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The problem is as follows:

A mixture of gases consisting of $\ce{CH4}$ and $\ce{C2H4}$ is let to pass over red hot $\ce{CuO}$. Then $\pu{0.6g}$ of $\ce{H2O}$ is collected along $\pu{1.185g}$ of $\ce{CO2}$. What will be the composition of the mixture if it is known that the combustion was complete?.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.& \ce{CH4}=51.8\%\,and\,\ce{C2H4}=48.2\,\%\\ 2.&\ce{CH4}=50\%\,and\,\ce{C2H4}=50\,\%\\ 3.&\ce{CH4}=33.3\%\,and\,\ce{C2H4}=66.6\,\%\\ 4.&\ce{CH4}=38.8\%\,and\,\ce{C2H4}=64.2\,\%\\ 5.&\ce{CH4}=36.3\%\,and\,\ce{C2H4}=63.7\,\%\\ \end{array}$

I'm not sure exactly how to proceed with this question. Should it be treated as if it is an analysis by combustion?.

What I did to solve this was to treat each gas in the mixture as reacting independently for the oxygen from interacting with the red hot $\ce{CuO}$ as follows:

$\ce{ CH4 + 2 O2 -> CO2 + 2 H2O}$

With: $x=\ce{ CH4}$

$x\,g\times\frac{1\,mol\,\ce{ CH4}}{16\,g\,\ce{ CH4}}\times \frac{2\,\,mol\,\ce{ H2O}}{1\,\,mol\,\ce{ CH4}}\times\frac{18\,g}{1\,mol\,\ce{H2O}}=\frac{18}{8}x\,g\,\ce{H2O}$

$x\,g\times\frac{1\,mol\,\ce{CH4}}{16\,g\,\ce{CH4}}\times \frac{1\,\,mol\,\ce{CO2}}{1\,\,mol\,\ce{CH4}}\times\frac{44\,g}{1\,mol\,\ce{CO2}}=\frac{44}{16}x\,g\,\ce{CO2}$

For:

$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O}$

With $y=\ce{C2H4}$

$y\,g\times\frac{1\,mol\,\ce{C2H4}}{28\,g\,\ce{C2H4}}\times \frac{2\,\,mol\,\ce{H2O}}{1\,\,mol\,\ce{C2H4}}\times\frac{18\,g}{1\,mol\,\ce{H2O}}=\frac{36}{28}y\,g\,\ce{H2O}$

$y\,g\times\frac{1\,mol\,\ce{C2H4}}{28\,g\,\ce{C2H4}}\times \frac{2\,\,mol\,\ce{CO2}}{1\,\,mol\,\ce{C2H4}}\times\frac{44\,g}{1\,mol\,\ce{CO2}}=\frac{88}{28}x\,g\,\ce{CO2}$

This is reduced to a system of equations as follows:

$\frac{44}{16}x+\frac{44}{14}y=1.185\,g\,\ce{CO2}$

$\frac{18}{8}x+\frac{18}{14}y=0.6\,g\,\ce{H2O}$

Solving this system yields:

$x=0.1024\,g\,\ce{CH4}$

$y=0.2874\,g\,\ce{C2H4}$

Then to calculate the composition of the mixture I would use the molar fraction:

$n_{\ce{CH4}}=\frac{0.1024}{16}=0.0064\,mol$

$n_{\ce{C2H4}}=\frac{0.2874}{28}=0.01026\,mol$

Then the percentage for composition for $\ce{CH4}$ would be:

$\frac{n_{\ce{CH4}}}{n_{\ce{CH4}}+n_{\ce{C2H4}}}=\frac{0.0064}{0.0064+0.01026}=0.3841$

For $\ce{C2H4}$:

$\frac{\ce{C2H4}}{n_{\ce{CH4}}+n_{\ce{C2H4}}}=\frac{0.01026}{0.0064+0.01026}=0.6159$

To which correspond each one as roughly $38.41\%$ and $61.59\%$ respectively for $\ce{CH4}$ and $\ce{C2H4}$. But none of this seem to check with any of the answers.

Could it be that my method was wrong or did I overlooked something?.

The only answer which is closer to mine is the forth alternative but the second number is off by a big margin thus I'm not very convinced on my result. What would be the recommended procedure in this case?

Can somebody help me here?.

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  • $\begingroup$ Let's put it this way: can a mixture consist of 90% of A and 90% of B? $\endgroup$ – Ivan Neretin Jan 23 at 8:19
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    $\begingroup$ BTW, just in case it makes you feel better, my numeric answer is similar to yours. $\endgroup$ – Ivan Neretin Jan 23 at 8:32
  • $\begingroup$ @IvanNeretin I'm still confused at what were you trying to say with the first comment?. It doesn't make sense that a mixture can be simultaneously 90% of two components. Wouldn't it be 90% of A and 10% of B or viceversa?. It's a relief that it seems that you arrived to the same answer as me. But I'm left pondering how did you got it there?. Mind sharing your method please?. :) $\endgroup$ – Chris Steinbeck Bell Jan 23 at 10:42
  • $\begingroup$ Why, the method is trivial, same as yours: from grams to moles and then solving a couple of linear equations. (Except that my x and y were the moles of CH4 and C2H4, since we never need their masses at all.) Now to the point: OK, indeed 90%+90% does not make any sense. What about 38.8%+64.2% (I am referring to the answer #4 from the book), does that make any more sense? $\endgroup$ – Ivan Neretin Jan 23 at 11:08
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    $\begingroup$ Well, books are written by humans, and humans make errors. An answer option where the percents do not add up to 100 is not necessarily bad per se (maybe the intention was to teach you to notice just that?), but the absence of the right answer certainly is. So it goes. $\endgroup$ – Ivan Neretin Jan 23 at 11:16
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This is clearly a homework question. Yet, since OP has given good effort to solve the question, I'd like to give OP some insight to solve these kinds of questions, taking current question as an example.

Suppose the mixture contains $x\;\pu {g}$ of $\ce{CH4}$ and $y\;\pu {g}$ of $\ce{C2H4}$. The combustion reactions of $\ce{CH4}$ and $\ce{C2H4}$ can be given as:

$$\ce{CH4 + 2 O2 -> CO2 + 2H2O} \tag{1}$$ $$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O} \tag{2}$$

Thus, the amount of $\ce{CO2}$ produced from the reaction $(1)$ is:

$$x\,\pu{g}\,\ce{CH4} \times\frac{\pu{1 mol}\,\ce{CH4}}{\pu{16 g}\,\ce{CH4}} \times \frac{\pu{1 mol}\,\ce{CO2}}{\pu{1 mol}\,\ce{CH4}}\times \frac{\pu{44 g} \,\ce{CO2}}{\pu{1 mol}\,\ce{CO2}}=\frac{11}{4}x\,\pu{g} \,\ce{CO2}$$

And, the amount of $\ce{H2O}$ produced from the reaction $(1)$ is:

$$x\,\pu{g}\,\ce{CH4} \times\frac{\pu{1 mol}\,\ce{CH4}}{\pu{16 g}\,\ce{CH4}} \times \frac{\pu{2 mol}\,\ce{H2O}}{\pu{1 mol}\,\ce{CH4}}\times \frac{\pu{18 g} \,\ce{H2O}}{\pu{1 mol}\,\ce{H2O}}=\frac{9}{4}x\,\pu{g} \,\ce{H2O}$$

Similarly, the amount of $\ce{CO2}$ produced from the reaction $(2)$ is:

$$y\,\pu{g}\,\ce{C2H4} \times\frac{\pu{1 mol}\,\ce{C2H4}}{\pu{28 g}\,\ce{C2H4}} \times \frac{\pu{2 mol}\,\ce{CO2}}{\pu{1 mol}\,\ce{C2H4}}\times \frac{\pu{44 g} \,\ce{CO2}}{\pu{1 mol}\,\ce{CO2}}=\frac{22}{7}y\,\pu{g} \,\ce{CO2}$$

And, the amount of $\ce{H2O}$ produced from the reaction $(2)$ is:

$$y\,\pu{g}\,\ce{C2H4} \times\frac{\pu{1 mol}\,\ce{C2H4}}{\pu{28 g}\,\ce{C2H4}} \times \frac{\pu{2 mol}\,\ce{H2O}}{\pu{1 mol}\,\ce{C2H4}}\times \frac{\pu{18 g} \,\ce{H2O}}{\pu{1 mol}\,\ce{H2O}}=\frac{9}{7}y\,\pu{g} \,\ce{H2O}$$

Since the mixture produced $\pu{1.185 g}$ of $\ce{CO2}$ and $\pu{0.6 g}$ of $\ce{H2O}$, we can derive two equations with two variables, $x$ and $y$:

$$\frac{11}{4}x+\frac{22}{7}y=1.185$$ and, $$\frac{9}{4}x+\frac{9}{7}y=0.6$$

Up to this point OP has taken correct path. Now, let's consider these two questions. These equations can be simplify as:

$$77 x + 88y=1.185\times 28$$ or $$ 7x + 8y=\frac{1.185\times 28}{11}=3.016 \tag {3}$$ and, $$ 7x + 4y=\frac{0.6\times 28}{9} = 1.867 \tag {4}$$

Thus, $(3)-(4)$ gives you:

$$4y = 3.016 - 1.867 = 1.149 $$ $$\therefore \; y = \frac{1.149}{4} = \pu{0.2873 g}\,\ce{C2H4}$$

And $(4)\times 2-(3)$ gives you:

$$7x = 1.867 \times 2 - 3.016 = \pu{0.718 g}\,\ce{CH4}$$ $$\therefore \; x = \frac{0.718}{7} = \pu{0.1026 g}\,\ce{CH4}$$

Thus, it is clear that you were still in the correct path on solving the same two equations for $x$ and $y$.

According to my solution, mass percentage of $\ce{CH4}$ is:

$$\frac{0.1026}{0.1026 + 0.2873} \times 100 = 26.31\% \; (w/w)$$

And mass percentage of $\ce{C2H4}$ is:

$$\frac{0.2873}{0.1026 + 0.2873} \times 100 = 73.69\% \; (w/w)$$

Since this is not the one of the given answers, let's check for mol% like what you have done:

$$n_{\ce{CH4}}=\frac{0.1026}{16}= \pu{0.0064 mol}$$

$$n_{\ce{C2H4}}=\frac{0.2873}{28}= \pu{0.0103 mol}$$

Thus, the percentage of composition for $\ce{CH4}$ would be:

$$\frac{0.0064}{0.0064+0.0103}\times 100 = 38.32\%\: \pu{mol}$$

Similarly, the percentage of composition for $\ce{C2H4}$ would be:

$$\frac{0.0103}{0.0064+0.0103}\times 100 = 61.68\%\: \pu{mol}$$

Accordingly, the most closest answer is (5), although both given values are $\approx 2\%$ off from calculated values. It is noteworthy that OP's calculations are correct to the end.

Note: This question is not well written. That's why I have to go through two calculations to see which $\%$ is they are talking about. Also, the sum of compositions given in answer choice (4) is not $100\%$.

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    $\begingroup$ I revisited my calculations by hand two times and I get to the same answer obtained before. Can you check your equation? I'm referring to $3$ and $4$. I've even compared with Wolfram Alpha's solving for equations and get the same value as in my post here When you multiply by $28$ you forgot to multiply in the numerator for both equations by $4$ and $7$!. Can you please recheck this?. $\endgroup$ – Chris Steinbeck Bell Jan 26 at 7:22
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    $\begingroup$ I did rechecked my work twice so I appreciate you could check this again. I agree that not all of the alternatives check in the sense that they do not add up to $100\%$ and this could be a hint of an error. But if your procedure validates mine, then it seems that the important aspect of this problem I was right isn't it?. $\endgroup$ – Chris Steinbeck Bell Jan 26 at 7:26
  • $\begingroup$ Your calculations are correct to the end. Thank you for finding my mental error. $\endgroup$ – Mathew Mahindaratne Jan 26 at 8:21
  • $\begingroup$ I'm glad that the error was not by my side. Can you please check this topic in solubility? Perhaps does it exist another way to solve the problem more intuitively?. $\endgroup$ – Chris Steinbeck Bell Jan 26 at 8:42

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