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Chapter 3: Problem 173

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 \(\mathrm{g}\) of the compound produced 0.213 \(\mathrm{g} \mathrm{CO}_{2}\) and 0.0310 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\) . In another experiment, it is found that 0.103 \(\mathrm{g}\) of the compound produces 0.0230 $\mathrm{g} \mathrm{NH}_{3} .$ What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\) . Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\) . Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Short Answer

Expert verified
The empirical formula of the compound is \(C_4H_3NO_4\).

Step by step solution

01

1. Calculate the Mass of Carbon and Hydrogen

After combustion, the compound produces CO2 and H2O. We have the masses of both CO2 and H2O, so we will use these to calculate the mass of carbon and hydrogen in the compound. To do this, we will apply the mass ratio of these elements to the respective compounds. Mass of Carbon: The mass ratio of carbon in CO2 is 12.01 g/mol (C) / 44.01 g/mol (CO2). We can find the mass of carbon using this ratio. Mass of Carbon = 0.213 g CO2 * (12.01 g/mol C / 44.01 g/mol CO2) = 0.0581 g Mass of Hydrogen: The mass ratio of hydrogen in H2O is 2.016 g/mol (H2) / 18.015 g/mol (H2O). Similarly, we find the mass of hydrogen. Mass of Hydrogen = 0.0310 g H2O * (2.016 g/mol H2 / 18.015 g/mol H2O) = 0.00345 g
02

2. Calculate the Mass of Nitrogen

The compound produces NH3 in another experiment. We have the mass of NH3, so we will use this mass to calculate the mass of nitrogen in the compound. As before, we will apply the mass ratio of the element to the compound. Mass of Nitrogen: The mass ratio of nitrogen in NH3 is 14.01 g/mol (N) / 17.03 g/mol (NH3). We find the mass of nitrogen using this ratio. Mass of Nitrogen = 0.0230 g NH3 * (14.01 g/mol N / 17.03 g/mol NH3) = 0.0188 g
03

3. Determine the Mass of Oxygen

We are given the mass of the compound. We can find the mass of oxygen by subtracting the masses of carbon, hydrogen, and nitrogen from the total mass of the compound. Mass of Oxygen = Total mass of the compound - (Mass of Carbon + Mass of Hydrogen + Mass of Nitrogen) Mass of Oxygen = 0.157 g - (0.0581 g + 0.00345 g + 0.0188 g) = 0.07665 g
04

4. Calculate Moles of Each Element

We will now use the molar masses of each element to calculate the moles of carbon, hydrogen, nitrogen, and oxygen in the compound. Moles of Carbon = 0.0581 g / 12.01 g/mol = 0.00483 mol Moles of Hydrogen = 0.00345 g / 1.008 g/mol = 0.00342 mol Moles of Nitrogen = 0.0188 g / 14.01 g/mol = 0.00134 mol Moles of Oxygen = 0.07665 g / 16.00 g/mol = 0.00479 mol
05

5. Calculate the Mole Ratios

To find the empirical formula, we need to determine the mole ratios of these elements. We will divide the moles of each element by the smallest value among them to find their ratios. Mole Ratio of Carbon = 0.00483 mol / 0.00134 mol = 3.61 ≈ 4 Mole Ratio of Hydrogen = 0.00342 mol / 0.00134 mol = 2.56 ≈ 3 Mole Ratio of Nitrogen = 0.00134 mol / 0.00134 mol = 1 Mole Ratio of Oxygen = 0.00479 mol / 0.00134 mol = 3.58 ≈ 4
06

6. Determine the Empirical Formula

Now that we have the mole ratios of each element in the compound, we can write the empirical formula of the compound using these ratios. Empirical Formula = C4H3N1O4 Hence, the empirical formula of the compound is \(C_4H_3NO_4\).

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Most popular questions from this chapter

Natural rubidium has the average mass of 85.4678 \(\mathrm{u}\) and is composed of isotopes \(^{85} \mathrm{Rb}(\mathrm{mass}=84.9117 \mathrm{u})\) and $^{87} \mathrm{Rb}\( . The ratio of atoms \)^{85} \mathrm{Rb} /^{87} \mathrm{Rb}$ in natural rubidium is \(2.591 .\) Calculate the mass of \(^{87} \mathrm{Rb}\) .

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