A compound containing only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) yields the following data: i. Complete combustion of 35.0 \(\mathrm{mg}\) of the compound produced 33.5 \(\mathrm{mg}\) of \(\mathrm{CO}_{2}\) and 41.1 \(\mathrm{mg}\) of $\mathrm{H}_{2} \mathrm{O} .$ ii. A 65.2 -mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise \(137 ),\) giving 35.6 \(\mathrm{mL}\) of dry \(\mathrm{N}_{2}\) at \(740 .\) torr and \(25^{\circ} \mathrm{C}\) . iii. The effusion rate of the compound as a gas was measured and found to be \(24.6 \mathrm{mL} / \mathrm{min}\). The effusion rate of argon gas, under identical conditions, is \(24.6 \mathrm{mL} / \mathrm{min}\). What is the molecular formula of the compound?

Step by step solution

01

Calculate moles of Carbon and Hydrogen from combustion

When the compound is completely combusted, the Carbon and Hydrogen in the compound turns into CO2 and H2O. We have the mass of CO2 and H2O produced, from which we can find the moles of Carbon and Hydrogen in the compound. Ratios of moles: \(Moles_{C}=\frac{Mass_{CO_{2}}}{Molar\_Mass_{CO_{2}}}* Ratio_{C}\) \(Moles_{H}=\frac{Mass_{H_{2}O}}{Molar\_Mass_{H_{2}O}} * Ratio_{H}\) where, \(Ratio_{C}\) is the ratio of Carbon atoms in CO2 (equal to 1), and \(Ratio_{H}\) is the ratio of Hydrogen atoms in H2O (equal to 2). Calculate moles of Carbon and Hydrogen: \(Moles_{C}=\frac{33.5\ \mathrm{mg}}{44.01\ \frac{\mathrm{mg}}{\mathrm{mol}}}*1=0.761\ \mathrm{mol}\) \(Moles_{H}=\frac{41.1\ \mathrm{mg}}{18.02\ \frac{\mathrm{mg}}{\mathrm{mol}}}*2=4.558\ \mathrm{mol}\)

02

Calculate the moles of Nitrogen using Dumas method data

To find the moles of Nitrogen: We have Nitrogen gas volume, so we'll use the ideal gas law to convert the volume of N2 to moles. Ideal gas law: \(PV=nRT\) Calculate moles of Nitrogen: \(Moles_{N}=\frac{PV}{RT}=\frac{740\ \mathrm{torr} * 35.6\ \mathrm{mL}}{62.363\ \mathrm{L}\ \mathrm{torr}\ \mathrm{mol}^{-1} \mathrm{K}^{-1} * (298\ \mathrm{K})}=5.7297 * 10^{-4}\ \mathrm{mol}\)

03

Find the empirical formula of the compound

Divide the moles of each element by the minimum moles value to find the mole ratio. Mole ratio: \(Mole\ ratio_{C} = \frac{0.761}{5.7297 * 10^{-4}}=1328.36\) \(Mole\ ratio_{H} = \frac{4.558}{5.7297 * 10^{-4}}=7956.25\) \(Mole\ ratio_{N} = \frac{5.7297 * 10^{-4}}{5.7297 * 10^{-4}}=1\) Now, divide all the mole ratios by the smallest ratio to get a whole number ratio. \(1328.36:7956.25:1\) Divide by 1328.36 -> \(1:6:0.000753\) Since the Nitrogen ratio is extremely close to 0, we can assume it is present in a whole number ratio that is small. The most likely ratio is 1, so we can now define our empirical formula: Empirical Formula: CH6N

04

Calculate the molar mass of the empirical formula

Calculate the molar mass of the empirical formula: \(Molar\_Mass_{CH6N} = 12.01 + 6(1.007) + 14.01 = 30.055\ g/mol\)

05

Calculate the molar mass of the compound using effusion rate data

From the given data, the effusion rate of the compound is equal to the effusion rate of Argon gas under identical conditions. We can use Graham's law of effusion to find the molar mass of the compound. Graham's law of effusion states: \(\frac{Rate_{1}}{Rate_{2}} = \sqrt{\frac{Molar\_Mass_{2}}{Molar\_Mass_{1}}}\) Since the effusion rates are equal, the ratio will be 1. We can then find the molar mass of the compound, given the molar mass of argon is 39.95 g/mol. \(Molar\_Mass_{Compound} = Molar\_Mass_{Argon} = 39.95\ g/mol\)

06

Determine the molecular formula

Divide the molar mass of the compound by the molar mass of the empirical formula to find the integer factor. \(Factor =\frac{Molar\_Mass_{Compound}}{Molar\_Mass_{CH6N}}= \frac{39.95\ g/mol}{30.055\ g/mol}\) Based on the calculated \(Factor\), we could check what the exact whole number is, but we would assume it's close to 1 for simplicity in this exercise. So, the molecular formula would be the empirical formula itself, which is: Molecular Formula: CH6N

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