This is a lengthy problem, but I really don't understand it. Any assistance given would be greatly appreciated.
An analysis was done on a rock sample to determine its rubidium content. The rubidium content of a portion of rock weighing .350g was extracted, and to the extracted sample was added an additional 29.45 micrograms of rubidium-87. the mass spectrum of this spiked sample showed a rubidium-87 peak that was 1.12 times as high as the peak of rubidium-85. assuming that the two isotopes react identically, what is the rubidium content of the rock (expressed in parts per million by mass)?
additional info:
rubidium-87 atomic mass=86.909u
% natural abundance=27.83% of all rubidium
rubidium-85 atomic mass=84.912u
% natural abundance=72.17% of all rubidium
I am particularly confused by the peaks aspect of this problem. Does that mean like something with mass or what? Please help. Thanks.
Sure, I'd be happy to help you understand this problem.
In this problem, we are given information about the isotopes of rubidium, their atomic masses, and their natural abundances. We are asked to find the rubidium content of a rock sample in parts per million by mass.
First, let's understand what the "peaks" in the mass spectrum mean. In a mass spectrum, different isotopes of an element are represented by peaks. The height of the peak represents the relative abundance of that isotope.
In this problem, the rubidium-87 peak is said to be 1.12 times as high as the rubidium-85 peak. This means that the abundance of rubidium-87 is 1.12 times the abundance of rubidium-85.
Now, let's break down the steps to solve the problem:
Step 1: Calculate the mass of the added rubidium-87
The additional rubidium-87 added to the sample is given as 29.45 micrograms. To convert this to grams, divide by 1,000,000 (since 1 microgram = 1/1,000,000 grams). So, the mass of rubidium-87 added is 0.00002945 grams.
Step 2: Calculate the mass of rubidium-85 in the sample
Since rubidium-85 is the more abundant isotope, we can assume that the mass of rubidium-85 is the original mass of the sample before any rubidium-87 was added. The mass of the sample is given as 0.350 grams.
Step 3: Calculate the relative abundance of rubidium-87 compared to rubidium-85
We are told that the rubidium-87 peak is 1.12 times as high as the rubidium-85 peak. Since mass is directly proportional to abundance, we can assume that the abundance of rubidium-87 is 1.12 times the abundance of rubidium-85.
Step 4: Calculate the individual masses of rubidium-85 and rubidium-87
To calculate the mass of rubidium-85, we can use the formula:
mass of rubidium-85 = (mass of sample) - (mass of added rubidium-87)
Substituting the values, we get:
mass of rubidium-85 = 0.350 g - 0.00002945 g = 0.34997055 g
For rubidium-87, we know that its atomic mass is 86.909u. We can use this value to calculate its mass:
mass of rubidium-87 = (mass of rubidium-87 added) / (abundance ratio of rubidium-87 to rubidium-85)
Substituting the values, we get:
mass of rubidium-87 = 0.00002945 g / 1.12 = 0.0000262955 g
Step 5: Calculate the rubidium content in parts per million by mass (ppm)
To calculate the rubidium content in ppm, we need to know the total mass of rubidium in the rock sample. This can be calculated by adding the masses of rubidium-85 and rubidium-87:
total rubidium mass = mass of rubidium-85 + mass of rubidium-87
Substituting the values, we get:
total rubidium mass = 0.34997055 g + 0.0000262955 g = 0.3500068455 g
Finally, to express this as parts per million by mass, we need to divide the rubidium mass by the mass of the sample and multiply by 1,000,000:
rubidium content in ppm = (total rubidium mass / mass of sample) * 1,000,000
Substituting the values, we get:
rubidium content in ppm = (0.3500068455 g / 0.350 g) * 1,000,000 = 1,000,024.557 ppm
Therefore, the rubidium content of the rock sample is approximately 1,000,024.557 ppm.