Question

A 0.320 g sample of a carboxylic acid is burned in oxygen, producing 0.571 g of co2 and 0.235 g of h2o. Determine the empirical formula of the carboxylic acid.

Answer 1

Answer:- Empirical formula of the carboxylic acid is $C_3H_6O_2$ .

Solution:- When the acid is burned then all it's carbon goes to form carbon dioxide and all the hydrogen goes to form water.

So, we convert the grams of carbon dioxide to moles and calculate the moles of carbon from it as one mol of carbon dioxide contains one mol of carbon.

$0.571gCO_2(\frac{1molCO_2}{44gCO_2})(\frac{1molC}{1molCO_2})$

= $0.0130molC$

Similarly, we convert the grams of water to moles and these moles are used to calculate the moles of hydrogen as each mol of water contains two moles of hydrogen.

$0.235gH_2O(\frac{1molH_2O}{18gH_2O})(\frac{2molH}{1molH_2O})$

= $0.0261molH$

Now, we convert the moles of C and H to grams and subtract the sum of their mass from the given mass of the acid to calculate the mass of oxygen present in the acid.

$0.0130molC(\frac{12gC}{1molC})$

= 0.156 g C

Similarly,  $0.0261molH(\frac{1gH}{1molH})$

= 0.0261 g H

mass of oxygen in acid = 0.320 g - (0.156 g + 0.0261 g)

= 0.320 g - 0.1821 g

= 0.1379 g

Let's convert the grams of oxygen to moles;

$0.1379gO(\frac{1molO}{16gO})$

= 0.00862 mol O

Next step is to calculate the mol ratio and for this we divide the moles ofch by the least one of them. We have least moles of oxygen, so let's divide the moles of each by the moles of oxygen.

$C=\frac{0.0130}{0.00862}=1.5$

$H=\frac{0.0261}{0.00862}=3.0$

$O=\frac{0.00862}{0.00862}=1.0$

Mol ratio sgould be whole number ratio, if it is not then we multiply all of them by 2, 3, 4 and so on to get the smallest whole number ratio. Here if we multiply by 2 then it works.

So, C = 2(1.5) = 3

H = 2(3.0) = 6

O = 2(1.0) = 2

Hence, the empirical formula of the carboxylic acid is $C_3H_6O_2$ .