A)
B) \[{{C}_{4}}H{{O}_{4}}\]
C) CHO
D) \[C{{H}_{2}}O\]
Correct Answer: D
Solution :
Percentage \[O%=100-(40+6.66)\]\[=53.34%\]Symbol | % | Atomic mass | Relative no. of atoms | Simple Reaction |
C | 40 | 12 | \[\frac{40}{12}=3.33\] | \[\frac{3.33}{3.33}=1\] |
H | 6.66 | 1 | \[\frac{6.66}{1}=6.66\] | \[\frac{6.66}{3.33}=2\] |
O | 53.34 | 16 | \[\frac{53.34}{16}3.33\] | \[\frac{3.33}{3.33}=1\] |
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