The spectrum shows three vertical bars corresponding to the molecular masses lab

Question

The spectrum shows three vertical bars corresponding to the molecular masses labeled on the x-axis. The height of each bar gives the fraction of all molecules having the corresponding mass.
Iodine has only a single, stable, naturally-occurring isotope (127I atomic mass 126.90447 amu).
Fluorine has only a single, stable, naturally-occurring isotope (19F atomic mass 18.98840322 amu).
There are exactly two stable naturally-occurring isotopes of element "X".

1. What is the atomic mass of the less massive isotope of element X?

2. What is the atomic mass of the more massive isotope of element X?

3. What is the percent natural abundance of the more massive isotope of element X?

4. What is the molar mass of the compound IX2F?

0.5 0.4 0.3 0.2 0.1 0.2569 303, 72955 Mass spectrum of IX2F 0.4999 305.72750 Mass /charge (amu/e) 0.2431 r----------- 307, 72545

Explanation / Answer

Atomic mass of Iodine = 126.90447 amu

Atomic mass of Fluorine = 18.98840322 amu

Total of atomic masses of Iodine and Flourine = 145.89287

SInce there are two isotopes of X and there are two atoms of X in one molecule of IX2F,

Four combinations are possible , but in mass specrum three vertical bars have been obtained because two combinations will appear at same amu/e.

Suppose Xa is lower atomic mass isotpoe while Xb is higher atomic mass isotope

Following four combinations are possible for the peaks appeared in mass spectum.

1. I Xa Xa F peak appeared at amu/e 303.72955

2. I Xa Xb F peak appeared at amu/e 305.72750

3. I Xb Xa F peak appeared at amu/e 305.72750

4. I Xb Xb F peak appeared at amu/e 307.72545

In case 1, the two X has lowest atomic masses . There masses can be obtained by substracting comined mass of I and F from the molecular mass of IX2F

Answer (1): Atomic mass of Xa ( less massive isotope of X) = (303.72955 - 145.89287) / 2 = 157.83668 / 2 = 78.91834 amu

Likewise,

Answer (2): Atomic mass of Xb ( more massive isotope of X) = ( 307.72545 - 145.89287 ) / 2 = 161.83258 / 2 = 80.91629 amu

Answer (3): The percent natural abundance can be obtained by taking simple ratio of fractional abundance peaks obtained. It will come as 1:2:1. Thus natural abundance of the more massive isotope of element X = 1

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