Question

For which compound is the empirical formula the same as the molecular formula?

Answer 1

Answer: Thus empirical and molecular formula is sa,e for $C_4H_8O$

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.  

1. The empirical formula is $CO_2H$

The empirical weight of $CO_2H$ = 1(12)+2(16)+1(1)= 45g.

The molecular weight = 90 g/mole

Now we have to calculate the molecular formula:

$n=\frac{\text{Molecular weight }{\text{Equivalent weight}=\frac{90}{45}=2$

The molecular formula will be=$2\times CO_2H=C_2O_4H_2$

2. The empirical formula is $CH_3O$

The empirical weight of $CH_3O$ = 1(12)+3(1)+1(16)= 31 g.

The molecular weight = 62 g/mole

Now we have to calculate the molecular formula:

$n=\frac{\text{Molecular weight }{\text{Equivalent weight}=\frac{62}{31}=2$

The molecular formula will be=$2\times CH_3O=C_2H_6O_2$

3. The empirical formula is $C_2H_4O$

The empirical weight of $C_2H_4O$ = 2(12)+4(1)+1(16)= 44 g.

The molecular weight = 88 g/mole

Now we have to calculate the molecular formula:

$n=\frac{\text{Molecular weight }{\text{Equivalent weight}=\frac{88}{44}=2$

The molecular formula will be=$2\times C_2H_4O=C_4H_8O_2$

4. The empirical formula is $C_4H_8O$

The empirical weight of $C_4H_8O$ = 4(12)+8(1)+1(16)= 72 g.

The molecular weight = 72 g/mole

Now we have to calculate the molecular formula:

$n=\frac{\text{Molecular weight }{\text{Equivalent weight}=\frac{72}{72}=1$

The molecular formula will be=$1\times C_4H_8O=C_4H_8O$