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Chapter 3: Problem 37

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

Short Answer

Expert verified
The average atomic mass of the element can be calculated using the weighted average formula: \((0.014 \times 203.973) + (0.241 \times 205.9745) + (0.221 \times 206.9759) + (0.524 \times 207.9766) \approx 207.1206\) u. Based on the average atomic mass, the element is identified as lead (Pb).

Step by step solution

01

Convert the Percentage Abundances to Decimal Values

First, we need to convert the given percent abundance values of each isotope to decimal values. This is done by dividing the percent value by 100. Isotope 1: \(1.40 \% \div 100 = 0.014\) Isotope 2: \(24.10 \% \div 100 = 0.241\) Isotope 3: \(22.10 \% \div 100 = 0.221\) Isotope 4: \(52.40 \% \div 100 = 0.524\) Now, we have decimal values for the abundance of each isotope.
02

Calculate the Weighted Average Atomic Mass

To calculate the average atomic mass, we will use the following formula: Average Atomic Mass = \((\text{Decimal Abundance of Isotope 1} \times \text{Mass of Isotope 1}) + (\text{Decimal Abundance of Isotope 2} \times \text{Mass of Isotope 2}) + (\text{Decimal Abundance of Isotope 3} \times \text{Mass of Isotope 3}) +(\text{Decimal Abundance of Isotope 4} \times \text{Mass of Isotope 4})\) Now, we have all the information to calculate the average atomic mass: Average Atomic Mass = \((0.014 \times 203.973) + (0.241 \times 205.9745) + (0.221 \times 206.9759) + (0.524 \times 207.9766)\) Average Atomic Mass ≈ \(2.8556 + 49.6395 + 45.6433 + 108.9822\) Average Atomic Mass ≈ \(207.1206\)
03

Identify the Element

Now that we have the average atomic mass of the element (approximately 207.1206 u), we can look it up on a periodic table to identify the element. The element closest to this atomic mass value is lead (Pb) with an atomic mass of approximately 207.2 u. So, the given isotopes are of the element lead (Pb).

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Most popular questions from this chapter

Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. A \(10.00-\mathrm{g}\) mixture of zinc and magnesium produces \(0.5171 \mathrm{~g}\) of hydrogen gas upon being mixed with an excess of hydrochloric acid. Determine the percent magnesium by mass in the original mixture.

Determine the molecular formula of a compound that contains \(26.7 \% \mathrm{P}, 12.1 \% \mathrm{~N}\), and \(61.2 \% \mathrm{Cl}\), and has a molar mass of \(580 \mathrm{~g} / \mathrm{mol}\)

One of relatively few reactions that takes place directly between two solids at room temperature is \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow\) \(\mathrm{Ba}(\mathrm{SCN})_{2}(s)+\mathrm{H}_{2} \mathrm{O}(t)+\mathrm{NH}_{3}(g)\) In this equation, the \(\cdot 8 \mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) indicates the presence of eight water molecules. This compound is called barium hydroxide octahydrate. a, Balance the equation. b. What mass of ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\) must be used if it is to react completely with \(6.5 \mathrm{~g}\) barium hydroxide octahydrate?

A compound contains \(47.08 \%\) carbon, \(6.59 \%\) hydrogen, and \(46.33 \%\) chlorine by mass; the molar mass of the compound is \(153 \mathrm{~g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If \(25.0 \mathrm{~g} \mathrm{Ag}_{2} \mathrm{O}\) is reacted with \(50.0 \mathrm{~g} \mathrm{C}_{10} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{SO}_{2}\), what mass of silver sulfadiazine, \(\mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{~N}_{4} \mathrm{SO}_{2}\), can be produced, assuming \(100 \%\) yield? $$ \mathrm{Ag}_{2} \mathrm{O}(s)+2 \mathrm{C}_{10} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{SO}_{2}(s) \longrightarrow 2 \mathrm{AgC}_{10} \mathrm{H}_{9} \mathrm{~N}_{4} \mathrm{SO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$
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