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Rina8888 [55]
3 years ago
9

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}}

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 ,

\longrightarrow\underline{\boxed{\bf Empirical \ formula = C_3H_2O_6 Ca  }}

Nataliya [291]3 years ago
5 0

The empirical formula of the compound is written as C₂H₂O₆Ca.

<h3>What is empirical formula?</h3>

The empirical formula of a compound is the simplest structural formula of the compound.

Fraction of the elements in the given compound;

Ca, = 24.7/40 = 0.62

H, = 1.2/1 = 1.2

C, = 14.8/12 = 1.23

O, = 59.3/16 = 3.7

Ca: = 0.62/0.62 = 1

H: = 1.2/0.62 = 2

C: = 1.23/0.62 = 2

O: = 3.7/0.62 = 6

Thus, the empirical formula of the compound is written as C₂H₂O₆Ca.

Learn more about empirical formula here: brainly.com/question/1603500

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