41 - (CHEMI/12, HL. Summer 2022.02) - Stoichiomeric Relationship 30g of an organic compound produces 44g CO_(2) and 18g H_(2)O as the only combustion products. Which of the following is the empirical formula for this compound? M_(r)CO_(2)=44M_(r)H_(2)O=18 A.) CH_(2) B. CH_(3) CHO D. CH_(2)O

To find the empirical formula of the compound, we need to determine the moles of carbon and hydrogen present in the compound.<br /><br />Given:<br />- 30g of the organic compound produces 44g of $CO_{2}$ and 18g of $H_{2}O$.<br /><br />First, let's calculate the moles of carbon in the compound:<br />- The molar mass of $CO_{2}$ is 44 $M_{r}$.<br />- The moles of $CO_{2}$ produced is 44g / 44 $M_{r}$ = 1 mole.<br /><br />Since each mole of $CO_{2}$ contains one mole of carbon, the moles of carbon in the compound is also 1 mole.<br /><br />Next, let's calculate the moles of hydrogen in the compound:<br />- The molar mass of $H_{2}O$ is 18 $M_{r}$.<br />- The moles of $H_{2}O$ produced is 18g / 18 $M_{r}$ = 1 mole.<br /><br />Since each mole of $H_{2}O$ contains two moles of hydrogen, the moles of hydrogen in the compound is 2 moles.<br /><br />Now, let's determine the empirical formula:<br />- The empirical formula is the simplest whole number ratio of the moles of each element in the compound.<br />- The ratio of carbon to hydrogen is 1:2.<br /><br />Therefore, the empirical formula for this compound is $CH_{2}$.<br /><br />So, the correct answer is:<br />A.) $CH_{2}$