Question

Q83.

1. Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If $200c{m}^3$of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.

$2{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)+5{\mathrm{O}}_2\to 4{\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

(ii) A gaseous compound of nitrogen and hydrogen contains $12.5\%$ hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is $37$.[$\mathrm{N}=14,\mathrm{H}=1$]. (ICSE 19)

Answer 1

i)

• The balanced equation is given as:

$\underset{\underset{2\mathrm{vol}}{\mathrm{ethene}}}{2{\mathrm{C}}_2{\mathrm{H}}_2\left(\mathrm{g}\right)}+\underset{\underset{5\mathrm{vol}}{\mathrm{oxygen}}}{5{\mathrm{O}}_2}\to \underset{\underset{4\mathrm{vol}}{\mathrm{carbon}\mathrm{dioxide}}}{4{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\underset{2\mathrm{vol}}{\mathrm{water}\mathrm{vapour}}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}$

• Calculate the volume of oxygen

Now we can see $2$ volume of acetylene requires$5$ volume of oxygen

Similarly, $1$volume of acetylene requires$\frac{5}{2}$volumes of oxygen

So $200c{m}^3$volume of acetylene will require $$\begin{gathered} =\frac{5}{2}\times 200 \\ =500c{m}^3 \end{gathered}$$of oxygen

• Calculate the voluime of acetylene required

Also $2$ volume of acetylene requires $4$volumes of ${\mathrm{CO}}_2$

Similarly, $1$volume of acetylene requires $\frac{4}{2}$volumes of ${\mathrm{CO}}_2$

So $200c{m}^3$volume of acetylene will require$$\begin{gathered} =\frac{4}{2}\times 200 \\ =400c{m}^3 \end{gathered}$$

• Thus the mixture required $500c{m}^3$of oxygen is required and $400c{m}^3$of ${\mathrm{CO}}_2$ is formed.

ii)

• First we form the composition table to find the empirical formula

Element	$\%age$	Atomic mass	Simplest ratio	Simplest whole number ratio
Nitrogen	$87.5\%$	$14$	$$\begin{gathered} \frac{\%age}{atomicmass}=\frac{87.5}{14} \\ =6.25 \end{gathered}$$	$$\begin{gathered} \frac{Simplestratio}{smallestsimplestratio}=\frac{6.25}{6.25} \\ =1 \end{gathered}$$
Hydrogen	$12.5\%$	$1$	$$\begin{gathered} \frac{\%age}{atomicmass}=\frac{12.5}{1} \\ =12.5 \end{gathered}$$	$$\begin{gathered} \frac{Simplestratio}{smallestsimplestratio}=\frac{12.5}{6.25} \\ =2 \end{gathered}$$

• Calculate the molecular formula

So the empirical formula thus become ${\mathrm{NH}}_2$

Now the empirical mass $$\begin{gathered} =14+2\times 1 \\ =16 \end{gathered}$$

Now we find $$\begin{gathered} n=\frac{molecularmass}{empiricalmass} \\ n=\frac{37}{16} \\ n=2.31 \end{gathered}$$

• Hence the molecular formula becomes ${\left({\mathrm{NH}}_2\right)}_{2.31}=\left({\mathrm{N}}_{2.31}{\mathrm{H}}_{4.62}\right)$

But this formula goes against the Avogadro's law hence does not exist.