Finding number of atoms per cm^3 of zinc?

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SUMMARY

The discussion focuses on calculating the number of atoms per cm³ of zinc, its atomic mass, and atomic volume. Using the density of zinc at 7.17 Mg/m³ and its atomic mass of 65.39 g/mol, the number of zinc atoms per cm³ is determined to be approximately 6.603 x 10²² atoms/cm³. The mass of a single zinc atom is calculated to be 1.0859 x 10⁻²² grams, and the atomic volume of zinc is derived as 6.603 x 10⁻²² cm³ per atom. These calculations confirm the relationships between density, atomic mass, and atomic volume in zinc.

PREREQUISITES
  • Understanding of atomic mass and density concepts
  • Familiarity with Avogadro's number (6.022 x 10²³)
  • Basic knowledge of unit conversions (g/m³ to g/cm³)
  • Proficiency in using scientific notation for calculations
NEXT STEPS
  • Learn about the implications of density in material science
  • Explore the concept of Avogadro's number in different contexts
  • Study the relationship between atomic mass and molecular weight
  • Investigate the properties of zinc and its applications in various industries
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Chemistry students, materials scientists, and anyone interested in atomic structure and properties of elements, particularly zinc.

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Homework Statement


Zinc has a density of 7.17 Mg/m^3. Calculate (a) the number of Zn atoms per cm^3, (b) the mass of a single Zn atom and (c) the atomic volume of Zn.


Homework Equations


atomic mass of zinc = 65.39 g/mol


The Attempt at a Solution



For part (a) I use the fact that zinc has atomic mass of 65.39 g/mol. So dividing 7,170,000 (g/m^3) by 65.39 (g/mol), you get 109,649.79 (mol/m^3). Then we divide that by 10^6 to get .10964979 mol/cm^3. Then we multiply that by avagadro's number to get 6.603*10^22 atoms/cm^3. Is this right?

For part (b) we know that one cm^3 of zinc has 7,170,000/10^6 grams of mass. AKA 7.17 grams of mass. And we also know from part (a) that one cm^3 of zinc has 6.603*10^22 atoms in it. So each atom has 7.17/6.603*10^22= 1.0859*10^-22 grams of mass. Is that right?

For part (c) we use the fact from part (a) once cm^3 of zinc has has 6.603*10^22 atoms in it. So that mean an individual zinc atom has a volume of 6.603*10^-22 cm^3. Is this right?
 
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