An element X has the following isotopic composition:
$^{200}$X: 90%
$^{199}$X: 8.0%
$^{202}$X: 2.0%
The weight

Question

An element X has the following isotopic composition:
$^{200}$X: 90%
$^{199}$X: 8.0%
$^{202}$X: 2.0%
The weighted average atomic mass of the naturally-occurring element X is closest to:

Accepted Answer

The weighted average atomic mass is calculated by summing the products of the atomic mass of each isotope and its fractional abundance.

$$	ext{Average Atomic Mass} = \sum (	ext{Mass}_i 	imes 	ext{Fractional Abundance}_i)$$

1.  **Convert Percentages to Decimals:**
    *   $^{200}$X: $90\% = 0.90$
    *   $^{199}$X: $8.0\% = 0.08$
    *   $^{202}$X: $2.0\% = 0.02$

2.  **Calculate Weighted Contributions:**
    *   $200 	imes 0.90 = 180.0$
    *   $199 	imes 0.08 = 15.92$
    *   $202 	imes 0.02 = 4.04$

3.  **Sum the Contributions:**
    $$180.0 + 15.92 + 4.04 = 199.96 	ext{ amu}$$

4.  **Round to Nearest Integer:**
    The value $199.96$ is closest to $200$ amu.

Step-by-Step Solution

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