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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]
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]
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