Question

x and y are two different element having their atomic mass ratio 1:2. the compound formed by containing x and y contains 50%x by weight. the emperical formula of compound is​

Answer 1

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Given that , x and y have their atomic mass ratio as , 1:2 . Also when they form a Compound then , the percentage of x in Compound is 50%. Therefore the percentage of y in the Compound should be (100-50)% = 50% . Now for finding the Empirical Formula , we follow the following steps .

$\large\red{\dashrightarrow}\underline{\sf Steps \ of \ Finding \ :- }$

• Divide the percentage of the elements of the Compound by their Atomic masses , to find the Relative number of atoms.
• Divide all the numbers obtaines by the smallest number obtained by previous step to find the Simplest Ratio . Say we got the nos. 1 , 2 & 3 in front of elements , A , B and C in the table. Then the Empirical formula will be , A₁B₂C₃ .

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Let us take the ratio of their weight be 1k : 2k . Now since we will find the ratio , k will eventually get cancelled at last. So we can neglect writing k here. We can simply write the masses are 1 and 2 respectively .

• Making a table :-

$\boxed{\begin{array}{c|c|c|c|c}\underline{\sf \red{ Element}} & \underline{\sf \red{ Percentage}} & \underline{\sf \red{ At. \ mass }} & \underline{\sf \red{ Relative\ no.} }& \underline{\sf \red{ Simplest\ Ratio }} \\\\\sf x & \sf 50\% &\sf 1 &\sf \dfrac{50}{1k}= 50 &\sf \dfrac{50/k}{25/k}= 2 \\\\\sf y &\sf 50\%&\sf 2 &\sf \dfrac{50}{2k}= 25 &\sf \dfrac{25/k}{25/k}= 1 \end{array}}$

Therefore we got the simple ratio as 2 & 1 respectively of x and y. Therefore the Empirical Formula of the Compound will be ,

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$\\\\\large\sf\dashrightarrow Compound =\boxed{\pink{\sf X_2Y}}$

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