Answer:
[tex] C_2H_2O_6Ca [/tex]
Explanation:
Here we are given that the percentage of ,
- Calcium = 24.7%
- Hydrogen = 1.20%
- Carbon = 14.8%
- Oxygen = 59.3%
And we need to find out the Empirical formula of the compound .
- Firstly divide the given percentage by the atomic mass of the compound . Here this is followed in the fourth column of the table , which will give the relative number of atoms .
- Secondly , divide the value obtained by the smallest value among them .Say if we got the relative number of atoms as 4 , 3 and 2 , then divide all three numbers by 2 . ( followed in 5th column below .)
- Finally write the Empirical formula of the compound by writing the ratio obtained as the subscript of respective compounds .
Let's make a table ,
[tex]\boxed{\begin{array}{c|c|c|c|c} \underline{\bf Element} & \underline{\bf Percentage} & \underline{\bf Atomic \ Weight } &\underline{\bf No. \ of \ atoms_{relative} } &\underline{\bf Ratio } \\\\ \sf Ca & \sf 24.7 &\sf 40 &\sf \dfrac{24.7}{40}=0.6 &\sf \dfrac{0.6}{0.6}=1 \\\\\sf H &\sf 1.20 &\sf 1 &\sf \dfrac{1.20}{1}=1.20&\sf \dfrac{1.20}{0.6}=2\\\\\sf C &\sf 14.8 &\sf 12&\sf\dfrac{14.8}{12}=1.23&\sf \dfrac{1.23}{0.6}=2\\\\\sf O &\sf 59.3 &\sf 16 &\sf\dfrac{59.3}{16}=3.7&\sf \dfrac{3.7}{0.6}=6 \end{array}}[/tex]
Therefore we got the ratio of Ca , H , C and O as 1:2:2:6 . Therefore the Empirical formula of the compund will be ,
[tex]\longrightarrow\underline{\boxed{\bf Empirical \ formula = C_3H_2O_6 Ca }}[/tex]
