- #1
zachary570
- 5
- 4
- Homework Statement
- In a given diffusion apparatus, 15.0 mL of HBr gas diffuses in 1.0 min. In the same apparatus and under the same conditions, 20.3 mL of an unknown gas diffuses in 1.0 min. The unknown gas is a hydrocarbon. Find its molecular formula.
- Relevant Equations
- Graham's law of effusion: ##\frac{rate_a}{rate_b} = \sqrt{\frac{M_b}{M_a}}##
Ideal Gas Law: ##PV = nRT##
Hello,
I am having some trouble finishing this problem from my textbook and would like some insight on the problem. We are looking for the molecular formula of a hydrocarbon. So, if we let x = number of C atoms and y = number of H atoms then we need to find ##C_xH_y##. From the problem statement we can find the rates of diffusion for both compounds:
$$rate_{HBr} = \frac{V}{t} = 15.0 \frac{mL}{min}$$
and
$$rate_{C_xH_y} = 20.3 \frac{mL}{min}$$
We know that the molar mass of HBr = 80.912 amu from the periodic table so we can use Graham's law to find the molar mass of the hydrocarbon.
$$\frac{rate_{HBr}}{rate_{C_xH_y}} = \sqrt{\frac{M_{C_xH_y}}{M_{HBr}}}$$
$$M_{C_xH_y} = M_{HBr} {\left(\frac{rate_{HBr}}{rate_{C_xH_y}}\right)}^2$$
$$M_{C_xH_y} = 80.912 amu {\left(\frac{15.0 \frac{mL}{min}}{20.3 \frac{mL}{min}}\right)}^2$$
$$M_{C_xH_y} = 44.17772817amu$$
So,
$$12.011x + 1.008y = 44.17772817$$
Here is where I am having trouble. I know that you need at least as many equations as you do unknowns to solve for them but I can't seem to find another equation. I thought using the ideal gas law, really Avogadro's Law, to find the number of moles in the apparatus. When I do that I get ##x + y = 1.353 moles## (assuming the initial n = 1) but this system gives a negative y value so it is clearly wrong. From the back of the book I know the answer should be ##C_3H_8##. If I were to just assume that the number of carbon atoms were as many as there could be then you get 3, from ##\lfloor \frac{44.17772817}{12.011} \rfloor##, and this gives the correct answer. But that doesn't feel like the right approach. If you assumed that there were only 2 carbon atoms then there would need to be 20 hydrogen atoms and the result would work just as well. Any advice on what I should be thinking about would be much appreciated.
Also, when googling this question the following video comes up. I used it to get to the point where I am now but I have two questions on their methods. (1) How did they know that ##rate = \frac{V}{t}##? I understand why it makes sense from what diffusion means but no such relation shows up in my textbook. I am using "Chemistry: A Molecular Approach, Fourth Edition by Nivaldo Tro" and the only relations given is that rate is inversely proportional to the square of molar mass as well as Graham's law when it discusses effusion and diffusion. I have looked it up and it seems to come from physics when dealing with liquids but I haven't learned that yet so how would I have known that without outside help? (2) When solving for x and y, she gets x as a function of y and then plugs that back into the original equation. The way she does it however seems to just be a rounding error. If she wouldn't have rounded then the expression would just be 58 = 58 which doesn't answer anything. Instead she uses it to solve for x and y. Isn't this wrong?
Thanks,
zachary570
I am having some trouble finishing this problem from my textbook and would like some insight on the problem. We are looking for the molecular formula of a hydrocarbon. So, if we let x = number of C atoms and y = number of H atoms then we need to find ##C_xH_y##. From the problem statement we can find the rates of diffusion for both compounds:
$$rate_{HBr} = \frac{V}{t} = 15.0 \frac{mL}{min}$$
and
$$rate_{C_xH_y} = 20.3 \frac{mL}{min}$$
We know that the molar mass of HBr = 80.912 amu from the periodic table so we can use Graham's law to find the molar mass of the hydrocarbon.
$$\frac{rate_{HBr}}{rate_{C_xH_y}} = \sqrt{\frac{M_{C_xH_y}}{M_{HBr}}}$$
$$M_{C_xH_y} = M_{HBr} {\left(\frac{rate_{HBr}}{rate_{C_xH_y}}\right)}^2$$
$$M_{C_xH_y} = 80.912 amu {\left(\frac{15.0 \frac{mL}{min}}{20.3 \frac{mL}{min}}\right)}^2$$
$$M_{C_xH_y} = 44.17772817amu$$
So,
$$12.011x + 1.008y = 44.17772817$$
Here is where I am having trouble. I know that you need at least as many equations as you do unknowns to solve for them but I can't seem to find another equation. I thought using the ideal gas law, really Avogadro's Law, to find the number of moles in the apparatus. When I do that I get ##x + y = 1.353 moles## (assuming the initial n = 1) but this system gives a negative y value so it is clearly wrong. From the back of the book I know the answer should be ##C_3H_8##. If I were to just assume that the number of carbon atoms were as many as there could be then you get 3, from ##\lfloor \frac{44.17772817}{12.011} \rfloor##, and this gives the correct answer. But that doesn't feel like the right approach. If you assumed that there were only 2 carbon atoms then there would need to be 20 hydrogen atoms and the result would work just as well. Any advice on what I should be thinking about would be much appreciated.
Also, when googling this question the following video comes up. I used it to get to the point where I am now but I have two questions on their methods. (1) How did they know that ##rate = \frac{V}{t}##? I understand why it makes sense from what diffusion means but no such relation shows up in my textbook. I am using "Chemistry: A Molecular Approach, Fourth Edition by Nivaldo Tro" and the only relations given is that rate is inversely proportional to the square of molar mass as well as Graham's law when it discusses effusion and diffusion. I have looked it up and it seems to come from physics when dealing with liquids but I haven't learned that yet so how would I have known that without outside help? (2) When solving for x and y, she gets x as a function of y and then plugs that back into the original equation. The way she does it however seems to just be a rounding error. If she wouldn't have rounded then the expression would just be 58 = 58 which doesn't answer anything. Instead she uses it to solve for x and y. Isn't this wrong?
Thanks,
zachary570