Respuesta :
The molecular formula of compound is [tex]Al(C_{2}H_{3}O_{2})_{3}[/tex].
Since, in 1 mole of [tex]Al(C_{2}H_{3}O_{2})_{3}[/tex] there are [tex]6.023\times 10^{23}[/tex] atoms of [tex]Al(C_{2}H_{3}O_{2})_{3}[/tex].
Thus, according to molecular formula, in 1 mole of [tex]Al(C_{2}H_{3}O_{2})_{3}[/tex] there are [tex]6\times 6.023\times 10^{23}=3.61\times 10^{24}[/tex] atoms of oxygen atoms.
1 atom of oxygen will be present in [tex]\frac{1}{3.61\times 10^{24}}[/tex] moles of [tex]Al(C_{2}H_{3}O_{2})_{3}[/tex] . Thus,
[tex]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}[/tex].
Molar mass of [tex]Al(C_{2}H_{3}O_{2})_{3}[/tex] is 236 g/mol, mass can be calculated as follows:
[tex]m=n\times M=0.7285 mol\times 236 g/mol=172 g[/tex]
Therefore, mass of [tex]Al(C_{2}H_{3}O_{2})_{3}[/tex] will be 172 g.
Answer : The mass of [tex]Al(C_2H_3O_2)_3[/tex] is, 148.48 grams.
Solution : Given,
Molar mass of [tex]Al(C_2H_3O_2)_3[/tex] = 204 g/mole
As we know that,
1 mole contains [tex]6.022\times 10^{23}[/tex] number of atoms
In the given compound [tex]Al(C_2H_3O_2)_3[/tex], there are 1 atom of aluminium, 6 atoms of carbon, 9 atoms of hydrogen and 6 atoms of oxygen.
As, [tex]6\times (6.022\times 10^{23})[/tex] number of atoms of oxygen present in 1 mole of [tex]Al(C_2H_3O_2)_3[/tex]
So, [tex]2.63\times 10^{24}[/tex] number of atoms of oxygen present in [tex]\frac{2.63\times 10^{24}}{6\times (6.022\times 10^{23})}=0.727[/tex] mole of [tex]Al(C_2H_3O_2)_3[/tex]
Now we have to calculate the mass of [tex]Al(C_2H_3O_2)_3[/tex].
Formula used :
[tex]\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[/tex]
Now put all the given values in this formula, we get
[tex]\text{Mass of }Al(C_2H_3O_2)_3=0.727mole\times 204g/mole=148.48g[/tex]
Therefore, the mass of [tex]Al(C_2H_3O_2)_3[/tex] is, 148.48 grams.
