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Chapter 3: Problem 33

An element consists of \(1.40 \%\) of an isotope with mass \(203.973\) amu, \(24.10 \%\) of an isotope with mass \(205.9745\) amu, \(22.10 \%\) of an isotope with mass \(206.9759 \mathrm{amu}\), and \(52.40 \%\) of an isotope with mass \(207.9766\) amu. Calculate the average atomic mass and identify the element.

Short Answer

Expert verified
The average atomic mass of the element with the given isotopic composition is approximately \(207.42 \mathrm{amu}\). The element is lead (Pb).

Step by step solution

01

Find the weighted mean of the atomic masses

To find the weighted mean, we will multiply the atomic mass of each isotope by its percentage abundance, and then add them together. In mathematical terms: Average atomic mass = (Percentage of isotope 1 × mass of isotope 1) + (Percentage of isotope 2 × mass of isotope 2) + (Percentage of isotope 3 × mass of isotope 3) + (Percentage of isotope 4 × mass of isotope 4) Substitute the given data: Average atomic mass = (1.40% × 203.973 amu) + (24.10% × 205.9745 amu) + (22.10% × 206.9759 amu) + (52.40% × 207.9766 amu) To convert percentages to decimals, divide by 100: Average atomic mass = (0.014 × 203.973 amu) + (0.241 × 205.9745 amu) + (0.221 × 206.9759 amu) + (0.524 × 207.9766 amu)
02

Calculate the average atomic mass

Now we will perform the multiplications and additions: Average atomic mass = (2.85562 amu) + (49.6404145 amu) + (45.7409479 amu) + (109.1832944 amu) Average atomic mass = 207.4202788 amu
03

Identify the element

The calculated average atomic mass is approximately 207.42 amu. To identify the element, we will refer to the periodic table and look for an element with an atomic mass close to 207.42 amu. In the periodic table, we can find that the element with an atomic mass close to 207.42 amu is lead (Pb) with an atomic mass of approximately 207.2 amu. The element with the given isotope composition and average atomic mass is lead (Pb).

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Most popular questions from this chapter

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{~g} \mathrm{~N}_{2}\) and \(5.00 \times 10^{2} \mathrm{~g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

Consider the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ Identify the limiting reagent in each of the reaction mixtures given below: a. 50 molecules of \(\mathrm{H}_{2}\) and 25 molecules of \(\mathrm{O}_{2}\) b. 100 molecules of \(\mathrm{H}_{2}\) and 40 molecules of \(\mathrm{O}_{2}\) c. 100 molecules of \(\mathrm{H}_{2}\) and 100 molecules of \(\mathrm{O}_{2}\) d. \(0.50 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}\). e. \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}_{2}\) f. \(1.0 \mathrm{~g} \mathrm{H}_{2}\) and \(0.25 \mathrm{~mol} \mathrm{O}_{2}\) g. \(5.00 \mathrm{~g} \mathrm{H}_{2}\) and \(56.00 \mathrm{~g} \mathrm{O}_{2}\)

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 g sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(\mathrm{g}) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

The empirical formula of styrene is \(\mathrm{CH}\); the molar mass of styrene is \(104.14 \mathrm{~g} / \mathrm{mol}\). What number of \(\mathrm{H}\) atoms are present in a \(2.00-\mathrm{g}\) sample of styrene?

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg} \mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction and assume that the NO produced in the third step is not recycled.
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