Question

Two elements $A$ and $B$ have atomic mass $75$ and $16$ respectively. They combine to give a compound having $75.8\%$ $A$. The empirical formula of the compound is:

• A. $AB$
• B. $A_{2}B$
• C. $A{B}_2$
• D. $A_2{B}_3$

Answer 1

The correct option is D $A_2{B}_3$

Let the total mass of the compound be $100g$
Then, mass of $A$ would be $75.8g$
Mass of $B$ would be $24.2g$
Now, moles of $A=\frac{75.8}{75}\approx 1$
moles of $B=\frac{24.2}{16}\approx 1.5$
Hence the required empirical formula = $A{B}_{1.5}=A_2{B}_3$.