Answer:
The empirical formula is:
![CH_3](https://img.qammunity.org/2020/formulas/physics/middle-school/3eykzixhzd4wzagvlw8qlv47jd4s591rjl.png)
Step-by-step explanation:
In a sample of 100 g:
![m_C=100g *0.8= 80g](https://img.qammunity.org/2020/formulas/chemistry/high-school/9p3ooyqppkn3ry0zn663e210e4iymniaz5.png)
![m_H=100g *0.2= 20g](https://img.qammunity.org/2020/formulas/chemistry/high-school/o51el5htho6aetw4qff0x7pr78px25xpqg.png)
To determine the molecular formula you need to calculate the moles of each element. To do that the atomic weights are needed:
![M_C=12g/mol](https://img.qammunity.org/2020/formulas/chemistry/high-school/9t92nuk8mort94gbgdgs3zzyou2ct4zjkm.png)
![M_H=1g/mol](https://img.qammunity.org/2020/formulas/chemistry/high-school/4lddyvm9hk5bkm8oci4o87e8plcflqmfp2.png)
So, the moles of each one:
![n_C=(80g)/(12g/mol)=6.66mol](https://img.qammunity.org/2020/formulas/chemistry/high-school/75iprv0bxcihylawrsripen61cukt2io5q.png)
![n_H=(20g)/(1g/mol)=20mol](https://img.qammunity.org/2020/formulas/chemistry/high-school/sks59sua1gou3jj8eipn4s0ud19jzvgt77.png)
Now, you need to take this values to non-fractional numbers (keeping the ratio):
![n_C=(6.66mol)/(6.66mol)=1](https://img.qammunity.org/2020/formulas/chemistry/high-school/k94c2n3czaf46qljktqxfh2bi5o7yvt458.png)
![n_H=(20mol)/(16.66mol)=3](https://img.qammunity.org/2020/formulas/chemistry/high-school/3xjfxd416dt5nsmfshq614z863pglmmpx2.png)
Therefore the empirical formula is:
![CH_3](https://img.qammunity.org/2020/formulas/physics/middle-school/3eykzixhzd4wzagvlw8qlv47jd4s591rjl.png)
The molecular formula probably is a multiple of the empirical due to the carbon's valence.