# Question db676

Nov 4, 2016

$1.3 \cdot {10}^{21} \text{atoms}$

#### Explanation:

The first thing to do here is to figure out how much magnesium you have in your sample by using the pigment's percent concentration by mass.

More specifically, you must use the fact that a 2.68%"m/m" concentration tells you that you get $\text{2.68 g}$ of magnesium for every $\text{100 g}$ of pigment.

This means that your sample will contain

2.0 color(red)(cancel(color(black)("g chlorophyll"))) * "2.68 g Mg"/(100color(red)(cancel(color(black)("g chlorophyll")))) = "0.0536 g Mg"

Now, in order to get the number of atoms of magnesium present in the sample, you must convert the mass to moles.

To do that, use magnesium's molar mass

0.0536 color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.305 color(red)(cancel(color(black)("g")))) = "0.002205 moles Mg"

Finally, you can convert the number of moles to atoms by using Avogadro's constant

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 mole" = 6.022 * 10^(23)"atoms}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's constant

You can thus say that your sample will contain

0.00205 color(red)(cancel(color(black)("moles Mg"))) * (6.022 * 10^(23)"atoms Mg")/(1color(red)(cancel(color(black)("mole Mg")))) = color(green)(bar(ul(|color(white)(a/a)color(black)(1.3 * 10^(21)"atoms Mg")color(white)(a/a)|)))#

The answer is rounded to two sig figs.