See bottom Patrick

A certain compound contains carbon, hydrogen , and oxygen. When 1.00 grams of this compound are burned in oxygen gas 1.47 grams of CO2 and 0.600 grams of water are formed. When 2.75 grams of the compound are dissolved in 10.0 grams of water the resulting solution freezes at –2.84oC.

1.Determine the empirical formula of the compound.

2.Calculate the molecular weight of the compound.

3.Determine the molecular formula of the compound.

4.Write a balanced equation for the combustion reaction described above and calculate the volume of oxygen gas required to complete the combustion. Assume the gas is measured at STP.



Chemistry - DrBob222, Sunday, January 16, 2011 at 5:32pm
I would convert 1.47 g CO2 to grams C, then to percent C. Same for H2O to hydrogen. Then add the two and subtract from 100 to obtain percent oxygen.
1.47 g CO2 x (atomic mass C/molar mass CO@) = 1.47 x (12/44) = about 40 but you should do it more accurately.

For H2O, that is
0.600 x (2/18) = about 6; again, confirm that.

then 100-6-40 = 54.

Now take a 100 g sample and you will have 40 g C, 6 g H, 54 g O. Convert those to moles.
40/12 = about 3.3
6/1 = about 6
54/16 = about 3.3
You can see the ratio of these elements is 1C, 2H, 1 O for the empirical formula of CH2O.
You should redo the math and get better numbers.

Chemistry - DrBob222, Sunday, January 16, 2011 at 5:38pm
2. Use the freezing point data to determine the approximate molecular weight of the compound.
delta T = Kf*m
Solve for m

m = moles/kg solvent
Solve for moles.

moles = grams/molar mass
Solve for molar mass.

I get about 170 or so but you need to do it more accurately than that.

Chemistry - DrBob222, Sunday, January 16, 2011 at 5:41pm
3. You want to see how many of the empirical units are in the molecular unit.
Empirical formula mass = CH2O = 12 + 2 + 16 = 30
Molecular weight from #2 = about 170
170/30 = 5.7 or so. Round to a whole number of 6 so the formula is
(CH2O)6 or you can re-write it as C6H12O6


Chemistry - DrBob222, Sunday, January 16, 2011 at 5:42pm
I will leave #4 for you. If you have a problem, post your work and tell us what you don't understand about it.

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Chemistry - Patrick, Monday, January 17, 2011 at 8:00pm
In number 2.Is the delta Tf = to -2.84?
I came up with 180 but confused with how -2.84 is delta Tf

See above.

To determine the empirical formula of the compound, you need to calculate the percentage composition of carbon, hydrogen, and oxygen in the compound.

For carbon dioxide (CO2):
- The molar mass of CO2 is 44 g/mol.
- Multiply the mass of CO2 (1.47 g) by the ratio of the atomic mass of carbon to the molar mass of CO2 [(12 g/mol) / (44 g/mol)].
- This gives you the mass of carbon in the compound, which is approximately 0.40 g.

For water (H2O):
- The molar mass of H2O is 18 g/mol.
- Multiply the mass of water (0.600 g) by the ratio of the number of hydrogen atoms in water to the molar mass of water [(2) / (18)].
- This gives you the mass of hydrogen in the compound, which is approximately 0.067 g.

Subtracting the masses of carbon and hydrogen from the initial mass of the compound (1.00 g) gives you the mass of oxygen, which is approximately 0.533 g.

Now, convert the masses of carbon, hydrogen, and oxygen to moles by dividing each by their respective atomic masses:
- Carbon: 0.40 g / 12 g/mol = approximately 0.033 mol
- Hydrogen: 0.067 g / 1 g/mol = approximately 0.067 mol
- Oxygen: 0.533 g / 16 g/mol = approximately 0.033 mol

Divide each of the mole values by the smallest mole value (0.033 mol) to obtain the mole ratios:
- Carbon: 0.033 mol / 0.033 mol = 1
- Hydrogen: 0.067 mol / 0.033 mol = 2
- Oxygen: 0.033 mol / 0.033 mol = 1

The empirical formula of the compound is CH2O.

To calculate the molecular weight of the compound, you need to use the freezing point depression data.

The freezing point depression (delta Tf) is given as -2.84 degrees Celsius.

Using the formula: delta Tf = Kf * m, where Kf is the freezing point depression constant and m is the molality of the solution, we can solve for m.

Since the molality (m) is moles of solute divided by mass of solvent in kg, we need to calculate the moles of solute first.

The mass of solute is given as 2.75 grams, and the molar mass of the empirical formula is approximately 30 g/mol.

Thus, the number of moles of solute is 2.75 g / 30 g/mol = approximately 0.0917 mol.

Now, we need to calculate the molality using the mass of solvent. The mass of solvent is given as 10.0 grams, which is equal to 0.010 kg.

Thus, the molality (m) = 0.0917 mol / 0.010 kg = approximately 9.17 mol/kg.

Now, rearrange the freezing-point depression equation to solve for the molecular weight.

m = moles / kg solvent
moles = grams / molar mass

Plugging in the values, we get:
9.17 mol/kg = (molecular weight) / 0.010 kg

Solve for the molecular weight:
(molecular weight) = 9.17 mol/kg * 0.010 kg = approximately 91.7 g/mol

The molecular weight of the compound is approximately 91.7 g/mol.

To determine the molecular formula of the compound, we compare the empirical formula mass (30 g/mol) to the actual molecular weight (91.7 g/mol).

Divide the molecular weight by the empirical formula mass to calculate the number of empirical formula units in the molecular formula:
91.7 g/mol / 30 g/mol = approximately 3.06

Round the number to the nearest whole number, which is 3.

The molecular formula is therefore (CH2O)3, or you can write it as C3H6O3.

For the balanced equation of the combustion reaction, you need to write the equation based on the given products.

From the information given, we know that 1.47 grams of CO2 is formed. Using the molar mass of CO2 (44 g/mol), we can calculate the number of moles of CO2 produced:
1.47 g CO2 / 44 g/mol = approximately 0.033 mol CO2

Since carbon in the compound forms CO2, there must be the same number of moles of carbon in the compound. Therefore, the number of moles of carbon is also 0.033 mol.

Since 1 mole of C reacts with 1 mole of O2 to form 1 mole of CO2 in the combustion reaction, the balanced equation for the combustion reaction can be written as:

C + O2 → CO2

Now, to calculate the volume of oxygen gas required to complete the combustion, we can use the molar volume of a gas at STP, which is 22.4 L/mol.

From the balanced equation, we can see that for every 1 mole of oxygen gas (O2), 1 mole of CO2 is produced.

Therefore, the volume of oxygen gas required would be equal to the number of moles of oxygen gas, which is also 0.033 mol, multiplied by the molar volume of a gas at STP:

0.033 mol O2 * 22.4 L/mol = approximately 0.7392 L of oxygen gas.