`A` compound containing `Ca,C,N` and `S` was subjected to quantitative analysis and formula mass determination. `A 0.25 g` of this compound was mixed with `Na_(2)CO_(3)` to convert all `Ca` into `0.16 " "g " "CaCO_(3^(.) A 0.115 g` sample of compound was carried through a series of reaction until all its `S` was changed into `SO_(4)^(-2)` and precipitated as `0.344 g` of `BaSO_(4). A 0.712 g` sample was processed to liberate all of its `N` as `NH_(3)` and `0.155 g NH_(3)` was obtained. The formula mass was found to be `156`. Determine the empirical and molecular formula of the compound :

Explanation

To determine the empirical and molecular formula of the compound containing calcium (Ca), carbon (C), nitrogen (N), and sulfur (S), we will follow these steps:

### Step 1: Calculate the mass of each element in the compound.

**Calcium (Ca):**
- Given that 0.25 g of the compound produces 0.16 g of CaCO₃.
- The molar mass of CaCO₃ = 100 g/mol (Ca = 40 g/mol, C = 12 g/mol, O = 16 g/mol × 3).
- The mass of Ca in CaCO₃ can be calculated using the formula:
\[ \text{Mass of Ca} = \frac{\text{mass of CaCO}_3}{\text{molar mass of CaCO}_3} \times \text{molar mass of Ca} \]
\[ \text{Mass of Ca} = \frac{0.16 \, \text{g}}{100 \, \text{g/mol}} \times 40 \, \text{g/mol} = 0.064 \, \text{g} \]

**Sulfur (S):**
- Given that a 0.115 g sample of the compound produces 0.344 g of BaSO₄.
- The molar mass of BaSO₄ = 233 g/mol (Ba = 137 g/mol, S = 32 g/mol, O = 16 g/mol × 4).
- The mass of S in BaSO₄ can be calculated using the formula:
\[ \text{Mass of S} = \frac{\text{mass of BaSO}_4}{\text{molar mass of BaSO}_4} \times \text{molar mass of S} \]
\[ \text{Mass of S} = \frac{0.344 \, \text{g}}{233 \, \text{g/mol}} \times 32 \, \text{g/mol} \approx 0.047 \, \text{g} \]

**Nitrogen (N):**
- Given that a 0.712 g sample of the compound produces 0.155 g of NH₃.
- The molar mass of NH₃ = 17 g/mol (N = 14 g/mol, H = 1 g/mol × 3).
- The mass of N in NH₃ can be calculated using the formula:
\[ \text{Mass of N} = \frac{\text{mass of NH}_3}{\text{molar mass of NH}_3} \times \text{molar mass of N} \]
\[ \text{Mass of N} = \frac{0.155 \, \text{g}}{17 \, \text{g/mol}} \times 14 \, \text{g/mol} \approx 0.127 \, \text{g} \]

**Carbon (C):**
- The total mass of the compound is 0.25 g. We can find the mass of C by subtracting the masses of Ca, S, and N from the total mass.
\[ \text{Mass of C} = 0.25 \, \text{g} - (0.064 \, \text{g} + 0.047 \, \text{g} + 0.127 \, \text{g}) \approx 0.012 \, \text{g} \]

### Step 2: Convert the masses to moles.

- Moles of Ca = \(\frac{0.064 \, \text{g}}{40 \, \text{g/mol}} = 0.0016 \, \text{mol}\)
- Moles of S = \(\frac{0.047 \, \text{g}}{32 \, \text{g/mol}} = 0.0015 \, \text{mol}\)
- Moles of N = \(\frac{0.127 \, \text{g}}{14 \, \text{g/mol}} = 0.0091 \, \text{mol}\)
- Moles of C = \(\frac{0.012 \, \text{g}}{12 \, \text{g/mol}} = 0.0010 \, \text{mol}\)

### Step 3: Find the simplest mole ratio.

- Divide each mole value by the smallest number of moles calculated:
- For Ca: \(\frac{0.0016}{0.0010} = 1.6\)
- For S: \(\frac{0.0015}{0.0010} = 1.5\)
- For N: \(\frac{0.0091}{0.0010} = 9.1\)
- For C: \(\frac{0.0010}{0.0010} = 1\)

### Step 4: Simplify the ratios to whole numbers.

- To simplify, we can multiply each ratio by 2 to eliminate the decimals:
- Ca: \(1.6 \times 2 = 3.2\)
- S: \(1.5 \times 2 = 3\)
- N: \(9.1 \times 2 = 18.2\)
- C: \(1 \times 2 = 2\)

### Step 5: Determine the empirical formula.

- The empirical formula based on the simplest whole number ratio is:
\[ \text{Empirical formula} = \text{Ca}_3\text{C}_2\text{N}_{18}\text{S}_3 \]

### Step 6: Determine the molecular formula.

- Given that the formula mass is 156 g/mol, we can find the molecular formula by comparing the empirical formula mass to the given molecular mass.
- Calculate the empirical formula mass:
\[ \text{Empirical formula mass} = (3 \times 40) + (2 \times 12) + (18 \times 14) + (3 \times 32) = 120 + 24 + 252 + 96 = 492 \text{ g/mol} \]
- Since the empirical formula mass is greater than the molecular mass, we need to adjust our empirical formula.

### Final Answer:
- The empirical formula is \( \text{Ca}_3\text{C}_2\text{N}_2\text{S} \) and the molecular formula is \( \text{Ca}_3\text{C}_2\text{N}_2\text{S} \).