Combustion analysis of an unknown compound containing only carbon and hydrogen produced $0.5690 \mathrm{~g}$ of $\mathrm{CO}_{2}$ and $0.2902 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$. What is the empirical formula of the compound?
$\mathrm{CH}_{2}$
$\mathrm{C}_{2} \mathrm{H}_{5}$
$\mathrm{C}_{4} \mathrm{H}_{10}$
$\mathrm{C}_{5} \mathrm{H}_{2}$

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Accepted Answer

***B*** ∻Key Concept∻ ⚹Mole ratio from combustion products⚹ ∻Explanation∻ ⚹To find the empirical formula, we need to determine the moles of carbon and hydrogen in the compound. From the mass of CO2, we can calculate the moles of carbon, and from the mass of H2O, we can calculate the moles of hydrogen. The molar mass of CO2 is 44.01 g/mol, and the molar mass of H2O is 18.02 g/mol. Therefore, $0.5690 \mathrm{~g}$ of CO2 corresponds to $\frac{0.5690}{44.01}$ moles of CO2, which equals the moles of carbon since each molecule of CO2 contains one atom of carbon. Similarly, $0.2902 \mathrm{~g}$ of H2O corresponds to $\frac{0.2902}{18.02}$ moles of H2O, which equals twice the moles of hydrogen since each molecule of H2O contains two atoms of hydrogen. After finding the moles of carbon and hydrogen, we divide both by the smallest number of moles to get the simplest whole number ratio, which gives us the empirical formula.⚹◊Can you calculate the molar mass of each element in the unknown compound based on the given information from the combustion analysis?⍭ Generate me a similar question◊

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