Question

the atomic masses of $^{35}\text{cl}$ and $^{37}\text{cl}$ are 34.9689 amu and 36.9659 amu, respectively. calculate the natural abundances of these two isotopes. the average atomic mass of chlorine is 35.452 amu.

the average atomic mass of magnesium is 24.305 amu. a sample of magnesium is found to contain 78.70% of $^{24}\text{mg}$ atoms (mass 23.98 amu), 10.13% of $^{25}\text{mg}$ atoms (mass 24.99 amu), and $^{26}\text{mg}$ atoms. calculate the mass of an $^{26}\text{mg}$ atom.

2.4 the mole
define
mole

avogadro’s number

molar mass

what is the atomic mass of sodium (na)? what is the molar mass of sodium (na)?

how are these two values similar? how are they different?

Explanation:

First Problem: Chlorine Isotope Abundances

Step1: Define variables for abundances

Let \(x\) = fraction of \(^{35}\text{Cl}\), so \(1-x\) = fraction of \(^{37}\text{Cl}\)

Step2: Set up average mass equation

Average atomic mass = (mass of \(^{35}\text{Cl} \times x\)) + (mass of \(^{37}\text{Cl} \times (1-x)\))
$$35.452 = 34.9689x + 36.9659(1-x)$$

Step3: Simplify and solve for x

$$35.452 = 34.9689x + 36.9659 - 36.9659x$$
$$35.452 - 36.9659 = 34.9689x - 36.9659x$$
$$-1.5139 = -1.997x$$
$$x = \frac{1.5139}{1.997} \approx 0.7581$$

Step4: Find second isotope abundance

Fraction of \(^{37}\text{Cl} = 1 - 0.7581 = 0.2419\)

Step5: Convert to percentages

\(^{35}\text{Cl}\): \(0.7581 \times 100 = 75.81\%\)
\(^{37}\text{Cl}\): \(0.2419 \times 100 = 24.19\%\)

Step1: Find fraction of \(^{26}\text{Mg}\)

Fraction = \(100\% - 78.70\% - 10.13\% = 11.17\% = 0.1117\)

Step2: Set up average mass equation

Let \(m\) = mass of \(^{26}\text{Mg}\)
$$24.305 = (23.98 \times 0.7870) + (24.99 \times 0.1013) + (m \times 0.1117)$$

Step3: Calculate known terms

$$23.98 \times 0.7870 \approx 18.872$$
$$24.99 \times 0.1013 \approx 2.531$$
$$18.872 + 2.531 = 21.403$$

Step4: Solve for m

$$24.305 - 21.403 = 0.1117m$$
$$2.902 = 0.1117m$$
$$m = \frac{2.902}{0.1117} \approx 25.98\text{ amu}$$

1. Mole: A standard unit for measuring the amount of substance, equal to the number of atoms in 12 grams of carbon-12.
2. Avogadro's number: The number of particles (atoms, molecules, ions) in one mole of a substance, equal to \(6.022 \times 10^{23}\) particles/mol.
3. Molar mass: The mass of one mole of a substance, numerically equal to the atomic/molecular formula mass in amu, with units of g/mol.
4. Sodium's atomic mass is the mass of one sodium atom, and its molar mass is the mass of one mole of sodium atoms. The values are numerically identical but have different units.

Answer:

Natural abundance of \(^{35}\text{Cl}\) is 75.81%, natural abundance of \(^{37}\text{Cl}\) is 24.19%

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Second Problem: Mass of \(^{26}\text{Mg}\)