Find emperical formula of a compound having composition c-19.57%, fe-15.2%, n-22.83%,k-42.39%
Answers
- E stands for ''Elements''
- % '' '' '' '' Percentage
- A.T '' '' Atomic Weight
- S.R '' '' Simplest ratio
- W.R '' '' Whole Ratio
→ E →% →A.T →%/A.T →S.R →W.R
C 19.57% 12 19.57/12 = 1.63 1.63/0.27 6
Fe 15.2% 56 15.2/56 = 0.27 0.27 /0.27 1
N 22.83% 14 22.83/14 = 1.63 1.63/0.27 6
K 42.39% 39 42.39/39 = 1.09 1.09/0.27 4
So
- Empirical formula = C₆ Fe N₆ K₄
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