Question

Deuterium, 2h (2.0140 u), is sometimes used to replace the principal hydrogen isotope 1h in chemical studies. the percent natural abundance of deuterium is 0.015%. part a if it can be done with 100% efficiency, what mass of naturally occurring hydrogen gas would have to be processed to obtain a sample containing 2.50Ã1021 2h atoms?

Answer 1

Deuterium is isotope of hydrogen with percent natural abundance 0.015%. Thus, if 100 gram of naturally occurring hydrogen gas is taken it will have 0.015 g of deuterium.

Or,

$0.015g ^{2}H\rightarrow 100 g ^{1}H$

or,

$1 g^{2}H\rightarrow 6666.67 g ^{1}H$

The number of deuterium atoms are given $2.50\times 10^{21}$.

Since, 1 mol is equal to $6.023\times 10^{23}$ atoms thus,

$1 atom\rightarrow \frac{1}{6.023\times 10^{23}}mol=1.66\times 10^{-24}mol$

Thus,  $2.50\times 10^{21}$ will be equal to $2.50\times 10^{21}\times 1.66\times 10^{-24}=0.00415 mol$

Molar mass of deuterium is 2.0140 u or 2.0140 g/mol thus, mass can be calculated as:

$m=n\times M=0.00415 mol\times 2.0140 g/mol=0.00836 g$

thus, mass of $^{2}H$ is 0.00836 g

Since, $1 g^{2}H\rightarrow 6666.67 g ^{1}H$

Thus, $0.00836 g ^{2}H\rightarrow 0.00836\times 6666.67=55.73 g ^{1}H$

Therefore, mass of naturally occurring hydrogen gas that has to be processed is 55.73 g.