Calculating the Number of Gold Atoms in a Gold Bracelet

How many gold atoms are in an 0.333 ounce, 18 k gold bracelet? (18 k gold is 75% gold by mass.)

Based on the given information, how can we determine the number of gold atoms in the gold bracelet?

Answer:

An 18 karat gold bracelet weighing 0.333 ounces contains approximately 2.16 × 10^22 atoms of gold.

In order to calculate the number of gold atoms in the gold bracelet, we need to follow a series of steps. First, we need to convert the weight of the bracelet into grams as the mass of gold is typically measured in grams. Since the bracelet is made of 18 karat gold, which is 75% gold by mass, we need to determine the amount of pure gold in the bracelet.

Given that the weight of the bracelet is 0.333 ounces, which is approximately 9.43 grams (1 ounce = 28.3495 grams), we can calculate that the mass of pure gold in the bracelet is approximately 7.0725 grams (0.333 ounces * 0.75).

Next, we need to consider the atomic mass of gold (Au), which is around 197 g/mol. By dividing the mass of gold in the bracelet by the atomic mass of gold, we can determine the number of moles of gold present.

Calculating 7.0725 grams Au ÷ 197 grams/mol Au gives us approximately 0.0359 moles of gold. Using Avogadro's number (6.022 × 10^23 atoms/mol), we can then find that there are about 2.16 × 10^22 atoms of gold in the 18 karat gold bracelet weighing 0.333 ounces.

Therefore, the final answer is that the gold bracelet contains approximately 2.16 × 10^22 atoms of gold.