# Question #9e257

##### 1 Answer

#### Explanation:

The idea here is that you need to use the definition of the *unified atomic mass unit*, *grams*.

So, the mass of a **single atom** is expressed in unified atomic mass units,

You can think about the unified mass unit as being equivalent to the mass of **one nucleon**, i.e. one *proton* **or** one *neutron*. More importantly, you should keep in mind that you have this relationship between the unified atomic mass unit and *grams*

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 u" = "1 g mol"^(-1))color(white)(a/a)|)))#

This will be the one of the two conversion factors that you'll use to go from **Avogadro's number**.

#color(blue)(|bar(ul(color(white)(a/a)"1 mol" = 6.022 * 10^(23)"atoms"color(white)(a/a)|))) -># Avogadro's number

So, you know that the **molar mass** of this new element, i.e. the mass of **one mole** of its atoms, will be

#293 color(red)(cancel(color(black)("u"))) * "1 g mol"^(-1)/(1color(red)(cancel(color(black)("u")))) = "293 g mol"^(-1)#

So, if **one mole** of atoms has a mass of

#33color(red)(cancel(color(black)("atoms"))) * overbrace((1color(red)(cancel(color(black)("mol"))))/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms")))))^(color(purple)("Avogadro's number")) * overbrace("293 g"/(1color(red)(cancel(color(black)("mole")))))^(color(blue)("molar mass")) = color(green)(|bar(ul(color(white)(a/a)1.6 * 10^(-20)"g"color(white)(a/a)|)))#

Simply put, if you could weigh out