The coinage metals -- copper, silver, and gold -- crystallize in a cubic closest packed structure. Use the density of gold (19.3 g/cm3) and its molar mass (197 g/mol) to calculate an approximate atomic radius for gold.
Work So Far:
{[1/(19.3x197)]x.74}/6.022e23
=3.2319e18
[(3.2319e18x3)/4pi]^(1/3)
=917190
I don't follow everything you've done.
There are 4 atoms to the face centered unit cell.
mass unit cell = (197 x 4/6.02E23) = ?grams.
volume unit cell = mass/density = ? cc
cubed root V = ? = a cm
Then a(2)^1/2 = 4r and solve for r = ? in cm. I obtained 1.44E-8 cm = 1.44 Angstroms. The internet gives 144.2 pm as the radius.
To calculate the approximate atomic radius of gold, we can use the given information and formula. Here are the step-by-step calculations:
1. Calculate the volume of one gold atom:
- Divide the molar mass of gold (197 g/mol) by the density of gold (19.3 g/cm3) to find the volume occupied by one mole of gold atoms:
Volume = 1 / (density x molar mass)
= 1 / (19.3 g/cm3 x 197 g/mol)
= 1 / 3781.0 g/cm3
= 2.644 x 10^-4 cm3/mol
2. Convert the volume from cm3/mol to cm3/atom:
- Divide the volume by Avogadro's number (6.022 x 10^23 atoms/mol) to find the volume occupied by one gold atom:
Volume per atom = Volume / Avogadro's number
= (2.644 x 10^-4 cm3/mol) / (6.022 x 10^23 atoms/mol)
= 4.396 x 10^-28 cm3/atom
3. Calculate the approximate atomic radius:
- To find the radius of a gold atom, we need to calculate the radius of a sphere that has the same volume as one gold atom.
- Use the formula for the volume of a sphere:
Volume = (4/3)πr^3, where r is the radius of the sphere.
- Rearrange the formula to solve for the radius, r:
r = (3 * Volume / (4π))^(1/3)
r = (3 * 4.396 x 10^-28 cm3 / (4π))^(1/3)
r = 9.17190 x 10^-9 cm
Therefore, the approximate atomic radius for gold is 9.17190 x 10^-9 cm.
To calculate the atomic radius of gold, we can use the formula for the volume of a sphere and relate it to the density and molar mass of gold.
The volume of a sphere can be expressed as V = (4/3)πr^3, where r is the radius of the sphere.
The density of gold is given as 19.3 g/cm^3. This means that a cubic centimeter of gold weighs 19.3 grams.
The molar mass of gold is given as 197 g/mol. This means that one mole of gold has a mass of 197 grams.
Now, we can calculate the volume of one atom of gold by dividing the mass of one atom of gold by the density of gold:
Volume of one atom = molar mass of gold / density of gold = 197 g/mol / 19.3 g/cm^3
To convert cm^3 to m^3 (since the atomic radius needs to be in meters), we need to multiply by 10^-6:
Volume of one atom = (197 g/mol / 19.3 g/cm^3) x (1 cm^3 / 10^-6 m^3) = 1.0198 x 10^-5 m^3/mol
Since one mole contains Avogadro's number (6.022 x 10^23) atoms, we can divide the volume by Avogadro's number to get the volume of one atom:
Volume of one atom = (1.0198 x 10^-5 m^3/mol) / (6.022 x 10^23 atoms/mol) = 1.693 x 10^-29 m^3/atom
Now, using the formula for the volume of a sphere, we can find the radius:
Volume of one atom = (4/3)πr^3
By rearranging the formula, we get:
r = [(3/(4π))(Volume of one atom)]^(1/3)
Plugging in the values:
r = [(3/(4π))(1.693 x 10^-29 m^3/atom)]^(1/3) ≈ 1.417 x 10^-10 m
So, the approximate atomic radius of gold is 1.417 x 10^-10 meters.