# Question 43768

Feb 2, 2016

$5.0 \cdot {10}^{49} \text{g}$

#### Explanation:

I think it's safe to say that the value given to you for the number of potassium atoms in that sample doesn't make any sense. The mass of ${10}^{71}$ atoms of anything is an astronomically large number!

Let's calculate, for the sake of argument, the mass of this sample. You know that your sample contains $5.5 \cdot {10}^{71}$ atoms of potassium, $\text{K}$.

Notice that one mole of potassium sulfide, $\text{K"_2"S}$, contains

• two moles of potassium atoms
• one mole of sulfur atoms

Your first step here will be to use Avogadro's number, which tells you the number of atoms you get per mole of an element, to determine how many moles of potassium are present in your hypothetical sample.

So, Avogadro's number tells you that one mole of an element contains $6.022 \cdot {10}^{23}$ atoms of that element. This means that you sample will contain

5.5 * 10^(71) color(red)(cancel(color(black)("atoms K"))) * overbrace("1 mole K"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms K")))))^(color(purple)("Avogadro's number")) = 9.133 * 10^(47)"moles K"

Now use the fact that one mole of potassium sulfide contains two moles of potassium to get the number of moles of potassium sulfide

9.133 * 10^(47)color(red)(cancel(color(black)("moles K"))) * ("1 mole K"_2"S")/(2color(red)(cancel(color(black)("moles K")))) = 4.567 * 10^(47)"moles K"_2"S"

Finally, use potassium sulfide's molar mass to determine how many grams would contain this many moles

4.567 * 10^(47)color(red)(cancel(color(black)("moles K"_2"S"))) * "110.26 g"/(1color(red)(cancel(color(black)("mole K"_2"S")))) ~~ 5.0 * 10^(49)"g"#

As a fun fact, the mass of ordinary matter in the observable Universe is often estimated at ${10}^{56} \text{g}$!

All in all, the important thing to remember here is the approach used to get from atoms to moles, and then to grams, by using Avogadro's number and the molar mass of the molecule.