Two elements A (Atomic Mass 12 g/mole) and B (Atomic Mass 16 g/mole) combine to yield a compound. The percentage mass of A in the compound is `27.3%`. The formula of the compound will be :

Options

A. `A_(2)B_(2)`
B. AB
C. `A_(2)B`
D. `AB_(2)`

Answer

To find the formula of the compound formed by elements A and B, we will follow these steps:

### Step 1: Write down the given data
- Atomic mass of A = 12 g/mole
- Atomic mass of B = 16 g/mole
- Percentage mass of A in the compound = 27.3%
- Percentage mass of B in the compound = 100% - 27.3% = 72.7%

### Step 2: Calculate the moles of A and B
To find the number of moles of each element in the compound, we can use the formula:
\[ \text{Moles of A} = \frac{\text{Mass % of A}}{\text{Atomic mass of A}} = \frac{27.3}{12} \]
Calculating this gives:
\[ \text{Moles of A} = \frac{27.3}{12} \approx 2.275 \]

Next, for B:
\[ \text{Moles of B} = \frac{\text{Mass % of B}}{\text{Atomic mass of B}} = \frac{72.7}{16} \]
Calculating this gives:
\[ \text{Moles of B} = \frac{72.7}{16} \approx 4.54 \]

### Step 3: Find the simplest whole number ratio
Now we will find the simplest whole number ratio of moles of A to moles of B:
- Moles of A = 2.275
- Moles of B = 4.54

To simplify this ratio, we can divide both values by the smaller number (2.275):
\[ \text{Ratio of A} = \frac{2.275}{2.275} \approx 1 \]
\[ \text{Ratio of B} = \frac{4.54}{2.275} \approx 2 \]

### Step 4: Write the empirical formula
From the ratio calculated, we find that:
- The ratio of A to B is approximately 1:2.

Thus, the empirical formula of the compound is:
\[ \text{Formula} = A_1B_2 \quad \text{or simply} \quad AB_2 \]

### Final Answer:
The formula of the compound is **AB₂**.

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