Final answer:
An 18 karat gold bracelet weighing 0.333 ounces contains approximately 2.16 × 10^22 atoms of gold. This calculation is based on converting the weight to grams, accounting for the 75% gold content of the bracelet, and using Avogadro's number.
Step-by-step explanation:
To calculate how many gold atoms are in an 0.333 ounce, 18 k (karat) gold bracelet, we must first convert the mass of the bracelet into grams since the mass of gold is typically measured in grams. We know that 18 karat gold is comprised of 75% gold by mass. Once the mass of pure gold is identified, we can use Avogadro's number to determine the amount of atoms in that mass.
First, 0.333 ounces is approximately 9.43 grams (1 ounce = 28.3495 grams). Since the bracelet is 18k gold or 75% gold, the mass of pure gold would be 75% of 9.43 grams, which equals approximately 7.0725 grams of gold.
We know from the periodic table that the atomic mass of gold (Au) is approximately 197 g/mol, which means that 1 mole of gold has a mass of about 197 grams. Using this, we can determine the number of moles of gold in our bracelet:
7.0725 grams Au × (1 mole Au / 197 grams Au) = 0.0359 moles Au
Avogadro's number tells us that 1 mole of any substance contains approximately 6.022 × 1023 atoms. So:
0.0359 moles Au × (6.022 × 1023 atoms/mole) = approximately 2.16 × 1022 atoms of gold.
Therefore, an 18 karat gold bracelet weighing 0.333 ounces, which corresponds to 75% gold by mass, contains approximately 2.16 × 1022 atoms of gold.