Finding number of atoms per cm^3 of zinc?

In summary, we can calculate the number of Zn atoms per cm^3, the mass of a single Zn atom, and the atomic volume of Zn using the given information. The atomic mass of zinc is 65.39 g/mol. For part (a), we divide the density of zinc (7.17 Mg/m^3) by its atomic mass to get 6.603*10^22 atoms/cm^3. For part (b), we divide the density by 10^6 to get the mass of a single atom (1.0859*10^-22 grams). And for part (c), we use the information from part (a) to calculate the atomic volume of Zn (6.
  • #1
robertjordan
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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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  • #2
Looks right.
 

Related to Finding number of atoms per cm^3 of zinc?

1. How do I calculate the number of atoms per cm^3 of zinc?

The formula for finding the number of atoms per cm^3 of a substance is:
Number of atoms per cm^3 = (density of the substance in g/cm^3) / (atomic weight of the substance in g/mol) x Avogadro's number (6.022 x 10^23 atoms/mol).

2. What is the density of zinc in g/cm^3?

The density of zinc is 7.14 g/cm^3.

3. What is the atomic weight of zinc in g/mol?

The atomic weight of zinc is 65.38 g/mol.

4. What is Avogadro's number and why is it used in this calculation?

Avogadro's number is a constant that represents the number of atoms in one mole of a substance. It is used in this calculation to convert the mass of zinc in grams to the number of atoms present.

5. Can this formula be used for other substances besides zinc?

Yes, this formula can be used for any substance as long as the density and atomic weight are known. However, it is important to note that the atomic weight and density may vary for different isotopes of an element.

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