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Chapter 2: Problem 58

Determine (a) the number of \(\mathrm{Kr}\) atoms in a 5.25 -mg sample of krypton (b) the molar mass, \(M,\) and identity of an element if the mass of a \(2.80 \times 10^{22}\) -atom sample of the element is \(2.09 \mathrm{g}\) (c) the mass of a sample of phosphorus that contains the same number of atoms as \(44.75 \mathrm{g}\) of magnesium

Short Answer

Expert verified
a) There are approximately \(3.78 \times 10^{20}\) atoms of Kr in the 5.25-mg sample. b) The molar mass of the element is 45 g/mol, and thus the element is Scandium (Sc). c) The mass of a phosphorus sample that contains the same number of atoms as a 44.75-g sample of magnesium is approximately 57 g.

Step by step solution

01

a) Determine the number of Kr atoms

First, convert the mass of krypton from milligrams to grams: \[5.25 \, \text{mg} = 5.25 \times 10^{-3} \, \text{g}\]. Since the molar mass of \(\mathrm{Kr}\) is 83.8 g/mol, we can calculate the number of moles by dividing the mass by its molar mass: \[ n = \frac{Mass}{Molar \, mass} = \frac{5.25 \times 10^{-3}\,g}{83.8 g/mol} \text{ mol} \]. To find the number of atoms, multiply the number of moles by Avogadro’s number: \[ \text{Number of atoms} = n \times (\text{Avogadro’s number}) = \left( \frac{5.25 \times 10^{-3}\,g}{83.8 g/mol} \right) \times 6.022 \times 10^{23} atoms/mol\]
02

b) Find the molar mass and identity of an element

We have the number of atoms and the mass of a sample. Let’s find the amount in moles using Avogadro’s number: \[ n = \frac{\text{Number of atoms}}{\text{Avogadro’s number}} = \frac{2.80 \times 10^{22}\,atoms}{6.022 \times 10^{23} \, atoms/mol} = 0.0465 \, mol\]. To determine the molar mass, we can now divide the mass by the number of moles. \[ Molar\;mass = \frac{Mass}{n} = \frac{2.09 \, g}{0.0465 \, mol} = 45 \, g/mol\]. The element with a molar mass near to 45 g/mol is Scandium (Sc).
03

c) Calculate the mass of a phosphorus sample

We know that the molar mass of Mg is 24.3 g/mol. Let’s first calculate the number of moles in 44.75 g of Mg: \[ n_{Mg} = \frac{44.75 g}{24.3 g/mol} = 1.84 mol \]. Since the number of atoms in a given mass is the same for all elements, the number of phosphorus atoms equals the number of magnesium atoms. Thus, the number of moles of Phosphorus, \(n_{P} = n_{Mg}= 1.84 mol\). Consequently, the mass of phosphorus required can be found using its molar mass of 30.97 g/mol: \[ Mass_{P} = n_{P} \times (\text{Molar mass of Phosphorus}) = 1.84 mol \times 30.97 g/mol\].

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Most popular questions from this chapter

From the densities of the lines in the mass spectrum of krypton gas, the following observations were made: \bullet Somewhat more than \(50 \%\) of the atoms were krypton-84. \(\bullet\) The numbers of krypton- 82 and krypton- 83 atoms were essentially equal. \(\bullet\) The number of krypton-86 atoms was 1.50 times as great as the number of krypton- 82 atoms. \(\bullet\) The number of krypton-80 atoms was \(19.6 \%\) of the number of krypton- 82 atoms. \(\bullet\) The number of krypton- 78 atoms was \(3.0 \%\) of the number of krypton- 82 atoms. The masses of the isotopes are \(^{78} \mathrm{Kr}, 77.9204 \mathrm{u} \quad^{80} \mathrm{Kr}, 79.9164 \mathrm{u} \quad^{82} \mathrm{Kr}, 81.9135 \mathrm{u}\) \(^{83} \mathrm{Kr}, 82.9141 \mathrm{u} \quad^{84} \mathrm{Kr}, 83.9115 \mathrm{u} \quad^{86} \mathrm{Kr}, 85.9106 \mathrm{u}\) The weighted-average atomic mass of \(\mathrm{Kr}\) is \(83.80 .\) Use these data to calculate the percent natural abundances of the krypton isotopes.

When \(10.0 \mathrm{g}\) zinc and \(8.0 \mathrm{g}\) sulfur are allowed to react, all the zinc is consumed, \(14.9 \mathrm{g}\) zinc sulfide is produced, and the mass of unreacted sulfur remaining is (a) \(2.0 \mathrm{g}\) (b) \(3.1 g\) (c) \(4.9 \mathrm{g}\) (d) impossible to predict from this information alone

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William Prout (1815) proposed that all other atoms are built up of hydrogen atoms, suggesting that all elements should have integral atomic masses based on an atomic mass of one for hydrogen. This hypothesis appeared discredited by the discovery of atomic masses, such as 24.3 u for magnesium and 35.5 u for chlorine. In terms of modern knowledge, explain why Prout's hypothesis is actually quite reasonable.
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