Question

the formula that does not have $\ce{ch_{2}o}$ as its empirical formula.
$\ce{c_{6}h_{10}o_{5}}$
$\ce{c_{3}h_{6}o_{3}}$
$\ce{c_{2}h_{4}o_{2}}$
$\ce{c_{6}h_{12}o_{6}}$
clear all

Explanation:

Step1: Recall Empirical Formula

The empirical formula is the simplest whole - number ratio of atoms in a compound. For a formula to have an empirical formula of $CH_2O$, the ratio of $C:H:O$ in the molecular formula should be reducible to $1:2:1$.

Step2: Analyze $C_6H_{10}O_5$

Find the greatest common divisor (GCD) of 6, 10, and 5. The GCD of 6, 10, and 5 is 1. So the ratio of $C:H:O$ is $6:10:5$, which cannot be reduced to $1:2:1$.

Step3: Analyze $C_3H_6O_3$

The GCD of 3, 6, and 3 is 3. Dividing each by 3, we get $C:H:O = 1:2:1$, so its empirical formula is $CH_2O$.

Step4: Analyze $C_2H_4O_2$

The GCD of 2, 4, and 2 is 2. Dividing each by 2, we get $C:H:O=1:2:1$, so its empirical formula is $CH_2O$.

Step5: Analyze $C_6H_{12}O_6$

The GCD of 6, 12, and 6 is 6. Dividing each by 6, we get $C:H:O = 1:2:1$, so its empirical formula is $CH_2O$.

Answer:

$C_6H_{10}O_5$