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.
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.
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.
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
I will leave #4 for you. If you have a problem, post your work and tell us what you don't understand about it.
In number 2.Is the delta Tf = to -2.84?
I came up with 180 but confused with how -2.84 is delta Tf
The chloride of an unknown metal is believed to have the formula . A 1.603 sample of the compound is found to contain .
To determine the empirical formula of the compound, we need to find the mole ratios of the elements present in it. These ratios can be obtained by comparing the number of moles of each element in the compound.
1. Calculate the number of moles of CO2 and H2O produced:
- Moles of CO2 = mass of CO2 / molar mass of CO2 = 1.47 g / 44.01 g/mol = 0.0334 mol
- Moles of H2O = mass of H2O / molar mass of H2O = 0.600 g / 18.02 g/mol = 0.0333 mol
2. Determine the number of moles of carbon and hydrogen in the original compound:
- Moles of carbon = moles of CO2 = 0.0334 mol
- Moles of hydrogen = moles of H2O * 2 (since there are 2 hydrogen atoms per water molecule) = 0.0333 mol * 2 = 0.0666 mol
3. Find the number of moles of oxygen by subtracting the moles of carbon and hydrogen from the total moles in the compound:
- Moles of oxygen = Total moles - (Moles of carbon + Moles of hydrogen) = 0.0334 + 0.0666 = 0.1000 mol
4. Divide the number of moles of each element by the smallest number of moles to get the mole ratios:
- Simplifying by dividing all moles by 0.0334, we find the empirical formula to be C1H2O3.
To calculate the molecular weight of the compound, we need to know the molecular formula. So, we'll move on to the next question to determine the molecular formula.
To determine the molecular formula of the compound, we need to compare the empirical formula weight with the given molar mass of the compound.
2. Calculate the empirical formula weight:
- Empirical formula weight = (atomic weight of carbon * number of carbon atoms) + (atomic weight of hydrogen * number of hydrogen atoms) + (atomic weight of oxygen * number of oxygen atoms)
- Empirical formula weight = (12.01 g/mol * 1) + (1.01 g/mol * 2) + (16.00 g/mol * 3) = 12.01 + 2.02 + 48.00 = 62.03 g/mol
2. Calculate the molecular weight:
- Molecular weight = empirical formula weight * n (where n is an integer)
- From the given mass of the compound and the empirical formula weight, we can calculate n as follows:
- n = mass of the compound / empirical formula weight = 1.00 g / 62.03 g/mol = 0.016 mol / mol
- Multiplying the empirical formula by the value of n, we get the molecular formula: C1H2O3 * 0.016 = C0.016H0.032O0.048
Since the molecule formula C0.016H0.032O0.048 is not a whole number ratio, we will multiply it by an appropriate integer to obtain a whole number ratio.
To write a balanced equation for the combustion reaction described above and calculate the volume of oxygen gas required, we'll need to use the stoichiometry of the reaction.
First, let's write the balanced equation:
C1H2O3 + O2 → CO2 + H2O
Since we're given the masses of CO2 and H2O produced, we'll calculate the number of moles of CO2 and H2O and use their mole ratios to find the number of moles of the compound.
1. Moles of CO2 = mass of CO2 / molar mass of CO2 = 1.47 g / 44.01 g/mol = 0.0334 mol
2. Moles of H2O = mass of H2O / molar mass of H2O = 0.600 g / 18.02 g/mol = 0.0333 mol
From the balanced equation, we see that the mole ratio of CO2 to the compound is 1:1. Therefore, the number of moles of the compound is also 0.0334 mol.
Next, let's calculate the number of moles of oxygen gas required for the combustion reaction:
From the balanced equation, we see that the mole ratio of oxygen gas to the compound is 1:1. Therefore, the number of moles of oxygen gas required is also 0.0334 mol.
To calculate the volume of oxygen gas at STP, we can use Avogadro's Law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules.
1 mol of any gas at STP occupies 22.4 liters. Therefore, 0.0334 mol of oxygen gas will occupy:
Volume = (0.0334 mol * 22.4 L/mol) = 0.748 L (rounded to three decimal places)
So, the volume of oxygen gas required to complete the combustion of 1.00 g of the compound is approximately 0.748 liters at STP.