The molar mass of Asprin=180 g/mol
The empirical formula of Aspirin=C9H8O4.
The molar mass of C=12 g/mol, H=1 g/mol and O=16 g/mol
the molar mass of C9H8O4=(9*12)+(8*1)+(4*16) g/mol=180 g/mol.
So the empirical formula is also molecular formula for Aspirin.
Therefore the molecular formula of Asprin=(C9H8O4)1.
hope that helps!
Answer: [tex]C_9H_8O_4[/tex]
Explanation:
Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound. Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.
The empirical formula is [tex]C_9H_8O_4[/tex]
The empirical weight of [tex]C_9H_8O_4[/tex] = 9(12)+8(1)+4(16)= 180g.
The molecular weight = 180 g/mole
Now we have to calculate the molecular formula:
[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{180}{180}=1[/tex]
The molecular formula will be=[tex]1\times C_9H_8O_4=C_9H_8O_4[/tex]
Thus molecular formula of aspirin will be [tex]C_9H_8O_4[/tex]
