$\overline{){\mathbf{mass}}{\mathbf{}}{\mathbf{percent}}{\mathbf{}}{\mathbf{gold}}{\mathbf{=}}\frac{\mathbf{mass}\mathbf{}\mathbf{gold}}{\mathbf{mass}\mathbf{}\mathbf{bracelet}}{\mathbf{\times}}{\mathbf{100}}}$

**Avogadro’s number = 6.022 x 10**^{23}** entities**

** ****entities = atoms, ions, molecules, particles, formula units**

**Calculate the # of atoms in 0.351-ounce 18 K gold bracelet****:**

Pure gold is usually too soft for jewelry, so it is often alloyed with other metals.

How many gold atoms are in an 0.351-ounce, 18 K gold bracelet? (18 K gold is 75% gold by mass.)

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Based on our data, we think this problem is relevant for Professor Mignerey's class at UMD.