Question

question 12

1. iron has three main isotopes. fe-54, fe-56 and fe-57. their relative abundances are 6.1%, 92.3% and 2.1% respectively. the average mass of iron is blank 1%

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5 points

Explanation:

Step1: Recall the formula for average atomic mass

The formula for average atomic mass ($A$) is the sum of the product of each isotope's mass ($m_i$) and its relative abundance ($f_i$, in decimal form). So, $A=\sum (m_i\times f_i)$.

Step2: Convert abundances to decimals

• For Fe - 54: Abundance $f_1 = 6.1\%=0.061$, mass $m_1 = 54$
• For Fe - 56: Abundance $f_2 = 92.3\% = 0.923$, mass $m_2=56$
• For Fe - 57: Abundance $f_3=2.1\% = 0.021$, mass $m_3 = 57$

Step3: Calculate each term

• Term 1: $m_1\times f_1=54\times0.061 = 3.294$
• Term 2: $m_2\times f_2=56\times0.923=51.688$
• Term 3: $m_3\times f_3=57\times0.021 = 1.197$

Step4: Sum the terms

$A=3.294 + 51.688+1.197=56.179$ (rounded to a reasonable decimal place, usually two or three. If we consider more precise calculation, but with the given data, this is the result)

Answer:

56.18 (or 56.179, depending on rounding. If we use more precise calculation, the average atomic mass of iron is approximately 55.85, but wait, maybe I made a mistake in the isotope masses? Wait, the actual isotope masses: Fe - 54 has a mass of approximately 53.9396, Fe - 56 is 55.9349, Fe - 57 is 56.9354. Oh, the problem might have used the mass number as the mass (approximate). Let's recalculate with the correct masses (more accurate):

Step1: Correct masses

• Fe - 54: $m_1 = 53.9396$, $f_1=0.061$
• Fe - 56: $m_2 = 55.9349$, $f_2 = 0.923$
• Fe - 57: $m_3=56.9354$, $f_3 = 0.021$

Step2: Recalculate terms

• Term1: $53.9396\times0.061\approx3.2903$
• Term2: $55.9349\times0.923\approx51.6279$
• Term3: $56.9354\times0.021\approx1.1956$

Step3: Sum

$3.2903 + 51.6279+1.1956\approx56.1138$. But the standard atomic mass of iron is about 55.845. Wait, maybe the problem has different abundances? Wait, the user's problem says abundances are 6.1%, 92.3%, 2.1%? Wait, actual abundances: Fe - 54: 5.845%, Fe - 56: 91.754%, Fe - 57: 2.119%, Fe - 58: 0.282%. Maybe the problem is simplified. Let's go with the problem's given data (using mass numbers as masses):

$A=(54\times0.061)+(56\times0.923)+(57\times0.021)$

$= 3.294+51.688 + 1.197$

$=56.179\approx56.18$

But if we use the correct atomic masses (not mass numbers), let's do that:

Fe - 54: mass = 53.9396126 u, abundance = 0.05845

Fe - 56: mass = 55.9349375 u, abundance = 0.91754

Fe - 57: mass = 56.9353940 u, abundance = 0.02119

Then average mass = $53.9396126\times0.05845+55.9349375\times0.91754 + 56.9353940\times0.02119$

Calculate each term:

$53.9396126\times0.05845\approx3.152$

$55.9349375\times0.91754\approx51.324$

$56.9353940\times0.02119\approx1.197$

Sum: $3.152 + 51.324+1.197\approx55.673$? No, that's not right. Wait, I think I messed up the abundances. The correct abundances (from IUPAC) for Fe are:

• $^{54}$Fe: 5.845%
• $^{56}$Fe: 91.754%
• $^{57}$Fe: 2.119%
• $^{58}$Fe: 0.282%

So with these, the average mass is:

$(53.9396\times0.05845)+(55.9349\times0.91754)+(56.9354\times0.02119)+(57.9333\times0.00282)$

$= 3.152+51.324 + 1.197+0.163$

$=55.836\approx55.84$

But the problem states three isotopes: Fe - 54, Fe - 56, Fe - 57 with abundances 6.1%, 92.3%, 2.1%. So using the problem's data (ignoring Fe - 58), and using mass numbers as masses (approximate):

Average mass = $(54\times0.061)+(56\times0.923)+(57\times0.021)$

$= 3.294+51.688+1.197 = 56.179\approx56.18$

So the answer, based on the problem's given data (using mass numbers as isotope masses), is approximately 56.18. If we use the actual atomic masses (not mass numbers), we need to use the precise masses. But since the problem gives the isotopes as Fe - 54, Fe - 56, Fe - 57 (mass numbers), we can assume the mass is the mass number for simplicity.

So the final answer, after calculation, is approximately 56.18 (or 56.179).