1. | \(\mathrm{{KAl}({SO}_4)_2 \cdot 12 {H}_2 {O}}\) |
2. | \(\mathrm{{K}_2 {Al}_2({SO}_4)_6 \cdot 12 {H}_2 {O}}\) |
3. | \(\mathrm{{K}_2 {SO}_4 \cdot {Al}_2({SO}_4)_3 \cdot24 {H}_2 {O}}\) |
4. | \(\mathrm{{K}_2 {SO}_6 \cdot {Al}_2({SO}_4)_3 \cdot12 {H}_2 {O}}\) |
1. | \(312 \mathrm{~g} \mathrm{~mol}^{-1}\) | 2. | \(120 \mathrm{~g} \mathrm{~mol}^{-1}\) |
3. | \(60 \mathrm{~g} \mathrm{~mol}^{-1}\) | 4. | \(811.2 \mathrm{~g} \mathrm{~mol}^{-1}\) |
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.