IsotopeRelative abundance (%)Atomic mass (u)^ | vedclass

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

Question

(A) The average atomic mass of carbon is calculated as follows:Average atomic mass of $C = (0.988 	imes 12) + (0.0118 	imes 13.1) + (0.0002 	imes 1

Vedclass · Accepted Answer

$16.0452$

Vedclass · Answer

(A) The average atomic mass of carbon is calculated as follows:Average atomic mass of $C = (0.988 	imes 12) + (0.0118 	imes 13.1) + (0.0002 	imes 14.1)$$= 11.856 + 0.15458 + 0.00282 = 12.0134 \ u$The molecular mass of $CH_4$ is given by the sum of the average atomic mass of carbon and $4$ times the atomic mass of hydrogen:Molecular mass of $CH_4 = 12.0134 + (4 	imes 1.008)$$= 12.0134 + 4.032 = 16.0454 \ u$Rounding to the nearest provided option,the correct value is $16.0452 \ u$.

Isotope	Relative abundance $(\%)$	Atomic mass $(u)$
$^{12}C$	$98.8$	$12$
$^{13}C$	$1.18$	$13.1$
$^{14}C$	$0.02$	$14.1$

Similar Questions

The present atomic weight scale is based on

The sulphate of a metal has the formula $M_2(SO_4)_3$. The formula for its phosphate will be

The molecular weight of a tribasic acid is $M$. What will be its equivalent weight?

Among $10^{-9} \ g$ (each) of the following elements,which one will have the highest number of atoms?
Element: $Pb, Po, Pr$ and $Pt$

In nature,carbon exists as a mixture of $C^{12}$ and $C^{13}$. The average atomic mass of carbon is $12.011$. What is the percentage abundance of $C^{12}$ in nature?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

IsotopeRelative abundance $(\%)$Atomic mass $(u)$$^{12}C$$98.8$$12$$^{13}C$$1.18$$13.1$$^{14}C$$0.02$$14.1$From the above data,what is the average molecular mass of $CH_4$ containing all isotopes of carbon,assuming hydrogen is only $^{1}_{1}H$? (Given that atomic mass of hydrogen $= 1.008 \ u$)