The percent composition by mass of an unknown chlorinated hydrocarbon was found to be 37.83% C, 6.35% H, and 55.83% Cl by mass. What is the empirical formula of this compound?

C2H4Cl

Explanation:

The first step in obtaining the empirical formula is to divide the percent by mass of each element by its relative atomic mass;

C- 37.83/12, H- 6.35/1, Cl- 55.83/35.5

3.15, 6.35, 1.57

Divide through by the lowest ratio

3.15/1.57, 6.35/1.57, 1.57/1.57

2, 4, 1

Hence, empirical formula is C2H4Cl

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

c. 75.7

i think that is the right