MCQ
Determine the empirical formula of Kelvar, used in making bullet proof vests, is $70.6\%\,C, 4.2\%\,H, 11.8\%\, N$ and $13.4\%\, O$
- A$C_7H_5NO_2$
- B$C_7H_5N_2O$
- C$C_7H_9NO$
- ✓$C_7H_5NO$
$C=70.6 g=\frac{70.6}{12}$ moles $=5.88$ moles
$H=4.2 g=\frac{4.2}{1}$ moles $=4.2$ moles
$N=11.8 g=\frac{11.8}{14}$ moles $=0.84$ moles
$O=13.4 g=\frac{13.4}{16}$ moles $=0.84$ moles
$C: H: N: O=5.88: 4.2: 0.84: 0.84$
$C: H: N: O=7: 5: 1: 1$
Therefore, empirical formula of Kelvar is $C_{7} H_{5} N O$
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(Atomic mass, $Ag =108, Br =80\, g\, mol ^{-1}$ )