Determine the empirical formula of Kevlar,use | vedclass

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(D) Assume the total mass of the compound is $100 \ g$. Number of moles of $C = \frac{70.6 \ g}{12 \ g/mol} = 5.88 \ mol$. Number of moles of $H = \fr

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$C_7H_5NO$

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

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

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