Complete combustion of 0.858 g of compound X gives 2.63 g of `CO_(2)` and 1.28 g of` H_(2)`O. The lowest molecular mass X can have

Options

A. (a)43 g
B. (b)86 g
C. (c )129 g
D. (d)172 g

Explanation

To find the lowest molecular mass of compound X, we will follow these steps:

### Step 1: Determine the number of moles of CO₂ and H₂O produced.
- Given the mass of CO₂ produced = 2.63 g
- Given the mass of H₂O produced = 1.28 g

**Molar mass of CO₂** = 12 (C) + 2 × 16 (O) = 44 g/mol
**Molar mass of H₂O** = 2 × 1 (H) + 16 (O) = 18 g/mol

Now, calculate the number of moles:
- Moles of CO₂ = mass of CO₂ / molar mass of CO₂ = 2.63 g / 44 g/mol = 0.0598 mol (approximately 0.06 mol)
- Moles of H₂O = mass of H₂O / molar mass of H₂O = 1.28 g / 18 g/mol = 0.0711 mol (approximately 0.07 mol)

### Step 2: Determine the number of moles of carbon (C) and hydrogen (H).
From the combustion reaction:
- Each mole of CO₂ contains 1 mole of carbon.
- Each mole of H₂O contains 2 moles of hydrogen.

Thus:
- Moles of carbon (C) = moles of CO₂ = 0.06 mol
- Moles of hydrogen (H) = 2 × moles of H₂O = 2 × 0.07 mol = 0.14 mol

### Step 3: Find the simplest mole ratio of C to H.
To find the simplest ratio:
- Ratio of C to H = moles of C : moles of H = 0.06 : 0.14

To simplify this ratio, divide both by the smallest number of moles (0.06):
- C : H = 0.06 / 0.06 : 0.14 / 0.06 = 1 : 2.33

To express this in whole numbers, we can multiply both parts by 3:
- C : H = 3 : 7

### Step 4: Write the empirical formula.
The empirical formula based on the ratio is C₃H₇.

### Step 5: Calculate the molecular mass of the empirical formula.
- Molecular mass of C₃H₇ = (3 × 12) + (7 × 1) = 36 + 7 = 43 g/mol.

### Conclusion:
The lowest molecular mass that compound X can have is **43 g/mol**.

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