Chapter 42, Problem 019 A periodic table might list the average atomic mass of m

Question

Chapter 42, Problem 019 A periodic table might list the average atomic mass of magnesium as being 24.312 u. That average value is the result of weighting the atomic masses of the magnesium isotopes according to their natural abundances on Earth.The three isotopes and their masses are 24Mg (23.98504 u), 25Mg (24.98584 u), and 26Mg (25.98259 u). The natural abundance of 24Mg is 78.99 % by mass (that is, 78.99 % of the mass of a naturally occurring sample of magnesium is due to the presence of 24Mg).What is the abundance of (a)25Mg and (b)26Mg? (a) Number Units (b) Number Units

Explanation / Answer

Let us first list the given details:

24Mg of mass 23.98504 u has a natural abundance of 78.99%

25Mg of mass 24.98584 u has a natural abundance of x%

26Mg of mass 25.98259 u has a natural abundance of y%

As all the three isotopes and their abundance contribute to the average atomic mass of magnesium,

x+y=100-78.99= 21.01......................................................................1

Now let us multiply the abundances(=percentage abdunace/100) with their respective atomic masses to find the average atomic mass.

For 24Mg, 23.98504*0.7899 =18.9457831 u

For 25Mg, 24.98584x u

For 26Mg, 25.98259y u

Adding the three abundances gives us the average atomic mass

Average atomic mass= (18.9457831+24.98584x+25.98259y) u

But it is given that the average atomic mass= 24.312 u

Thus, 24.312= (18.9457831+24.98584x+25.98259y)

5.3662169=24.98584x+25.98259y..........................................................2

Equation 1 and 2 and a pair of simulataneous equations that can be solved to find the value of x and y

5.3662169=24.98584(0.2101-y)+25.98259y

5.3662169=(-24.98584+25.98259)y+5.249524984

y=0.11707

Thus, x=0.2101-y= 0.2101-0.11707

x=0.09303

X and y are the abundance values of 25Mg and 26Mg respectively

Therefore,

a) Percentage Abundance of 25Mg = 11.707%

a) Percentage Abundance of 26Mg = 9.303%

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