In a eudiometer, 20 ml of a gaseous hydrocarbon A is mixed with 150 ml of oxygen gas. The combustion of A is initiated by an electric spark. After cooling and condensing the vapour, the volume of the gaseous mixture is 110 ml, of which 80 ml causes turbidity of limewater. The remaining volume of the gas makes a glowing splint to catch fire.
determine the formula of A
There may be an easier (better?) way to do this. If so I would like to see it. Here is what I did. The fact that the remaining gas lighted a glowing splint means the remaining gas is O2 which merans the hydrocarbon was the limiting reagent.
...........CxHy + O2 ==> CO2 + H2O
I.........20 mL....150 mL..0...........0
C.......-20...........-x...................
After combustion total volume is 110 mL.That will be the CO2 + the unused O2 gas.
Volume CO2 from the turbidity test is 80
Volume of O2 left uncombusted = 110-80 = 30
Volume of O2 used = 150-30 = 120 mL
Ratio hydrocarbon to O2 is 1 to 6 (that's 20 mL HC used 120 mL O2). Note: I never used a eudiometer (EVER) in my career so I'm assuming the volume readings are just that and I've made no corrections for the condensation of the water vapor. and how that might affect the volume reading. I then wrote balanced equations for the first four saturated hydrocarbons (alkanes) and found no 1:6 ratios. Then I tried the first few alkenes. You should balance the first 4 to confirm that. C4H8 did it; i.e.,
C4H8 + 6O2 ==> 4CO2 + 4H2O
I think the unknown HC is C4H8.
To determine the formula of hydrocarbon A, we can use the information provided about the gaseous mixture before and after combustion.
Step 1: Calculate the volume of oxygen gas used in the combustion:
Initial volume of oxygen = Total volume of gaseous mixture (before condensation) - Volume causing turbidity in limewater
Initial volume of oxygen = 110 ml - 80 ml
Initial volume of oxygen = 30 ml
Step 2: Calculate the volume of carbon dioxide produced:
Volume of carbon dioxide = Total volume of gaseous mixture (before condensation) - Remaining volume after ignition
Volume of carbon dioxide = 110 ml - Remaining volume after ignition
Volume of carbon dioxide = 110 ml - 30 ml
Volume of carbon dioxide = 80 ml
Step 3: Calculate the moles of carbon dioxide produced:
Moles of carbon dioxide = Volume of carbon dioxide / 22.4 liters (1 mole of any gas at STP)
Moles of carbon dioxide = 80 ml / 22.4
Moles of carbon dioxide = 3.57 moles
Step 4: Determine the moles of carbon in the hydrocarbon:
Since one mole of hydrocarbon reacts with one mole of oxygen to produce one mole of carbon dioxide, the moles of carbon in the hydrocarbon are also 3.57 moles.
Step 5: Calculate the empirical formula of hydrocarbon A:
The empirical formula represents the simplest whole-number ratio of the atoms in the compound. In this case, we have 3.57 moles of carbon. To get the simplest whole-number ratio, we need to divide by the smallest moles present.
Dividing 3.57 moles by 3.57 gives us 1 mole of carbon (C).
Therefore, the empirical formula of hydrocarbon A is C.
Since we only have carbon as the empirical formula, we can deduce that the formula of hydrocarbon A is Cn, where n represents the number of carbon atoms in the compound.
To determine the formula of hydrocarbon A, we need to analyze the given information and work through a series of steps. Here's how you can do it:
Step 1: Determine the balanced equation for the combustion reaction.
First, we need to write a balanced chemical equation for the combustion of hydrocarbon A with oxygen. Let's assume the formula of hydrocarbon A is CxHy, where x and y are integers representing the number of carbon and hydrogen atoms.
The balanced equation for the combustion of hydrocarbon A would be:
CxHy + (x + y/4)O2 → xCO2 + y/2H2O
Step 2: Calculate the moles of gaseous hydrocarbon A.
Using the ideal gas law, we can find the number of moles of hydrocarbon A. The equation is:
n = V/P, where n is the number of moles, V is the volume in liters, and P is the pressure in atm.
Given that the initial volume of hydrocarbon A is 20 ml, we need to convert it to liters:
20 ml = 20/1000 = 0.02 L
We can assume the pressure is constant since it's not specified. Therefore, we can use PV = nRT as a proportional relationship.
Step 3: Calculate the moles of oxygen gas.
Given that the volume of oxygen gas is 150 ml, let's convert it to liters:
150 ml = 150/1000 = 0.15 L
Again, assuming constant pressure, we can use the same proportional relationship.
Step 4: Calculate the moles of carbon dioxide (CO2) and water (H2O) produced.
From the balanced equation in Step 1, we can see that for every mole of hydrocarbon A burned, we get x moles of CO2 and y/2 moles of H2O.
Using the equation n = V/P and the volume of the gaseous mixture after combustion (110 ml = 0.11 L), we can calculate the moles of CO2 and H2O.
Step 5: Determine the ratio of x and y.
From the volume of the gaseous mixture that causes turbidity of limewater (80 ml = 0.08 L) and the volume that makes a glowing splint catch fire (30 ml = 0.03 L), we can determine the mole ratio and hence the ratio of x and y.
Since CO2 is the gas responsible for turbidity of limewater, we can conclude that the volume ratio of CO2 to hydrocarbon A is 0.08 L/0.02 L = 4.
Since H2O is the gas responsible for making a glowing splint catch fire, we can conclude that the volume ratio of H2O to hydrocarbon A is 0.03 L/0.02 L = 1.5.
Using these ratios, we can calculate the values of x and y.
Step 6: Determine the empirical formula.
To find the empirical formula, we need to divide the number of atoms of carbon and hydrogen by their greatest common divisor (GCD).
Step 7: Determine the molecular formula.
Given that the hydrocarbon A is a gaseous compound, its molecular formula may be the same as its empirical formula.
By following these steps, you should be able to determine the empirical and molecular formula of hydrocarbon A.