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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.