Respuesta :
Answer:
The empirical formula of this substance is:
[tex]C_3H_6O_2[/tex]
Explanation:
To find the empirical formula of this substance we need the molecular weight of the elements Carbon, Hydrogen and Oxygen, we can find this information in the periodic table:
- C: 12.01 g/mol
- H: 1.00 g/mol
- O: 15.99 g/mol
With the information in this exercise we can suppose in 100 g of the substance we have:
C: 48.64 g
H: 8.16 g
O: 43.2 g (100 g - 48.64g - 8.16g= 43.2 g)
Now, we need to divide these grams by the molecular weight:
[tex]C=\frac{48.64g}{12.01 g/mol} =4.05 mol\\H=\frac{8.16g}{1.00g/mol}= 8.16 mol\\O=\frac{43.2g}{15.99 g/mol} = 2.70 mol[/tex]
We need to divide these results by the minor result, in this case O=2.70 mol
[tex]C=\frac{4.05mol}{2.70mol}= 1.5\\H= \frac{8.16mol}{2.70 mol} = 3.02 \\O=\frac{2.70mol}{2.70mol} = 1[/tex]
We need to find integer numbers to find the empirical formula, for this reason we multiply by 2:
[tex]C= 1.5*2=3\\H= 3.02*2= 6.04 \\O= 1*2=2[/tex]
This numbers are very close to integer numbers, so we can find the empirical formula as subscripts in the chemical formula:
[tex]C_3H_6O_2[/tex]