Question

A compound contains 4.07% hydrogen,24.27% carbon and 71.65% chlorine. Its molar mass is 98.96g Determine its empirical and molecular formulas

Answer 1

Explanation:

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

Answer 2

Answer:

Explanation:

%         mass   ratio1 ratio2

4.07         1          4       2

24.27       12        2       1

71.65        35.5     2       1

CH2Cl   =    49.5    emperical formula

(CH2Cl   =    49.5)*2 = C2H4Cl2  = molecular formula