1. | \(\mathrm{KAl}\left(\mathrm{SO}_4\right)_2 \cdot 12 \mathrm{H}_2 \mathrm{O}\) |
2. | \(\mathrm{K}_2 \mathrm{Al}_2\left(\mathrm{SO}_4\right)_6 \cdot 12 \mathrm{H}_2 \mathrm{O}\) |
3. | \(\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \cdot24 \mathrm{H}_2 \mathrm{O}\) |
4. | \(\mathrm{K}_2 \mathrm{SO}_6 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \cdot12 \mathrm{H}_2 \mathrm{O}\) |
An organic compound contains 80 % (by wt.) carbon and the remaining percentage of hydrogen. The empirical formula of this compound is:
[Atomic wt. of C is 12, H is 1]
1. | CH3 | 2. | CH4 |
3. | CH | 4. | CH2 |
The percentages of C, H, and N in an organic compound are 40%, 13.3%, and 46.7% respectively.
The empirical formula of the compound is:
1. | C3H13N3 | 2. | CH2N |
3. | CH4N | 4. | CH6N |
A compound contains C, H, and O. If C = 40% and H = 6.67% and rest is oxygen, then the empirical formula of the compound will be:
1.
2.
3.
4. CHO
Percentage of C, H & N of a compound is given as follows:
C=40%, H=13.33%, N=46.67%
The empirical formula of the compound will be:
1.
2.
3.
4.