Question

0.90 g of an orgamic compound containing only carbon, oxgen& nitrogen on combustion gives 1.1 g of CO2 & 0.3 g of water.What is %C, %H and % N in the organic compound?

Answer 1

Task:

0.90 g of an organic compound containing only carbon, oxygen& nitrogen on combustion gives 1.1 g of CO2 & 0.3 g of water. What is %C, %H and % N in the organic compound?

Solution:

"containing only carbon, oxygen& nitrogen" – Maybe you meant hydrogen instead of oxygen?

The chemical equation for combusting reaction is

$$4 \mathrm{C}_{x} \mathrm{H}_{y} \mathrm{N}_{z} + (4 x + y) \mathrm{O}_{2} = 4 x \mathrm{CO}_{2} + 2 y \mathrm{H}_{2} \mathrm{O} + 2 z \mathrm{N}_{2}$$

Let's find the amount of $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$

$$n (\mathrm{mol}) = \frac{m (g)}{M W (g / \mathrm{mol})}$$ $$\begin{array}{l} \mathrm{M W} (\mathrm{CO}_{2}) = \mathrm{M W} (\mathrm{C}) + 2 \mathrm{M W} (\mathrm{O}) = 12 + 2 \cdot 16 = 44 \, \mathrm{g} / \mathrm{mol} \\ \mathrm{M W} (\mathrm{H}_{2} \mathrm{O}) = 2 \mathrm{M W} (\mathrm{H}) + \mathrm{M W} (\mathrm{O}) = 2 \cdot 1 + 16 = 18 \, \mathrm{g} / \mathrm{mol} \\ n (\mathrm{CO}_{2}) = 1.1 / 44 = 0.025 \, \mathrm{mol} \\ n (\mathrm{H}_{2} \mathrm{O}) = 0.3 / 18 = 0.017 \, \mathrm{mol} \\ \end{array}$$

The number of moles of C is equal to the number of moles of $\mathrm{CO}_{2}$

$$n (C) = n (\mathrm{CO}_{2}) = 0.025 \, \mathrm{mol}$$

The number of moles of H is twice the number of moles of $\mathrm{H}_{2} \mathrm{O}$

$$n (H) = 2 \cdot n (\mathrm{H}_{2} \mathrm{O}) = 2 \cdot 0.017 = 0.034 \, \mathrm{mol}$$

The mass of C is

$$m (C) = n (C) \cdot M W (C) = 0.025 \cdot 12 = 0.300 \, \mathrm{g}$$

The mass of H is

$$m (H) = n (H) \cdot M W (H) = 0.034 \cdot 1 = 0.034 \, \mathrm{g}$$

The mass of N is

$$m (N) = m (\text{compound}) - m (C) - m (H) = 0.9 - 0.3 - 0.034 = 0.566 \, \mathrm{g}$$

The yield of C, N, H is

$$\begin{array}{l} w (C) = m (C) / m_{\text{compound}} \cdot 100\% \\ w (C) = 0.300 / 0.9 \cdot 100\% = 33.3\% \\ w (H) = m (H) / m_{\text{compound}} \cdot 100\% \\ w (H) = 0.034 / 0.9 \cdot 100\% = 3.8\% \\ w (N) = m (N) / m_{\text{compound}} \cdot 100\% \\ w (N) = 0.566 / 0.9 \cdot 100\% = 62.9\% \\ \end{array}$$

Answer: $w (C) = 33.3\%$; $w (H) = 3.8\%$; $w (N) = 62.9\%$