Two isotopes of Boron occur in nature. Boron-10 has a mass of 10.013 amu and an abundance of 19.61%. Boron-11 has an abundance of 80.39%. The molar mass of a natural sample of boron is 10.811 g/mol. What is the mass of a boron-11 atom?
Please show how work is done (and no this is not homework, this is review for my final). Thanks.
2006-12-21 05:14:53 · 4 answers · asked by Anitec 2 in Science & Mathematics ➔ Chemistry
10.81=1.96354+(.8039)x
11.01 is the right answer but for some reason I still can't get that to work. What exactly do you do from here?
2006-12-21 05:37:35 · update #1
Molar Mass(B)=Mass(B10)(abun(B10)/100)+ Mass(B11)(abun(B11)/100)
10.81=(10.013)(.1961)+(.8039)(x)
10.81=1.96354+(.8039)x
x=11.004g
I hope this helps!!
2006-12-21 05:23:35 · answer #1 · answered by smart-crazy 4 · 0⤊ 0⤋
easy
molar mass is the same (barring the units) as the atomic mass. So, average atomic mass of Boron is 10.811 amu's.
And now this is made of Boron 10, at 10.013 amu's, and Boron 11. So you've got a simple equation:
(10.013)*(19.61%) + x * (80.39%) = 10.811
x * (80.39%) = 8.8475
x = 11.0057
Hope this helps
2006-12-21 14:11:24 · answer #2 · answered by AntoineBachmann 5 · 0⤊ 0⤋
(10.013 x 19.61) + (y x 80.39) / 100 = 10.811
Solve for y.
By "mass of a boron atom" I expect the question means relative atomic mass.
2006-12-21 13:19:28 · answer #3 · answered by Gervald F 7 · 0⤊ 0⤋
OK!
(B-10 amu)*(0.1961) + (B-11 amu)*(0.8939) = (B natural) (1.0000)
(10.013)*(0.1961) + (B-11 amu)*(0.8939) = (10.811) (1.0000)
Solve for (B-11 amu)
2006-12-21 13:17:30 · answer #4 · answered by Jerry P 6 · 0⤊ 1⤋