a sample has 10.06% carbon, 89.10% chlorine and the rest hydrogen. whats the empirical formula of the compound

Question

b) also if the molar mass of this compound is 119.5 determine the molecular formula...


i am a newb to chemistry.. can someone do this for me.. so i can learn how to do it. thanks :)

Answer 1

10.06 + 89.10 + x = 100

x = 0.84 % = H

10.06 g / 12.011 = 0.837 C

89.10 g / 35.453 = 2.51 Cl

0.84 / 1.00794 = 0.833 H

we divide for the smallest number and we get

0.837/0.833 = 1 C

2.51 / 0.833 = 3 Cl

0.833/0.833 =1 H

C Cl3 H ( mass = 119.5 g/mol)

the compound is CHCl3

Answer 2

10.06% + 89.10% = 99.16% So H=0.86%
Atomic weights: Cl=35.5 C=12 H=1

10.06/12 = 0.84
89.10/35.5 = 2.51
0.86/1.00 = 0.86

Divide by the smallest:


0.84/0.84 = 1
2.51/0.84 = 3
0.86/0.84 = 1

So the empirical formula is CHC3, formula weight 119.5