A particular sample of hydrogen has an average atomic mass of 1.00788 amu. This sample contains two isotopes of hydrogen: 1H (isotopic mass = 1.00783 amu) and 2H (isotopic mass = 2.01410 amu). The 2H isotope is called deuterium and is sometimes written as D. Calculate the percent abundance of deuterium in this sample of hydrogen.

 

Report your answer to 4 decimal places without the % sign.

To calculate the percent abundance of deuterium (\(^2H\) or D) in this sample of hydrogen, we'll use the formula for the average atomic mass:

\[
\text{Average atomic mass} = (\text{fractional abundance of } ^1H) \times (\text{mass of } ^1H) + (\text{fractional abundance of } ^2H) \times (\text{mass of } ^2H)
\]

Let's define:
- \( x \) as the fractional abundance of deuterium (\(^2H\)).
- Then, the fractional abundance of \(^1H\) will be \( 1 - x \).

The given dataTo calculate the percent abundance of deuterium (\(^2H\) or D) in this sample of hydrogen, we'll use the formula for the average atomic mass:

\[
\text{Average atomic mass} = (\text{fractional abundance of } ^1H) \times (\text{mass of } ^1H) + (\text{fractional abundance of } ^2H) \times (\text{mass of } ^2H)
\]

Let's define:
- \( x \) as the fractional abundance of deuterium (\(^2H\)).
- Then, the fractional abundance of \(^1H\) will be \( 1 - x \).

The given 
- Average atomic mass of the sample = 1.00788 amu
- Isotopic mass of \(^1H\) = 1.00783 amu
- Isotopic mass of \(^2H\) = 2.01410 amu

Using the formula:

\[
1.00788 = (1 - x)(1.00783) + x(2.01410)
\]

Now, we'll solve for \( x \):

\[
1.00788 = 1.00783 - 1.00783x + 2.01410x
\]

\[
1.00788 = 1.00783 + 1.00627x
\]

Subtract 1.00783 from both sides:

\[
0.00005 = 1.00627x
\]

Now, divide both sides by 1.00627:

\[
x = \frac{0.00005}{1.00627} \approx 0.0000497
\]

The percent abundance of deuterium is:

\[
\text{Percent abundance of } ^2H = x \times 100 \approx 0.00497 \%
\]

Thus, the fractional abundance of deuterium, reported to four decimal places, is **0.0049**.

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Do you want more details on any step or have any questions?

Here are 5 related questions:
1. How would the calculation change if the isotopic masses were different?
2. What is the significance of knowing the percent abundance of isotopes?
3. Can you calculate the percent abundance of \(^1H\) based on the result?
4. How would this calculation be done for an element with more than two isotopes?
5. What are the real-world applications of isotope abundance calculations?

**Tip:** Always double-check your subtraction and decimal placement when working with very small numbers to ensure precision.

Learn how to calculate the percent abundance of deuterium (D) in a sample of hydrogen with an average atomic mass of 1.00788 amu. Detailed step-by-step solution using isotopic masses and fractional abundances.

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