Determine the empirical formula of Kevlar,use | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $13.4 \% \, O$.

Similar Questions

On analysis,a saturated hydrocarbon is found to contain $83.70\%$ carbon and $16.30\%$ hydrogen. The empirical formula will be (at. wt. of $C=12$, at. wt. of $H=1$).

$0.765 \ g$ of an acid gives $0.535 \ g$ of $CO_2$ and $0.13 \ g$ of $H_2O$. The ratio of the percentage of $C$ and $H$ is:

The molecular formula of feldspar is

An organic compound on analysis gave $C = 48 \,g$, $H = 8 \,g$ and $N = 56 \,g$. The volume of $1.0 \,g$ of the compound was found to be $200 \,mL$ at $NTP$. What is the molecular formula of the compound?

When equal weights of the two fertilizers,urea and ammonium sulphate are taken,urea contains

Vedclass Test Series

Exam Paper Generator

Online Exam Module