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

**Avogadroâ€™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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Our tutors have indicated that to solve this problem you will need to apply the Mass Percent Formula concept. You can view video lessons to learn Mass Percent Formula. Or if you need more Mass Percent Formula practice, you can also practice Mass Percent Formula practice problems.

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