A hydrocarbon contains 10.5 , g of carbon and | vedclass

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

Question

(D) Step $1$: Determine the empirical formula.$Moles \, of \, C = \frac{10.5 \, g}{12 \, g/mol} = 0.875 \, mol$$Moles \, of \, H = \frac{1 \, g}{1 \,

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Step $1$: Determine the empirical formula.$Moles \, of \, C = \frac{10.5 \, g}{12 \, g/mol} = 0.875 \, mol$$Moles \, of \, H = \frac{1 \, g}{1 \, g/mol} = 1 \, mol$Ratio $C:H = 0.875:1 \approx 7:8$. Thus,the empirical formula is $C_7H_8$.Step $2$: Determine the molar mass $(M)$ using the ideal gas equation $PV = nRT = \frac{w}{M}RT$.$P = 1 \, atm, V = 1 \, L, w = 2.4 \, g, R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}, T = 127 + 273 = 400 \, K$.$M = \frac{wRT}{PV} = \frac{2.4 	imes 0.0821 	imes 400}{1 	imes 1} = 78.8 \, g/mol \approx 79 \, g/mol$.Step $3$: Compare empirical mass with molar mass.Empirical mass of $C_7H_8 = (7 	imes 12) + (8 	imes 1) = 92 \, g/mol$.Since the calculated molar mass $(79 \, g/mol)$ does not match the empirical formula mass or its multiples,the correct hydrocarbon is not listed among the options.

$A$ hydrocarbon contains $10.5 \, g$ of carbon and $1 \, g$ of hydrogen. If the volume of $2.4 \, g$ of this hydrocarbon is $1 \, L$ at $127 \, ^\circ C$ and $1 \, atm$ pressure,then the hydrocarbon is:

Similar Questions

By combustion of $0.2475 \ g$ of an organic compound,$0.4950 \ g$ of $CO_2$ and $0.2025 \ g$ of $H_2O$ are obtained. Calculate the percentage of $C, H,$ and $O$. Given atomic masses: $C = 12, H = 1, O = 16$.

The empirical formula of a compound is $CH_2O$ and its molecular weight is $120$. The molecular formula of the compound is:

The percentage of $N_2$ in urea is about

An organic compound contains $69 \%$ carbon and $4.8 \%$ hydrogen,the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when $0.20 \ g$ of this substance is subjected to complete combustion.

What is the formula called that represents the simplest ratio of atoms present in a molecule of a substance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module