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Chapter 3: Problem 24

A cylindrical piece of pure copper \(\left(d=8.92 \mathrm{~g} / \mathrm{cm}^{3}\right)\) has diameter \(1.15 \mathrm{~cm}\) and height \(4.00\) inches. How many atoms are in that cylinder? (Note: the volume of a right circular cylinder of radius \(r\) and height \(h\) is \(V=\pi r^{2} h .\) )

Short Answer

Expert verified
Answer: The cylindrical piece of pure copper contains approximately \(8.03 \times 10^{23}\) atoms.

Step by step solution

01

Calculate the volume of the cylinder

To find the volume of the cylinder, we use the formula \(V = \pi r^2h\), where \(r\) is the radius, and \(h\) is the height. We are given the diameter, which is twice the radius, so \(r = \frac{1.15}{2}\) cm. The height is given as 4.00 inches, which we need to convert to cm. We know that 1 inch is equal to 2.54 cm, so the height in cm is \(4.00 \times 2.54 = 10.16\) cm. Now we can find the volume: \(V = \pi \left(\frac{1.15}{2}\right)^2 (10.16) \approx 9.485 \mathrm{~cm}^{3}\).
02

Calculate the mass of the copper cylinder

Now that we have the volume, we can use the density to find the mass of the copper. The density formula is: \(Density = \frac{Mass}{Volume}\) Rearranging to solve for mass, we get: \(Mass = Density \times Volume\) The density of copper is given as \(8.92 \mathrm{~g} / \mathrm{cm}^{3}\), so the mass of the cylinder is: \(Mass \approx 8.92 \times 9.485 \approx 84.698 \mathrm{~g}\)
03

Determine the number of moles of copper

To find the number of moles of copper in the cylinder, we need to divide the mass by the molar mass of copper. The molar mass of copper is approximately 63.546 g/mol. Therefore, the number of moles in the cylinder is: \(Moles = \frac{84.698}{63.546} \approx 1.333 \mathrm{~mol}\)
04

Calculate the number of atoms using Avogadro's number

Finally, we can use Avogadro's number to find the number of atoms in the copper cylinder. Avogadro's number is approximately \(6.022 \times 10^{23}\) atoms/mol. So, the number of atoms in the cylinder is: \(Atoms \approx 1.333 \times 6.022 \times 10^{23} \approx 8.03 \times 10^{23}\) atoms The cylindrical piece of pure copper contains approximately \(8.03 \times 10^{23}\) atoms.

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Most popular questions from this chapter

Chlorine has two isotopes, Cl-35 and Cl-37. Their abundances are \(75.53 \%\) and \(24.47 \%\), respectively. Assume that the only hydrogen isotope present is \(\mathrm{H}-1\). (a) How many different HCl molecules are possible? (b) What is the sum of the mass numbers of the two atoms in each molecule? (c) Sketch the mass spectrum for \(\mathrm{HCl}\) if all the positive ions are obtained by removing a single electron from an HCl molecule.

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