Answer:
H O
Explanation:
Here we are given that the percentage of ,
- Oxygen = 94.10%
- Hydrogen = 5.90%
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 . ( follówed 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 O & \sf 94.10 &\sf 16 &\sf \dfrac{94.10}{16}=5.88 &\sf \dfrac{5.88}{5.88}=1 \\\\\sf H &\sf 5.90 &\sf 1 &\sf \dfrac{5.90}{1}=5.90&\sf \dfrac{5.90}{5.88}\approx 1 \end{array}}[/tex]
Hence from this table we see that the ratio of oxygen and hydrogen is 1:1 . Therefore the Empirical formula of the compóund will be ,
[tex]\longrightarrow\underline{\boxed{\bf Empirical \ formula = H_1 O_1 = H \ O }}[/tex]
