Question

Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.

Answer 1

The molecular formula of compound is $Al(C_{2}H_{3}O_{2})_{3}$.

Since, in 1 mole of  $Al(C_{2}H_{3}O_{2})_{3}$ there are $6.023\times 10^{23}$ atoms of  $Al(C_{2}H_{3}O_{2})_{3}$.

Thus, according to molecular formula, in 1 mole of  $Al(C_{2}H_{3}O_{2})_{3}$ there are $6\times 6.023\times 10^{23}=3.61\times 10^{24}$ atoms of oxygen atoms.

1 atom of oxygen will be present in $\frac{1}{3.61\times 10^{24}}$ moles of $Al(C_{2}H_{3}O_{2})_{3}$ . Thus,

$2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}$.

Molar mass of $Al(C_{2}H_{3}O_{2})_{3}$ is 236 g/mol, mass can be calculated as follows:

$m=n\times M=0.7285 mol\times 236 g/mol=172 g$

Therefore, mass of $Al(C_{2}H_{3}O_{2})_{3}$ will be 172 g.

Answer 2

Answer : The mass of $Al(C_2H_3O_2)_3$ is, 148.48 grams.

Solution : Given,

Molar mass of $Al(C_2H_3O_2)_3$ = 204 g/mole

As we know that,

1 mole contains $6.022\times 10^{23}$ number of atoms

In the given compound $Al(C_2H_3O_2)_3$, there are 1 atom of aluminium, 6 atoms of carbon, 9 atoms of hydrogen and 6 atoms of oxygen.

As, $6\times (6.022\times 10^{23})$ number of atoms of oxygen present in 1 mole of $Al(C_2H_3O_2)_3$

So, $2.63\times 10^{24}$ number of atoms of oxygen present in $\frac{2.63\times 10^{24}}{6\times (6.022\times 10^{23})}=0.727$ mole of $Al(C_2H_3O_2)_3$

Now we have to calculate the mass of $Al(C_2H_3O_2)_3$.

Formula used :

$\text{Mass of }Al(C_2H_3O_2)_3=\text{Moles of }Al(C_2H_3O_2)_3\times \text{Molar mass of }Al(C_2H_3O_2)_3$

Now put all the given values in this formula, we get

$\text{Mass of }Al(C_2H_3O_2)_3=0.727mole\times 204g/mole=148.48g$

Therefore, the mass of $Al(C_2H_3O_2)_3$ is, 148.48 grams.