Question 9e257

Apr 2, 2016

$1.6 \cdot {10}^{- 20} \text{g}$

Explanation:

The idea here is that you need to use the definition of the unified atomic mass unit, $u$, to express the mass of $33$ atoms of this new element in grams.

So, the mass of a single atom is expressed in unified atomic mass units, $u$. One unified atomic mass unit is defined as the mass of single, unbound carbon-12 atom in its ground state.

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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{1 u" = "1 g mol}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This will be the one of the two conversion factors that you'll use to go from $u$ to grams, the other being Avogadro's number.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mol" = 6.022 * 10^(23)"atoms} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ 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 $\text{293 g}$, it follows that $33$ atoms will have a mass of

$33 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atoms"))) * overbrace((1color(red)(cancel(color(black)("mol"))))/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms")))))^(color(purple)("Avogadro's number")) * overbrace("293 g"/(1color(red)(cancel(color(black)("mole")))))^(color(blue)("molar mass")) = color(green)(|bar(ul(color(white)(a/a)1.6 * 10^(-20)"g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Simply put, if you could weigh out $33$ atoms of this new element, you'll find that they have a mass of $1.6 \cdot {10}^{- 20} \text{g}$.