# Question #8cba1

Oct 2, 2017

$5.51 \cdot {10}^{26}$

#### Explanation:

You can;t go from grams to atoms without going through *moles8 first, so start by converting the mass of calcium to moles.

To do that, use the molar mass of calcium.

$\text{36,700" color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078 color(red)(cancel(color(black)("g")))) = "915.71 moles Ca}$

Now, in order to go from moles to atoms, use the fact that, by definition, $1$ mole of calcium must contain $6.022 \cdot {10}^{23}$ atoms of calcium $\to$ this is known as Avogadro's constant.

In other words, for every $6.022 \cdot {10}^{23}$ atoms of calcium you have in your sample, you have $1$ mole of calcium.

This implies that your sample will contain

$915.71 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Ca"))) * (6.022 * 10^(23)color(white)(.)"atoms Ca")/(1color(red)(cancel(color(black)("mole Ca")))) = color(darkgreen)(ul(color(black)(5.51 * 10^(26)color(white)(.)"atoms Ca}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the mass of calcium.