H4O3.

Now, let's move on to finding the molar mass of the compound. We can do this using the freezing-point depression method. We know that 10.56 g of the compound was dissolved in 25.0 g of water, and the freezing point was depressed by -5.20 °C. By using the equation:

ΔT = Kf * m

Where ΔT is the freezing point depression, Kf is the freezing point depression constant (which is specific for water and equals 1.86 °C/m), and m is the molality of the solution (moles of solute per kilogram of solvent). In this case, we have 10.56 g of solute (the compound) and 25.0 g of solvent (water).

First, we need to find the molality of the solution:

Moles of the compound = 10.56 g / molar mass of the compound
Moles of water = 25.0 g / 18 g/mol (molar mass of water)

Molality (m) = Moles of compound / kg of water = (moles of compound) / 0.025 kg

Now, let's find ΔT:

ΔT = Kf * m

And let's substitute the known values:

-5.20 °C = (1.86 °C/m) * (moles of compound / 0.025 kg)

Now, we just need to solve for the moles of the compound:

moles of compound = (-5.20 °C * 0.025 kg) / 1.86 °C/m

Now that we have the moles of the compound, we can calculate the molar mass:

Molar mass of the compound = 10.56 g / moles of compound

Finally, to find the molecular formula of the compound, we need to know the molar mass of the empirical formula. By comparing the molar mass of the empirical formula to the molar mass we calculated, we can determine the ratio:

Molar mass of the molecular formula = molar mass of the empirical formula * n

Where n is the number of empirical formula units in the molecular formula.

So, after all these calculations, we will finally have the empirical formula, molar mass, and molecular formula of the compound! Phew, that was quite a journey!

To determine the empirical formula, molar mass, and molecular formula of the compound, we will follow these steps:

Step 1: Calculate the number of moles of water used:
To do this, we can use the formula:

Moles = Mass / Molar mass

We are given that the mass of water used is 25.0 g and the molar mass of water is 18.015 g/mol.

Moles of water = 25.0 g / 18.015 g/mol = 1.386 mol

Step 2: Calculate the freezing point depression:
The freezing point depression is the difference between the freezing point of the solution and the freezing point of the pure solvent. In this case, the freezing point depression is given as -5.20°C.

Step 3: Calculate the molality (m) of the solution:
Molality (m) is defined as the number of moles of solute divided by the mass of the solvent (in kilograms). In this case, we can calculate it as:

m = moles of solute / mass of solvent (in kg)

m = 1.386 mol / (25.0 g / 1000 g/kg) = 55.44 mol/kg

Step 4: Calculate the change in freezing point, ΔTf:
The change in freezing point, ΔTf, can be calculated using the formula:

ΔTf = Kf • m

Kf is the freezing point depression constant for water, which is 1.86°C•kg/mol.

ΔTf = 1.86°C•kg/mol • 55.44 mol/kg = 103.03°C

Step 5: Calculate the molality (m) of the solute in the original solution:
We can use the formula:

ΔTf = Kf • m

Rearranging the formula, we get:

m = ΔTf / Kf

m = 103.03°C / 1.86°C•kg/mol = 55.44 mol/kg

Step 6: Calculate the moles of solute in the original solution:
Using the formula:

Moles of solute = molality x mass of solvent (in kg)

Moles of solute = 55.44 mol/kg x 0.025 kg = 1.386 mol

Step 7: Calculate the empirical formula:
Since we know the mass percentages of carbon and hydrogen, we can calculate the moles of carbon and hydrogen in the compound using these percentages.

Given mass % of carbon = 31.57% and mass % of hydrogen = 5.30%

Moles of carbon = mass % of carbon x mass of compound / molar mass of carbon
Moles of hydrogen = mass % of hydrogen x mass of compound / molar mass of hydrogen

We need to determine the empirical formula, which means finding the simplest whole-number ratio of the atoms in the compound. To do this, we divide the moles of each element by the smallest number of moles obtained.

Step 8: Calculate the molar mass:
To calculate the molar mass, we need to determine the empirical formula and use it to find the mass of one mole of the compound.

Step 9: Calculate the molecular formula:
To find the molecular formula, we need to compare the molar mass of the compound with the molar mass calculated in the previous step.

Let's perform the calculations:

- Step 7:
Moles of carbon = 31.57% x mass of compound / molar mass of carbon
Moles of carbon = 0.3157 x mass of compound / 12.01 g/mol

Moles of hydrogen = 5.30% x mass of compound / molar mass of hydrogen
Moles of hydrogen = 0.053 x mass of compound / 1.008 g/mol

Dividing both equations by the smallest number of moles obtained (0.053):

Moles of carbon = 0.3157 / 0.053 = 5.96
Moles of hydrogen = 0.053 / 0.053 = 1

Therefore, the empirical formula is CH6.

- Step 8:
To calculate the molar mass, we need to determine the empirical formula's mass. The empirical formula CH6 consists of 1 carbon atom and 6 hydrogen atoms.

Molar mass of carbon = 12.01 g/mol
Molar mass of hydrogen = 1.008 g/mol

Molar mass of CH6 = (1 x molar mass of carbon) + (6 x molar mass of hydrogen)
Molar mass of CH6 = (1 x 12.01 g/mol) + (6 x 1.008 g/mol) = 18.09 g/mol

Therefore, the molar mass of the compound is 18.09 g/mol.

- Step 9:
To find the molecular formula, we need to compare the molar mass of the compound with the molar mass of the empirical formula (18.09 g/mol).

The molar mass of the compound is given as an unknown value. Without knowing the specific molar mass of the compound, we cannot determine the molecular formula.

To determine the empirical formula, molar mass, and molecular formula of the compound, we need to follow a step-by-step process. Let's break it down:

Step 1: Calculate the moles of carbon and hydrogen in the compound.
- Given that the percent composition of carbon is 31.57%, assuming a 100g sample, there would be 31.57g of carbon.
- Given that the percent composition of hydrogen is 5.3%, assuming a 100g sample, there would be 5.3g of hydrogen.

To find the moles, we need to divide the mass of each element by its molar mass:
- The molar mass of carbon (C) is 12.01 g/mol.
- The molar mass of hydrogen (H) is 1.01 g/mol.

Moles of Carbon = (31.57g C) / (12.01 g/mol C)
Moles of Hydrogen = (5.3g H) / (1.01 g/mol H)

Step 2: Determine the moles of oxygen in the compound.
- The compound contains only carbon, hydrogen, and oxygen. To find the moles of oxygen, we need to subtract the moles of carbon and hydrogen from the total moles present in the compound. This is based on the law of conservation of mass.

Moles of Oxygen = Total Moles - (Moles of Carbon + Moles of Hydrogen)

Step 3: Determine the empirical formula.
- Now that we have the moles of each element, we need to find their ratio and simplify it to obtain the empirical formula.
- Divide the moles of each element by the smallest number of moles attained in step 2.
- If all the numbers are close to whole numbers, we can multiply them to get a whole number ratio.

Step 4: Calculate the molar mass of the empirical formula.
- To determine the molar mass of the empirical formula, we need to find the sum of the atomic masses for each element in the empirical formula.
- Multiply the atomic mass of each element by the number of atoms in the empirical formula and sum them up.

Step 5: Determine the molecular formula.
- The molar mass determined experimentally is needed to find the molecular formula.
- Divide the molar mass of the compound by the molar mass of the empirical formula obtained in step 4.
- If the value obtained is a whole number, multiply the subscript in the empirical formula by this value to get the molecular formula.

Let's perform the calculations now:

Step 1: Calculate the moles of carbon and hydrogen.
Moles of Carbon = (31.57g C) / (12.01 g/mol C) = 2.627 moles of C
Moles of Hydrogen = (5.3g H) / (1.01 g/mol H) = 5.25 moles of H

Step 2: Determine the moles of oxygen in the compound.
Assuming a 100g sample, the mass of oxygen would be:
100g - (31.57g C + 5.30g H) = 63.13g of oxygen

Moles of Oxygen = (63.13g O) / (16.00 g/mol O) = 3.945 moles of O

Step 3: Determine the empirical formula.
Divide the moles of each element by the smallest number of moles (2.627 in this case):
C: 2.627 moles / 2.627 = 1 mole
H: 5.25 moles / 2.627 ≈ 2 moles
O: 3.945 moles / 2.627 ≈ 1.5 moles

Multiplying by 2 to convert all values to whole numbers, we get the empirical formula:
C2H4O3

Step 4: Calculate the molar mass of the empirical formula.
Molar mass of C2H4O3 = (2 x 12.01 g/mol C) + (4 x 1.01 g/mol H) + (3 x 16.00 g/mol O) ≈ 88.08 g/mol

Step 5: Determine the molecular formula.
To find the molecular formula, we need to calculate the ratio of the molar mass determined experimentally to the molar mass of the empirical formula.
Molar mass = 88.08 g/mol (determined experimentally)

Molecular formula = (88.08 g/mol) / (empirical formula mass ≈ 88.08 g/mol)
Since the ratio is approximately 1, the empirical formula is also the molecular formula.

Therefore, the empirical formula and molecular formula of the compound are both C2H4O3.