What is the empirical formula for a compound that is 24.7 % calcium, 1.20 % hydrogen, 14.8 % carbon, and 59.3 % oxygen? *

Chemistry

2 answers:

goldfiish [28.3K]3 years ago

6 0

Answer:

$C_2H_2O_6Ca$

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 ,

$\boxed{\begin{array}{c|c|c|c|c} \underline{\bf Element} & \underline{\bf Percentage} & \underline{\bf Atomic \ Weight } &\underline{\bf No. \ of \ atoms_{relative} } &\underline{\bf Ratio } \\\\ \sf Ca & \sf 24.7 &\sf 40 &\sf \dfrac{24.7}{40}=0.6 &\sf \dfrac{0.6}{0.6}=1 \\\\\sf H &\sf 1.20 &\sf 1 &\sf \dfrac{1.20}{1}=1.20&\sf \dfrac{1.20}{0.6}=2\\\\\sf C &\sf 14.8 &\sf 12&\sf\dfrac{14.8}{12}=1.23&\sf \dfrac{1.23}{0.6}=2\\\\\sf O &\sf 59.3 &\sf 16 &\sf\dfrac{59.3}{16}=3.7&\sf \dfrac{3.7}{0.6}=6 \end{array}}$