Question

Element A reacts with oxygen to form a compound ${A_2}{O_3}$​. If $0.359$gram of A reacts to give $0.559$ gram of the compound, then atomic weight of A will be:
A) $51$
B) $43.08$
C) $49.7$
D) $47.9$

Answer 1

The relative molecular mass of a compound is that the sum of the atomic weights of the atoms within the molecules that form these compounds. Follow some steps to find out the atomic weight of compound, which are given as follows:
To find atomic weight first determine the molecular formula of a compound.
Use a periodic table to determine atomic mass of each element.
Multiply each element's atomic mass by the number of atoms of that element in the molecule.
Add all values together for each different atom in the molecule; value represents atomic mass of the element.

Complete answer:
Step 1: First we have to write the given data in Question
Let atomic mass of element A be $= a{\text{ }}g/mol$
Atomic mass of oxygen $= {\text{ }}16{\text{ }}g/mol$
So molar mass of the compound ${A_{2}}{O_3}{\text{ }} = {\text{ }}\left( {2a + 16x{\text{ }}3} \right)g/mol$
One molecule of the compound ${A_{2}}{O_3}$ ​ Contains $2$ atoms.
So the ratio of mass of A and ${A_{2}}{O_3}$ in a molecule = $\dfrac{{2a}}{{2a + 48}}$
Step 2: Now, it is given that $0.359$ grams of A reacts to give $0.559$ grams of the compound. So we can write
=>$\dfrac{{2a}}{{2a + 48}} = \dfrac{{0.359}}{{0.559}}$
=> $\dfrac{a}{{a + 24}} = \dfrac{{359}}{{559}}$
⇒$1 + (24/a) = \dfrac{{559}}{{359}} = 1.55$
⇒$(24/a) = 0.55$
On solving, $a = 43.6$

So, the correct option is (B) .

Note:
While writing the atomic mass of a compound always remembers, no subscript after an element symbol represents one atom. Use the correct number of significant figures for your answer.