How many gold atoms are in an 0.375 ounce, 18 k gold bracelet?

What is the mass and purity of the gold in the bracelet?

Calculate the number of gold atoms in the bracelet.

Answer:

2.42×10^22 atoms of gold.

When dealing with a scenario involving the number of gold atoms in a gold bracelet, we first need to determine the mass and purity of the gold present. In this case, the bracelet weighs 0.375 ounces and is made of 18 karat gold, which is 75% pure gold by mass.

To calculate the mass of gold in the bracelet, we first convert 0.375 ounces to grams. Since 1 ounce is equal to 28.34952 grams, 0.375 ounces is equal to 10.631 grams. Given that the gold is 75% pure, we calculate the mass of pure gold in the bracelet by multiplying the total mass by the purity fraction: 10.631 grams x 0.75 = 7.973 grams of gold.

Next, we need to determine the amount of substance, in this case, the number of moles of gold in the bracelet. Using the formula n(Au) = m(Au) ÷ M(Au), where M(Au) is the molar mass of gold (197 g/mol), we calculate that n(Au) = 7.973 g ÷ 197 g/mol = 0.0404 mol of gold.

Finally, to find the number of gold atoms present in the bracelet, we use Avogadro's number (6.023×10^23 atoms/mol). N(Au) = 0.0404 mol x 6.023×10^23 1/mol = 2.42×10^22 atoms of gold. Therefore, there are 2.42×10^22 gold atoms in the 18 k gold bracelet weighing 0.375 ounces.