Mass of AgBr formed = 3.36 g
Molar mass of AgBr=108+80=188 g mol−1
Moles of AgBr formed = 1883.36≈0.01787 mol
Since 1 mole of AgBr contains 1 mole of Br, moles of Br in the compound = 0.01787 mol.
Mass of Br=0.01787×80=1.43 g
Percentage of Br in compound (X)=2.01.43×100=71.5%
Given percentage of C=26.7%
Percentage of H=100%−(71.5%+26.7%)=1.8%
Now, we find the molar ratio of the elements in the compound:
Moles of C=1226.7=2.225
Moles of H=11.8=1.8
Moles of Br=8071.5=0.893
Dividing by the smallest value (0.893) to get the simplest ratio:
C:H:Br=0.8932.225:0.8931.8:0.8930.893
C:H:Br≈2.5:2:1
Multiplying by 2 to obtain whole numbers, we get the ratio 5:4:2.
The empirical formula of the compound is C5H4Br2.
The total number of carbon atoms in the empirical formula is 5.
Answer: 5