Let X = mass Br-81
Note: If Br-79 is 50.69%, then Br-81 must be 100-50.69 = 49.31%.
(78.918*0.5069) + (X*0.4931) = Y where Y = the value you look up on the periodic table. Solve for X.
Bromine has two naturally occurring isotopes. Bromine-79 has a mass of 78.918 amu and is 50.69% abundant. Using the atomic mass reported on the periodic table, determine the mass of bromine-81, the other isotope of bromine.
5 answers
4.010
Br-81= 80.92 amu
Process:
[convert percent into decimal form, if using a scientific calculator ignore this step]
(78.918*50.69%)+(x*49.31%)=79.904 [x is the mass of Br-81]
(40.0)+(49.31%x)=70.904 [combine first the values in parenthesis]
49.31%x=79.904-40.0 [move 40 to get the value of x]
x=39.904/49.31% [divide 49.31% to both sides]
x=80.92476171 [use 4 significant figures for the final answer]
Process:
[convert percent into decimal form, if using a scientific calculator ignore this step]
(78.918*50.69%)+(x*49.31%)=79.904 [x is the mass of Br-81]
(40.0)+(49.31%x)=70.904 [combine first the values in parenthesis]
49.31%x=79.904-40.0 [move 40 to get the value of x]
x=39.904/49.31% [divide 49.31% to both sides]
x=80.92476171 [use 4 significant figures for the final answer]
If Br-79 is 50.69%, then Br-81 must be 100-50.69 = 49.31%.
(78.918*50.69) + (X*49.31) = 79.90*100
(4000.35342)+(x*49.31)=7990
(x*49.31)=7990-4000.35342
(x*49.31)=3989.64658
(x)=3989.64658/49.31
x=80.9094
(78.918*50.69) + (X*49.31) = 79.90*100
(4000.35342)+(x*49.31)=7990
(x*49.31)=7990-4000.35342
(x*49.31)=3989.64658
(x)=3989.64658/49.31
x=80.9094
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