Determine the empirical formula of Kevlar,use | vedclass

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(D) Assume the mass of the compound is $100 \ g$.Moles of $C = \frac{70.6 \ g}{12 \ g/mol} = 5.88 \ mol$.Moles of $H = \frac{4.2 \ g}{1 \ g/mol} = 4.2

Vedclass · Accepted Answer

$C_7H_5NO$

Vedclass · Answer

(D) Assume the mass of the compound is $100 \ g$.Moles of $C = \frac{70.6 \ g}{12 \ g/mol} = 5.88 \ mol$.Moles of $H = \frac{4.2 \ g}{1 \ g/mol} = 4.2 \ mol$.Moles of $N = \frac{11.8 \ g}{14 \ g/mol} = 0.84 \ mol$.Moles of $O = \frac{13.4 \ g}{16 \ g/mol} = 0.84 \ mol$.Divide each by the smallest value $(0.84)$:$C : H : N : O = \frac{5.88}{0.84} : \frac{4.2}{0.84} : \frac{0.84}{0.84} : \frac{0.84}{0.84} = 7 : 5 : 1 : 1$.Therefore,the empirical formula of Kevlar is $C_7H_5NO$.

Determine the empirical formula of Kevlar,used in making bulletproof vests,which contains $70.6 \% \, C, \, 4.2 \% \, H, \, 11.8 \% \, N$ and $13.4 \% \, O$.

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