An inorganic substance has the following composition:
`N = 35% H = 5%, O = 60 %`
On being heated, it yielded a gaseous compound containing `N = 63.63%` and `O = 36.37%`. Suggest a formula for each substance and equation for the chemical change.

Explanation

To solve the problem step by step, we will first analyze the composition of the inorganic substance and then determine the formula for both the original substance and the gaseous compound produced upon heating.

### Step 1: Determine the composition of the inorganic substance
Given the percentages:
- Nitrogen (N) = 35%
- Hydrogen (H) = 5%
- Oxygen (O) = 60%

Assuming we have 100 grams of the substance:
- Mass of Nitrogen = 35 grams
- Mass of Hydrogen = 5 grams
- Mass of Oxygen = 60 grams

### Step 2: Calculate the number of moles of each element
Using the molar masses:
- Molar mass of Nitrogen (N) = 14 g/mol
- Molar mass of Hydrogen (H) = 1 g/mol
- Molar mass of Oxygen (O) = 16 g/mol

Calculating moles:
- Moles of Nitrogen = \( \frac{35 \text{ g}}{14 \text{ g/mol}} = 2.5 \text{ moles} \)
- Moles of Hydrogen = \( \frac{5 \text{ g}}{1 \text{ g/mol}} = 5 \text{ moles} \)
- Moles of Oxygen = \( \frac{60 \text{ g}}{16 \text{ g/mol}} = 3.75 \text{ moles} \)

### Step 3: Find the simplest mole ratio
The mole ratio of N:H:O is:
- N: 2.5
- H: 5
- O: 3.75

To simplify, we can divide each by the smallest number of moles (2.5):
- N: \( \frac{2.5}{2.5} = 1 \)
- H: \( \frac{5}{2.5} = 2 \)
- O: \( \frac{3.75}{2.5} = 1.5 \)

This gives us the ratio:
- N:H:O = 1:2:1.5

To convert to whole numbers, multiply by 2:
- N: 2
- H: 4
- O: 3

Thus, the empirical formula of the inorganic substance is \( \text{N}_2\text{H}_4\text{O}_3 \).

### Step 4: Analyze the gaseous compound produced upon heating
The gaseous compound has the following composition:
- Nitrogen (N) = 63.63%
- Oxygen (O) = 36.37%

Assuming we have 100 grams of the gaseous compound:
- Mass of Nitrogen = 63.63 grams
- Mass of Oxygen = 36.37 grams

### Step 5: Calculate the number of moles of the gaseous compound
Calculating moles:
- Moles of Nitrogen = \( \frac{63.63 \text{ g}}{14 \text{ g/mol}} \approx 4.55 \text{ moles} \)
- Moles of Oxygen = \( \frac{36.37 \text{ g}}{16 \text{ g/mol}} \approx 2.27 \text{ moles} \)

### Step 6: Find the simplest mole ratio for the gaseous compound
The mole ratio of N:O is:
- N: 4.55
- O: 2.27

Dividing by the smallest number of moles (2.27):
- N: \( \frac{4.55}{2.27} \approx 2 \)
- O: \( \frac{2.27}{2.27} = 1 \)

Thus, the empirical formula of the gaseous compound is \( \text{N}_2\text{O} \).

### Step 7: Write the balanced chemical equation
The decomposition reaction can be written as:
\[ \text{N}_2\text{H}_4\text{O}_3 \xrightarrow{\text{heat}} \text{N}_2\text{O} + 2 \text{H}_2\text{O} \]

### Summary of Results
- Formula of the inorganic substance: \( \text{N}_2\text{H}_4\text{O}_3 \)
- Formula of the gaseous compound: \( \text{N}_2\text{O} \)
- Balanced chemical equation:
\[ \text{N}_2\text{H}_4\text{O}_3 \xrightarrow{\text{heat}} \text{N}_2\text{O} + 2 \text{H}_2\text{O} \]