# Chemistry

Boron has only two naturally occurring isotopes. The mass of boron-10 is 10.01294 amu and the mass of boron-11 is 11.00931 amu.

Use the atomic mass of boron to calculate the relative abundance of boron-10.

Atomic mass of boron is 10.81

How do I solve this?

1. 10.81 = (10.01294x) + (11.00931y)
y=1-x
10.81 = (10.01294x) + (11.00931(1-x))
10.81 = 10.01294x + 11.00931 + -11.00931x
10.81 = 10.01294x + 11.00931 + -11.00931x
10.81 = -0.99637x + 11.00931
10.81-11.00931 = -0.99637x + 11.00931-11.00931
-0.2831 = -0.99637x
-0.2831/-0.99637 = 0.99637/-0.99637x
0.28413139697 = x
Boron 10 = 28.41%
Boron 11 = 71.59%(subtract 100%- 28.41%)

posted by R.C

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