Question

The atomic weight of aluminum is $26.982{\text{ }}u$ , how many aluminum atoms are there in a $4.55g$ sample of aluminum ?

Answer 1

To figure the quantity of molecules in a recipe, the heaviness of an example, its nuclear mass from the occasional table and a consistent known as Avogadro's number are required. The way in to this issue is the right translation of what nuclear weight implies.

Complete step by step answer:
Notice that aluminum is said to have a nuclear mass of $26.982{\text{ }}u$ . The unit utilized here is the bound together nuclear mass unit , $u$ .
The bound together nuclear mass unit is characterized as $1/12th\;$ of the mass of an impartial and neutral $carbon - 12{\text{ }}atom$ and is roughly equivalent to the mass of one nucleon, as in the mass of one proton or one neutron. Additionally, you should realize that you have
$1{\text{ }}ux = x1{\text{ }}g/mol$
Molar mass of aluminum $\left( {Al} \right)$ is
$\left( {Al} \right)$ = $26.9815386{\text{ }}g/mol.$
So $1{\text{ }}mole$ of aluminum has a mass of $26.9815386{\text{ }}grams$ .
This implies that the mass of one mole of aluminum molecules will be $26.982{\text{ }}g$ .
As you probably are aware, one mole of any component contains $6.022*{10^{23}}$atoms of that component.
This implies that on the off chance that you know the number of moles of aluminum you have in $4.55{\text{ }}g$, you can utilize Avogadro's number to locate the quantity of molecules.
$4.55g*{\text{ }}(1{\text{ }}mole{\text{ }}Al/\;26.982g) = 0.168631{\text{ }}moles{\text{ }}Al$
Consequently, you will have
$0.168631moles{\text{ }}of{\text{ }}Al*{\text{ }}(6.022*1023atoms{\text{ }}of{\text{ }}Al/{\text{ }}1mole{\text{ }}Al) = 1.02*1023atoms{\text{ }}of{\text{ }}Al$

Note:
The principal thing that you need to do here is to sort out the mass of the example.
Next, you should change the mass of the example over to moles by utilizing the molar mass of aluminum, which is equivalent to $26.982{\text{ }}g{\text{ }}mol - 1$ .
Finally , to convert the number of moles to atoms , use the fact that $1\;$mole of aluminium must contain $6.022*{10^{23}}$ atoms of aluminium → this is known as Avogadro’s constant.