How many atoms of copper (cu) are in a pure copper coin weighing 12.0g
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How many atoms of copper (cu) are in a pure copper coin weighing 12.0g

[From: ] [author: ] [Date: 11-09-15] [Hit: ]
546. This means we have a conversion factor of (63.546g / 1 mol) or (1 mol / 63.546g). Since were given grams and we want moles, well use the conversion factor that places grams on the bottom.......
This requires a two-step conversion. We must first use the molar mass of copper to convert the given grams to moles. Then we can use Avogadro's # (6.022 x 10^23) to convert the moles to individual atoms.

The molar mass of copper (found on the periodic table) is 63.546. This means we have a conversion factor of (63.546g / 1 mol) or (1 mol / 63.546g). Since we're given grams and we want moles, we'll use the conversion factor that places grams on the bottom. Now, grams will cancel and we'll be left with moles:

12.0g x 1 mol / 63.546g = 0.188839581mol

Now, we are limited to 3 significant figures because of the limiting 12.0g, but this is just an intermediate step, so we should wait to round off until the end.

Now that we know how many moles we have, we can use Avogadro's # to convert moles to atoms. Avogadro's # tells that that for every 1 mole, we have 6.022 x 10^23 of whatever we're measuring. This gives us a conversion factor of (1 mole / 6.022 x 10^23 atoms) or (6.022 x 10^23 atoms / 1 mole). Since we have moles and we want atoms, we'll use the conversion factor with moles on the bottom. So:

0.188839581 mol x (6.022 x 10^23 atoms) / 1 mol = 1.137191955 x 10^23

Remember from before that we are limited to 3 significant figures. Since our calculations are complete, we can now round down to: 1.14 x 10^23

That is our final answer.

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