If the mass of Br-81 is 80.9163 amu what is the mass of Br 79 I know the answer is 78.9183 but how does one find the answer?

Answer 1

Problems like these are somewhat tricky. Yes, the mass of the Br-79 isotope is 78.9183. Bromine has only two naturally occurring isotopes Br-79 and Br-81. Br-81 is 49.31% abundant. Knowing this we can determine Br-79's percent abundance. We determine Br-79's percentage abundance by taking 100.00% and subtracting 49.31%. We can do this safely because all % are out of 100 and we know that there are only two Bromine isotopes. When the math problem has been worked out, we conclude that Br-79 is 50.69% abundant. 100.00% - 49.31% = 50.69% If we start to do weighted averages (the way these problems are solved), we end up with this: Isotope - Weight amu - Percentage Abundance Br-79 - X amu - 50.69% Br-81 - 80.9163 amu - 49.31% Amu means Atomic Mass Unit. We need to solve for x Now, when we begin our weighted average problems we must convert our percentages to decimal form. To do this we divide each percent by 100. (Simply move the decimal place two points to the left.) 50.69/100 = .0569 and 49.31/100 = .4931 With this information, we can set up an equation. Letting X = the Amu weight of Br-79. Also, recall that our average atomic mass for Br = 79.904 (from periodic table). This equation looks like this, (Mass in Amu of Br-81) * (percentage abundance in decimal of Br-81) + (X mass in amu of Br-79) * (percentage abundance in decimal of Br-79) = 79.904. Or simply (80.9163)(.4931) + (X)(.5069) = 79.904. To solve this we need to get x by itself. Step 1 Multiply 80.9163 by .4931 Our equation now looks like: (39.89982753) + (X)(.5069) = 79.904 Step 2 Subtract 39.89982753 from each side. Remember what we do to one side must be done to the other. 39.89982753 + (X)(.5069) = 79.904 -39.89982753 - 39.89982753 Now our equation is: (X)(.5069) = 40.00417247 Step 3 Divide each side by .5069 to get x by itself. The equation then looks like this (X)(.5069) = 40.00417247 (.5069) = (.5069) Then X = 78.91925916 Depending on the calculator used, and where a person rounds the answers may vary slightly. This problem if it came from where I think it came needs to be rounded to 4 sig figs. If done properly when rounded the answer is 78.91 amu.