A hydrocarbon contains 85.7 % C. If 42 , mg o | vedclass

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

Question

(B) $1$. Calculate the molar mass of the hydrocarbon:Given $42 \, mg = 0.042 \, g$ contains $3.01 	imes 10^{20}$ molecules.Number of moles $n = \frac

Vedclass · Accepted Answer

$C_6H_{12}$

Vedclass · Answer

(B) $1$. Calculate the molar mass of the hydrocarbon:Given $42 \, mg = 0.042 \, g$ contains $3.01 	imes 10^{20}$ molecules.Number of moles $n = \frac{3.01 	imes 10^{20}}{6.022 	imes 10^{23}} \approx 0.5 	imes 10^{-3} \, mol = 5 	imes 10^{-4} \, mol$.Molar mass $M = \frac{	ext{mass}}{	ext{moles}} = \frac{0.042 \, g}{5 	imes 10^{-4} \, mol} = 84 \, g/mol$.$2$. Determine the number of atoms of $C$ and $H$:Mass of $C = 85.7 \% 	ext{ of } 84 = 0.857 	imes 84 \approx 72 \, g$.Number of $C$ atoms $= \frac{72 \, g}{12 \, g/mol} = 6$.Mass of $H = 84 - 72 = 12 \, g$.Number of $H$ atoms $= \frac{12 \, g}{1 \, g/mol} = 12$.$3$. Conclusion:The molecular formula is $C_6H_{12}$.

$A$ hydrocarbon contains $85.7 \% C$. If $42 \, mg$ of the compound contains $3.01 \times 10^{20}$ molecules,the molecular formula of the compound will be

Similar Questions

$A$ hydrocarbon has $C = 85.72\%$ and the remaining is $H$. The hydrocarbon is:

An organic compound containing $C$, $H$, and $N$ has $20 \%$ nitrogen by mass. What is the minimum possible molecular weight of this compound?

The formula of a substance is $(CO)_x$. If its vapor density is $70$,find the value of $x$.

When a crystalline salt $Na_2SO_4 \cdot xH_2O$ is heated,it loses $55.9\%$ of its weight. What is the formula of the crystalline salt?

How much copper can be obtained from $100 \, g$ of copper sulphate $(CuSO_{4})$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module