Question

An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.

Answer 1

Hint: Think about how a reaction of combustion occurs and what may be the reactants and products of the reaction. First find the empirical formula of the given reactant and then predict the reaction of combustion. Then you may directly use the % of mass formula.

Complete answer:
We know that an organic compound will produce carbon dioxide and water on combustion. For combustion reactions to occur, oxygen is necessary.
Now, let us calculate the mass of $\text{C}{{\text{O}}_{\text{2}}}$ produced:
It is given that the mass of the compound taken is 0.20 g and the percentage of carbon in the compound is 69%. Let us substitute these values in our formula,
 % of M =$\dfrac{\left( \text{atomic mass of M} \right)}{\left( \text{molecular mass of M-compound} \right)}\times \dfrac{\left( \text{mass of M-compound formed} \right)}{\left( \text{mass of substance taken} \right)}\times 100$.
To calculate the molecular mass of $C{{O}_{2}}$
\[\begin{align}
  & \text{Molecular mass of }C{{O}_{2}}=(1\times 12)+(2\times 16) \\
 & \text{Molecular mass of }C{{O}_{2}}=44 \\
\end{align}\]
Thus, putting the known values in the formula above, we get:
 $\text{ }\!\!\%\!\!\text{ of C = }\dfrac{12}{44}\times \dfrac{\left( \text{mass of C}{{\text{O}}_{2}}\text{ formed} \right)}{\left( \text{mass of compound} \right)}\times 100$
Now, putting the values given in the problem in the formula:
$69=\dfrac{12}{44}\times \dfrac{\left( \text{mass of C}{{\text{O}}_{2}}\text{ formed} \right)}{\left( 0.20g \right)}\times 100$
Rearranging the equation by cross-multiplication to find the mass of $C{{O}_{2}}$
Therefore, mass of $C{{O}_{2}}$ formed = $\dfrac{69\times 44\times \left( 0.20g \right)}{12\times 100}=0.506g$
There is 0.506 g of carbon dioxide formed.
Now, let us calculate the mass of water formed. It is given that the mass of the compound is 0.20 g and the percentage of hydrogen is 4.8%.
When we substitute the known values in the formula, we get:
Percentage of hydrogen = $\dfrac{2}{18}\times \dfrac{\left( \text{mass of water formed} \right)}{\left( \text{mass of compound} \right)}\times 100$
Now, substituting the values that are given in the question:
$4.8=\dfrac{2}{10}\times \dfrac{\left( \text{Mass of water formed} \right)}{\left( 0.20g \right)}\times 100$
Rearranging the equation by cross-multiplication to find the mass of ${{H}_{2}}O$
Therefore, Mass of ${{H}_{2}}O$ formed = $\dfrac{4.8\times 18\times \left( 0.20g \right)}{2\times 100}=0.0864g$
0.0864g of water is formed.
So, 0.506 g of carbon dioxide and 0.0864 g of water is formed when 0.20 g of the given substance is subjected to complete combustion.

Note: You should know that the combustion is the reaction in which the hydrocarbon is converted into carbon dioxide and water in the presence of oxygen. If there is no sufficient amount of oxygen, then partial combustion happens and it results in the formation of carbon monoxide.