mass of Fe atom = 56 grams/mol
how many cm^3/mol ?
56 g / 7.874 g/cm^3 = 7.11 cm^3/mol
how many atoms is that?
7.11 cm^3/mol / (9.38*10^-24 cm^3/atom)
= .76*10^24 atoms/mol
= 7.6 *10^23
well, not too bad :)
Iron crystallizes in a body–centered cubic structure. The volume of one Fe atom is 9.38 x 10–24 cm3, and the density of Fe is 7.874 g/cm3. Find an approximate value for Avogadro’s number.
2 answers
Damon's method looks MUCH simpler than what I would do but here is mine.
volume Fe atom = 9.38E-24 cc.
V = (4/3)*pi*r^3 and solve for r = radius
r = about 1.31E-8 cm
For a body centered cubic, a = edge length. 4r = a(3)^1/2 and solve for a.
a = about 3.03E-8 cm
volume of unit cell = a^3 = about 2.77E-23 cc.
mass unit cell = volume x density = 2.77E-23 x 7.874 = 2.18E-22
u.c. = unit cell
mass u.c. = 2.18E-22 atoms/unit cell x 55.85/N and solve for N.
That's about 5.1E23 compared to the accepted value of 6.02E23.
volume Fe atom = 9.38E-24 cc.
V = (4/3)*pi*r^3 and solve for r = radius
r = about 1.31E-8 cm
For a body centered cubic, a = edge length. 4r = a(3)^1/2 and solve for a.
a = about 3.03E-8 cm
volume of unit cell = a^3 = about 2.77E-23 cc.
mass unit cell = volume x density = 2.77E-23 x 7.874 = 2.18E-22
u.c. = unit cell
mass u.c. = 2.18E-22 atoms/unit cell x 55.85/N and solve for N.
That's about 5.1E23 compared to the accepted value of 6.02E23.
1 answer