Question

calculate the percentage of each element in the following compounds:- CH20,and NH4N3

Answer 1

Answer on Question #43508 - Chemistry - Inorganic Chemistry

Question:

Calculate the percentage of each element in the following compounds: $\mathrm{CH}_2\mathrm{O}$ and $\mathrm{NH}_4\mathrm{N}_3$.

Solution:

Relative molecular mass of $\mathrm{CH}_2\mathrm{O}$:

$$\mathrm{M}_r(\mathrm{CH}_2\mathrm{O}) = \mathrm{A}_r(\mathrm{C}) + 2 \cdot \mathrm{A}_r(\mathrm{H}) + \mathrm{A}_r(\mathrm{O}) = 12 + 2 \cdot 1 + 16 = 30,$$

where $\mathrm{A}_r(\mathrm{C})$, $\mathrm{A}_r(\mathrm{H})$ and $\mathrm{A}_r(\mathrm{O})$ relative atomic masses of C, H and O, respectively.

Percentage of each element:

$$\%(\mathrm{C}) = \mathrm{A}_r(\mathrm{C}) / \mathrm{M}_r(\mathrm{CH}_2\mathrm{O}) \cdot 100 = 12 / 30 \cdot 100 = 40.0\%$$ $$\%(\mathrm{H}) = 2 \cdot \mathrm{A}_r(\mathrm{H}) / \mathrm{M}_r(\mathrm{CH}_2\mathrm{O}) \cdot 100 = 2 \cdot 1 / 30 \cdot 100 = 6.7\%$$ $$\%(\mathrm{O}) = \mathrm{A}_r(\mathrm{O}) / \mathrm{M}_r(\mathrm{CH}_2\mathrm{O}) \cdot 100 = 16 / 30 \cdot 100 = 53.3\%$$

Relative molecular mass of $\mathrm{NH}_4\mathrm{N}_3$:

$$\mathrm{M}_r(\mathrm{NH}_4\mathrm{N}_3) = 4 \cdot \mathrm{A}_r(\mathrm{N}) + 4 \cdot \mathrm{A}_r(\mathrm{H}) = 4 \cdot 14 + 4 \cdot 1 = 60,$$

Percentage of each element:

$$\%(\mathrm{N}) = 4 \cdot \mathrm{A}_r(\mathrm{N}) / \mathrm{M}_r(\mathrm{NH}_4\mathrm{N}_3) \cdot 100 = 4 \cdot 14 / 60 \cdot 100 = 93.3\%$$ $$\%(\mathrm{H}) = 4 \cdot \mathrm{A}_r(\mathrm{H}) / \mathrm{M}_r(\mathrm{NH}_4\mathrm{N}_3) \cdot 100 = 4 \cdot 1 / 60 \cdot 100 = 6.7\%$$

Answer:

In $\mathrm{CH}_2\mathrm{O}$:

$$\%(\mathrm{C}) = 40.0\%,$$ $$\%(\mathrm{H}) = 6.7\%,$$ $$\%(\mathrm{O}) = 53.3\%.$$

In $\mathrm{NH}_4\mathrm{N}_3$:

$$\%(\mathrm{N}) = 93.3\%,$$ $$\%(\mathrm{H}) = 6.7\%.$$

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