Select all the correct answers.

Which compounds have the empirical formula [tex]$CH_2O$[/tex]?

A. [tex]$C_2H_4O_2$[/tex]
B. [tex][tex]$C_3H_6O_3$[/tex][/tex]
C. [tex]$CH_2O_2$[/tex]
D. [tex]$C_5H_{10}O_5$[/tex]
E. [tex][tex]$C_8H_{12}O_6$[/tex][/tex]



Answer :

To determine which compounds have the empirical formula [tex]\(CH_2O\)[/tex], we need to examine and simplify each given molecular formula to see if it corresponds to the ratio of carbon, hydrogen, and oxygen atoms in [tex]\(CH_2O\)[/tex]. The ratio of atoms in the empirical formula [tex]\(CH_2O\)[/tex] is 1:2:1 for carbon, hydrogen, and oxygen, respectively.

Let's go through each compound and reduce it to its simplest ratio form:

### 1. [tex]\(C_2H_4O_2\)[/tex]
- Carbon atoms: 2
- Hydrogen atoms: 4
- Oxygen atoms: 2

Simplify the ratio:
[tex]\[ \frac{C_2}{C_2} : \frac{H_4}{C_2} : \frac{O_2}{C_2} = 1 : 2 : 1 \][/tex]
This corresponds to the ratio 1:2:1. Thus, this compound has the empirical formula [tex]\(CH_2O\)[/tex].

### 2. [tex]\(C_3H_6O_3\)[/tex]
- Carbon atoms: 3
- Hydrogen atoms: 6
- Oxygen atoms: 3

Simplify the ratio:
[tex]\[ \frac{C_3}{C_3} : \frac{H_6}{C_3} : \frac{O_3}{C_3} = 1 : 2 : 1 \][/tex]
This corresponds to the ratio 1:2:1. Thus, this compound has the empirical formula [tex]\(CH_2O\)[/tex].

### 3. [tex]\(CH_2O_2\)[/tex]
- Carbon atoms: 1
- Hydrogen atoms: 2
- Oxygen atoms: 2

Simplify the ratio:
[tex]\[ \frac{C_1}{C_1} : \frac{H_2}{C_1} : \frac{O_2}{C_1} = 1 : 2 : 2 \][/tex]
This corresponds to the ratio 1:2:2. Thus, this compound does not have the empirical formula [tex]\(CH_2O\)[/tex].

### 4. [tex]\(C_5H_{10}O_5\)[/tex]
- Carbon atoms: 5
- Hydrogen atoms: 10
- Oxygen atoms: 5

Simplify the ratio:
[tex]\[ \frac{C_5}{C_5} : \frac{H_{10}}{C_5} : \frac{O_5}{C_5} = 1 : 2 : 1 \][/tex]
This corresponds to the ratio 1:2:1. Thus, this compound has the empirical formula [tex]\(CH_2O\)[/tex].

### 5. [tex]\(C_8H_{12}O_6\)[/tex]
- Carbon atoms: 8
- Hydrogen atoms: 12
- Oxygen atoms: 6

Simplify the ratio:
[tex]\[ \frac{C_8}{C_2} : \frac{H_{12}}{C_2} : \frac{O_6}{C_2} = 4 : 6 : 3 \][/tex]
Next step:
[tex]\[ \frac{C_8}{C_4} : \frac{H_{12}}{C_4} : \frac{O_6}{C_4} = 2 : 3 : 1.5 \][/tex]

Since [tex]\(8, 12,\text{ and } 6\)[/tex] are not equally divisible by a common number to match [tex]\(1\text{ and }2\)[/tex], this compound does not have the empirical formula [tex]\(CH_2O\)[/tex].

### Conclusion
The compounds that have the empirical formula [tex]\(CH_2O\)[/tex] are:

- [tex]\(C_2H_4O_2\)[/tex]
- [tex]\(C_3H_6O_3\)[/tex]
- [tex]\(C_5H_{10}O_5\)[/tex]

Thus, the correct compounds are:

- [tex]\(C_2H_4O_2\)[/tex]
- [tex]\(C_3H_6O_3\)[/tex]
- [tex]\(C_5H_{10}O_5\)[/tex]