A compound made up of two elements A and B has `A = 70%, B = 30%`. Their relative number of moles in the compound is 1.25 and 1.88, calculate :
Molecular formula of the compound , if its molecular mass is found to be 160.

Answer

To find the molecular formula of the compound made up of elements A and B, we will follow these steps:

### Step 1: Calculate the Atomic Mass of A and B
The atomic mass can be calculated using the formula:
\[ \text{Atomic Mass} = \frac{\text{Percentage of Element}}{\text{Relative Number of Moles}} \]

#### For Element A:
- Percentage of A = 70%
- Relative number of moles of A = 1.25

\[ \text{Atomic Mass of A} = \frac{70}{1.25} = 56 \text{ g/mol} \]

#### For Element B:
- Percentage of B = 30%
- Relative number of moles of B = 1.88

\[ \text{Atomic Mass of B} = \frac{30}{1.88} \approx 15.96 \text{ g/mol} \approx 16 \text{ g/mol} \]

### Step 2: Find the Simplest Ratio of Moles
To find the simplest ratio of A and B, we will divide the relative number of moles by the smallest number of moles.

#### For A:
\[ \text{Simplest Ratio for A} = \frac{1.25}{1.25} = 1 \]

#### For B:
\[ \text{Simplest Ratio for B} = \frac{1.88}{1.25} \approx 1.504 \]

### Step 3: Convert to Whole Numbers
To convert the simplest ratio into whole numbers, we can multiply both ratios by 2 (to eliminate the decimal).

- For A: \(1 \times 2 = 2\)
- For B: \(1.504 \times 2 \approx 3\)

Thus, the simplest whole number ratio of A to B is 2:3.

### Step 4: Write the Empirical Formula
From the simplest ratio, we can write the empirical formula of the compound as:
\[ \text{Empirical Formula} = A_2B_3 \]

### Step 5: Calculate the Molecular Mass
Now, we need to verify if the molecular mass of the compound matches the given molecular mass of 160 g/mol.

\[ \text{Molecular Mass} = (2 \times \text{Atomic Mass of A}) + (3 \times \text{Atomic Mass of B}) \]
\[ = (2 \times 56) + (3 \times 16) \]
\[ = 112 + 48 = 160 \text{ g/mol} \]

### Conclusion
Since the calculated molecular mass matches the given molecular mass, the molecular formula of the compound is:
\[ \text{Molecular Formula} = A_2B_3 \]