I am doing a process flow question and trying to calculate the grams of $\text{C}_{10}\text{H}_{14}$, $\text{C}_4\text{H}_6\text{O}_3$, and $\text{C}_{12}\text{H}_{16}\text{O}$ produced during a reaction. 134 grams of $\text{C}_{10}\text{H}_{14}$ and 134g of $\text{C}_4\text{H}_6\text{O}_3$ were used as reactants. The reaction also produced 68g of acetic acid. I defined my stream variables in terms of the elements ($\text{CHO}$), since the reaction is of unknown stoichiometry. I created a system of equations as follows, where $x$ is the grams of $\text{C}_{10}\text{H}_{14}$ produced, $y$ is the grams of $\text{C}_4\text{H}_6\text{O}_3$ produced, and $z$ is the grams of $\text{C}_{12}\text{H}_{16}\text{O}$ produced.

For $\text{C}$: $12.98 = 10x/134 + 4y/102 + 12z/176$

For $\text{H}$: $17.347 = 14x/134 + 6y/102 + 16z/176$

For $\text{O}$: $1.67 = 0x + 3y/102 + z/176$

I combined these into linear matrix equation $$\left[\begin{matrix} \dfrac{10x}{134} & \dfrac{4y}{102} & \dfrac{12z}{176} \\ \dfrac{14x}{134} & \dfrac{6y}{102} & \dfrac{16z}{176} \\ 0 & \dfrac{3y}{102} & \dfrac{1z}{176} \end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}12.98 \\ 17.347 \\ 1.67\end{matrix}\right]$$

Given that $Ax=b$, and $x=A^{-1}b$, I then took the inverse matrix of $A$ and multiplied it by the matrix $b$ (in my calculator). This resulted in the solution that $x = -16.46$, $y= 18.7$, and $z=197.64$. While this does satisfy the equations, the textbook gave the answers: $x=1.6$, $y=22.9$, and $z=175.5$, which also satisfies the equations, and makes sense unlike my answer where I have a negative gram amount produced.

I believe my error has something to do with setting a basis, though I'm not sure what that means. Also, the book has the same equations in its solution that I have, but the answer comes out differently (it does not explain how the three equations were solved, other than that it was simultaneous.)

How do I get the correct answer and where is my mistake? Also if you could explain setting a basis that would be great as well.


1 Answer 1


Your question has issues. First you say "$\text{C}_{10}H_{14}$, $\text{C}_4\text{H}_6\text{O}_3$, and $\text{C}_{12}\text{H}_{16}\text{O}$ produced during a reaction", then you say "134 grams of $\text{C}_{10}H_{14}$ and 134g of $\text{C}_4\text{H}_6\text{O}_3$", so $\text{C}_{10}H_{14}$, and $\text{C}_4\text{H}_6\text{O}_3$ are reactants or products?

Lets assume they are reactants, you say also acetic acid is produced so the reaction would be:

$\text{C}_{10}H_{14}+\text{C}_4\text{H}_6\text{O}_3 \rightarrow \text{C}_{12}\text{H}_{16}\text{O}+\text{C}_2\text{H}_4\text{O}_2$

All stochiometric coefficients are equal to 1. So if you assume 68 grams of acetic acid where produced, because $\text{MW}(\text{C}_2\text{H}_4\text{O}_2)=60$, more than one mole was produced. But because 134 g of $\text{C}_{10}H_{14}$ were used as reactant, and $\text{MW}(\text{C}_{10}H_{14})=134$, the maximum extent would be 1 mol, so the data are inconsistent.

Also in your original equations, for example:

For $C$: $12.98 = 10x/134 + 4y/102 + 12z/176$

You didn't specify the units. Based on the rhs it appear there are mol quantities, if that's the case, the third term would need to have a negative side. And if lhs are mols produced (from acetic acid), the data don't match,

mol C produced (acetic acid) = $2\cdot68/60=2.26$

  • $\begingroup$ This answer correctly points out the serious deficiencies in the question. Absent further corrections from the OP, I would vote to close this question. $\endgroup$ Oct 12, 2018 at 20:31

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