The molecular formula of a compound containg only carbon and hydrogen is to be determined. When a sample of the compound is burned on oxygen gas, 7.2 grams of water and 7.2 liters of carbon dioxide gas are produced(measured at STP). When 0.600grams of the compound are dissolved in 100.0 grams of CHCl3 the mixture frezes at
-64.0degrees C. The normal freezing point of CHCl3 is -63.5 degrees C and its molal freezing point depression constant is 4.68C/m.
1) What is the empirical formula of the compound?
2) Calculate the molecular weight of the compound.
3) Determine the molecular formula of the compound.
4) Wtite a balance equation for the combustion of the compound and calculate the mass of oxygen gas that would be required for the combustion described above.
Convert grams H2O and L CO2 to moles H and moles C.
moles H2O = grams/molar mass = 7.2/18.015 = 0.40 and moles H = 2x that = 0.80
moles CO2 = 7.2L/22.4 = 0.32 and moles C = 0.32
Take the ratio of the two to each other. The easy way to do that is to divide the smaller number by itself, then divide the other number by the same small number.
That should give you 1 for C and 2.5 for H. Normally you would round to whole numbers but we can't throw away as much as 1/2 so we note that 1:2.5 is the same as 2.0 to 5.0 in whole numbers. The empirical formula is C2H5.
2). delta T = Kf*molality
You know delta T and Kf, solve for molality.
molality = moles/kg solvent
You know m and kg solvent, solve for moles.
moles = grams/molar mass
You know moles and grams, solve for molar mass. I get approximately 56 but you need to confirm that.
3) empirical formula mass of C2H5 is about 29. Confirm that.
molar mass is about 56 from the freezing point data. The question actually is, how many of the C2H5 units will fit into the 56.
56/29 = 1.93; round that to the nearest whole number or 2.0 so the molecular formula is (C2H5)2 which can be rewritten as C4H10.
I'll leave 4 for you to do. This is a simple stoichiometry problem. If you need help with that here is a solved stoichiometry example. Just follow the steps. Post your work if you get stuck.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the empirical formula of the compound, we need to calculate the mole ratio between carbon and hydrogen based on the given information.
1) First, let's calculate the number of moles of water produced during combustion. The molar mass of water is 18 g/mol, so 7.2 grams of water would be equal to 7.2 g / 18 g/mol = 0.4 moles of water.
Since water is composed of two hydrogen atoms per molecule, the number of moles of hydrogen in the compound is also 0.4 moles.
2) Next, let's calculate the number of moles of carbon dioxide produced during combustion. The molar mass of carbon dioxide is 44 g/mol, so 7.2 liters of carbon dioxide at STP can be converted to moles using the ideal gas law. At STP, 1 mole of gas occupies 22.4 liters.
So, the number of moles of carbon dioxide is 7.2 L / 22.4 L/mol = 0.32 moles of carbon dioxide.
Since carbon dioxide is composed of one carbon atom per molecule, the number of moles of carbon in the compound is also 0.32 moles.
3) Now, we can calculate the empirical formula by dividing the number of moles of each element by the smallest number of moles.
Dividing 0.32 moles of carbon by 0.32 gives us 1 mole of carbon.
Dividing 0.4 moles of hydrogen by 0.32 gives us 1.25 moles of hydrogen.
However, empirical formulas must have whole number ratios, so we need to multiply each element by a factor that will give us the nearest whole number ratio.
Multiplying both the number of moles of carbon and hydrogen by 4 gives us 4 moles of carbon and 5 moles of hydrogen.
Therefore, the empirical formula of the compound is CH4.
4) To calculate the molecular weight of the compound, we need to know the molecular weight of the empirical formula, which is the sum of the atomic weights of the constituent atoms.
The atomic weight of carbon (C) is 12.01 g/mol, and the atomic weight of hydrogen (H) is 1.01 g/mol.
So the molecular weight of CH4 is (12.01 g/mol × 1) + (1.01 g/mol × 4) = 16.05 g/mol.
5) To determine the molecular formula of the compound, we need the molecular weight of the compound. The molecular weight has been calculated as 16.05 g/mol in the previous step.
The molecular formula is a multiple of the empirical formula, so we need to find the ratio between the molecular weight and the empirical formula weight.
The empirical formula weight is (12.01 g/mol × 1) + (1.01 g/mol × 4) = 16.05 g/mol.
Dividing the molecular weight (16.05 g/mol) by the empirical formula weight (16.05 g/mol) gives us a ratio of 1.
Therefore, the molecular formula of the compound is also CH4.
6) The balanced equation for the combustion of the compound can be written as follows:
CnHm + (n+ m/4)O2 → nCO2 + m/2H2O
Since we have determined that the empirical formula is CH4, the balanced equation becomes:
CH4 + 2O2 → CO2 + 2H2O
7) To calculate the mass of oxygen gas required for the combustion, we can use stoichiometry based on the balanced equation.
From the equation, we can see that 1 mole of CH4 requires 2 moles of O2. The molar mass of O2 is 32 g/mol.
So, the moles of O2 required for combustion can be calculated as: (0.600 g CH4 / 16.05 g/mol) × (2 mol O2 / 1 mol CH4) = 0.0748 mol O2.
To convert the moles of O2 to grams, we multiply by the molar mass: 0.0748 mol O2 × 32 g/mol = 2.39 g of O2.
Therefore, the mass of oxygen gas required for the combustion described above is 2.39 grams.