# How to determine the empirical formula for terephthalic acid from its combustion products?

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only $\ce{C}$, $\ce{H}$, and $\ce{O}$. In one experiment, $0.4953~\mathrm{g}$ of the acid was burnt in excess oxygen. The gases produced were passed through $\ce{CaCl2}$ and $\ce{NaOH}$. It was found that the mass of the $\ce{CaCl2}$ increased by $0.1613~\mathrm{g}$ and the mass of the $\ce{NaOH}$ increased by $1.0495~\mathrm{g}$. What is the empirical formula of terephthalic acid?

We have: $$\ce{C_{x}H_{y}O_{z} + O2 -> CO2 + H2O ->[\ce{CaCl2 + NaOH}] CaCl2 + NaOH}$$

The anhydrous $\ce{CaCl2}$ absorbs the $\ce{H2O}$ produced, and the $\ce{CO2}$ is absorbed from $\ce{NaOH}$? So the products are $\ce{CaCl2. 2H2O}$ and $\ce{Na2CO3}$

I may have just figured the empirical formula out, and was wondering what the final answer is so I can compare and see where I went right/wrong. I realised that instead of $\ce{Na2CO3}$, $\ce{NaHCO3}$ would be produced.

Working out: Via conservation of mass, the $\ce{CO2}$ and $\ce{H2O}$ must also be $0.4953~\mathrm{g}$.

Therefore: conservation of mass (mass of $\ce{CO2 + H2O}$) - increase in mass of $\ce{CaCl2 + NaOH}$

$0.4953 - (0.1613 + 1.0495) = 0.7155~\mathrm{g}$ <--- mass of $\ce{CaCl2}$ and $\ce{NaOH}$

Mass of $\ce{CaCl2.2H2O} = 0.7155 + 0.1613 = 0.8768~\mathrm{g}$

Mass of $\ce{NaHCO3} = 0.7155 + 1.0495 = 1.765~\mathrm{g}$

     C:               H:              O
1.765x(12/84):     0.8768x(4/147.1):

(m) 0.252142857:     0.023842284:   0.4953-(0.252142857+0.023842284)= 0.219314859

(n) m/12       :     m/1         :    m/16

0.021011905 : 0.023842284     :  0.013707179

(Divide by lowest value)

( 1.53      :     1.74        :    1 ) x4

6         :       7         : 4


Therefore $\ce{C6H7O4}$

You are on the right track, but if you google terephthalic acid, you will know that your formula is not correct.

One mistake you are making is using the final mass of the $\ce{CaCl2}$ and $\ce{NaOH}$ and not the change in mass. For example, let's take the $\ce{CaCl2}$:

First, we don't know what the initial mass of $\ce{CaCl2}$ is, so we cannot calculate the final mass the way you are trying to do. We are only given the mass of the terephthalic acid (0.4953 g). The calcium chloride and sodium hydroxide are not part of this mass.

However, we know that the calcium chloride increased in mass by 0.1613 g, presumably by absorbing water (you are correct that the calcium chloride is absorbing water). So, the mass of water produced by the combustion reaction is 0.1613 g.

Similarly, we only know the change in mass of the sodium hydroxide (1.0495 g), which occurred by absorbing carbon dioxide. So the mass of carbon dioxide is 1.4095 g.

The second mistake is to assume that all of the mass of the carbon dioxide and water produced must come from the original mass of the terephthalic acid. Not so! We have excess oxygen added contributing mass to the reactant side, and this oxygen is also in the products. $$\ce{C}_x\ce{ H}_y\ce{ O}_z\ce{ +}\left(\frac{4x+2y-z}{4}\right)\ce{ O2 -> }x\ce{CO2 +}\frac{y}{2}\ce{ H2O}$$

We need to do more maths to figure out how much of the leftover mass in the original sample is oxygen. Here is the beginning of the setup, now that you have some sense of the mass of water and carbon dioxide produced:

1. Mass of water produced $\ce{->}$ moles of water produced $\ce{->}$ moles of hydrogen in the water produced (and in the original sample!) $\ce{->}$ grams of hydrogen (in sample)

2. Mass of carbon dioxide produced $\ce{->}$ mass of carbon (in sample)

3. Mass of oxygen in sample (by subtraction) $\ce{->}$ moles of oxygen in sample

You are doing everything correctly after this.

I am afraid I could not follow your reasoning and unfortunately can not pinpoint you to your error therefore. The result you obtain, does not resemble the empirical formula of terephthalic acid. As a general tip: These problems are of entirely mathematical nature, so it is important to use proper equations and keep the units at all times. I am going to tackle this problem from beginning to end.

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only $\ce{C}$, $\ce{H}$, and $\ce{O}$. In one experiment, $0.4953~\mathrm{g}$ of the acid was burnt in excess oxygen. The gases produced were passed through $\ce{CaCl2}$ and $\ce{NaOH}$. It was found that the mass of the $\ce{CaCl2}$ increased by $0.1613~\mathrm{g}$ and the mass of the $\ce{NaOH}$ increased by $1.0495~\mathrm{g}$. What is the empirical formula of terephthalic acid?

First of all it is important to formulate the correct reaction equation: $$\ce{C_{x}H_{y}O_{z} + \frac{\left(2x+\frac{y}{2}-z\right)}{2}\,O2 -> x\,CO2 + \frac{y}{2}\,H2O}$$

The information about calcium chloride and sodium hydroxide is only a way to tell you the mass of the product. It is completely irrelevant to the question itself. However, the correct equation would then be: $$\ce{C_{x}H_{y}O_{z} + \frac{\left(2x+\frac{y}{2}-z\right)}{2}\,O2 ->[\ce{CaCl2, NaOH}] x\,NaHCO3 + \frac{y}{2n}\,CaCl2.n\,H2O}$$

First of all, we need to gather all the information we are given in the question or that we know otherwise: \begin{align} m(\ce{C_{$x$}H_{$y$}O_{$z$}}) &= 0.4953~\mathrm{g} &M(\ce{C_{$x$}H_{$y$}O_{$z$}}) &= ?\\ m(\ce{H2O}) &= 0.1613~\mathrm{g} & M(\ce{H2O}) &= 18~\mathrm{\frac{g}{mol}}\\ m(\ce{CO2}) &= 1.0495~\mathrm{g} & M(\ce{CO2}) &= 44~\mathrm{\frac{g}{mol}}\\ \end{align}

We can go ahead and calculate the number of moles of carbon dioxide and water produced. We therefore also know the relative number of moles of carbon and hydrogen in terephthalic acid. The basic formula that needs to be applied here is $$n=\frac{m}{M}.$$

How many moles of carbon are in carbon dioxide, what is the mathematical relationship? What is it for hydrogen and water?

$n(\ce{C}) : n(\ce{CO2}) = 1$
$n(\ce{H}) : n(\ce{H2O}) = 2$

How many moles of carbon dioxide and water have been produced? How does that relate to carbon and hydrogen?

$n(\ce{CO2})=\frac{m(\ce{CO2})}{M(\ce{CO2})}= 0.0239~\mathrm{mol} \therefore n(\ce{C})= 0.0239~\mathrm{mol}$
$n(\ce{H2O})=\frac{m(\ce{H2O})}{M(\ce{H2O})}= 0.0090~\mathrm{mol} \therefore n(\ce{H})= 0.0180~\mathrm{mol}$

Calculate the mass of hydrogen and carbon in the original sample with $$m=n\cdot M$$ and use it to find the mass of oxygen in the sample.

$m(\ce{C}) = 0.2868~\mathrm{g}$
$m(\ce{H}) = 0.0180~\mathrm{g}$
$m(\ce{O}) = m(\ce{C_{$x$}H_{$y$}O_{$z$}}) - m(\ce{C}) - m(\ce{H}) = 0.1906~\mathrm{g}$

How many moles of oxygen are present in the original sample?

$n(\ce{O}) = 0.119~\mathrm{mol}$

Now you only have to relate the number of moles for each element to integer ratios and you will find the empirical formula.

$x:y:z = n(\ce{C}) : n(\ce{H}) : n(\ce{O}) = 0.0239~\mathrm{mol} : 0.0180~\mathrm{mol} : 0.119~\mathrm{mol}$
$x:y:z = 4:3:2$
Therefore the empirical formula is $\ce{C4H3O2}$, which is half of the sum formula $\ce{C8H6O4}$, see wikipedia for more information.