Question

100 point question❤️


A compound contains 4.07 % of hydrogen , 24.27% carbon and 71.65 % of chlorine , its molar mass is 98.96. find out its empirical and molecular formuale.


solve it !!​

Answer 1

Answer:

In 100g of sample of the compound, 4.07g of hydrogen,

24.27g of carbon and 71.65g of chlorine are present.

Moles of hydrogen= 4.07g/ 1g = 4.0  

Moles of carbon= 24.27g/ 12g = 2.0  

Moles of chlorine= 71.65g/ 35g= 2.0  

Since 2.0 is the smallest value, so by dividing each of the mole values obtained by this smallest value we will get a ratio of 2:1:1 for H:C:Cl.  

Thus, the empirical formula of the compound is CH2Cl.  

For CH2Cl, empirical formula mass= 12+ (2×1) +35 = 49g.  

Molar mass/ empirical formula = 98.96g/ 49g = 2=n  

Therefore, Empirical formula = CH2Cl  n=2  Hence, molecular formula= C2H4Cl2.  

Answer 2

Answer:

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Step -1

Divide the given percentages of  atoms with their molecular masses

H---4.07/1      =4.07

C---24.27/12  =2.02

Cl--71.65/35.5=2.01

step 2

divide all values with the lowest value obtained.

 H---4.07/2.01=2

 C---2.02/2.01=1

 Cl---2.01/2.01=1

therefore the empirical formula is CH2Cll

WEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5

Given molecular weight =98.96 which is double of empirical weight .therefore molecular formula is C2H4Cl2