Problem 52 - Chemistry: Principles and Reactions

Question

Dimethylhydrazine, the fuel used in the Apollo lunar descent module, has a molar mass of \(60.10 \mathrm{~g} / \mathrm{mol}\). It is made up of carbon, hydrogen, and nitrogen atoms. The combustion of \(2.859 \mathrm{~g}\) of the fuel in excess oxygen yields \(4.190 \mathrm{~g}\) of carbon dioxide and \(3.428 \mathrm{~g}\) of water. What are the simplest and molecular formulas for dimethylhydrazine?

Step-by-Step Solution

Answer

Answer: The empirical formula for dimethylhydrazine is CH4N, and the molecular formula is C2H8N2.

Step 1: Determine the Amount of each Element in the Products

We're given the mass of carbon dioxide (4.190 g) and water (3.428 g) produced from the combustion of 2.859 g of dimethylhydrazine fuel. With this information, we can determine the amount of carbon, hydrogen, and oxygen in the products. - Carbon: From carbon dioxide (CO2), we know that the mass of carbon is 12.01 g/mol, and the mass of oxygen is 16.00 g/mol. \(4.190 \mathrm{~g}\) CO2 contains \((4.190 \mathrm{~g}) \times (\frac{12.01}{12.01 + 2\times16.00}) = 1.151 \mathrm{~g} \mathrm{C}\) - Hydrogen: From water (H2O), we know that the mass of hydrogen is 1.008 g/mol, and the mass of oxygen is 16.00 g/mol. \(3.428 \mathrm{~g}\) H2O contains \((3.428 \mathrm{~g}) \times (\frac{2 \times 1.008}{2 \times 1.008 + 16.00}) = 0.383 \mathrm{~g} \mathrm{H}\) - Nitrogen: We can find the mass of nitrogen from combustion, using the mass of fuel, mass of carbon, and mass of hydrogen. The mass of nitrogen would be the difference in mass: \(2.859 \mathrm{~g} - 1.151 \mathrm{~g} - 0.383 \mathrm{~g} = 1.325 \mathrm{~g} \mathrm{N}\)

Step 2: Calculate the Empirical Formula

Now we have the masses of carbon, hydrogen, and nitrogen present in the fuel. We divide these mass values by their respective molar masses (in g/mol) to get the moles of C, H, and N, and find the ratio among the elements. - Carbon: \((1.151 \mathrm{~g}) \times (\frac{1}{12.01 \mathrm{~g/mol}}) = 0.0958 \mathrm{mol} \mathrm{C}\) - Hydrogen: \((0.383 \mathrm{~g}) \times (\frac{1}{1.008 \mathrm{~g/mol}}) = 0.381 \mathrm{mol} \mathrm{H}\) - Nitrogen: \((1.325 \mathrm{~g}) \times (\frac{1}{14.01 \mathrm{~g/mol}}) = 0.0946 \mathrm{mol} \mathrm{N}\) We find the ratio by dividing all mole values by the smallest value: C: \(\frac{0.0958}{0.0946} = 1.01\) (round to 1) H: \(\frac{0.381}{0.0946} = 4.03\) (round to 4) N: \(\frac{0.0946}{0.0946} = 1\) So the empirical formula is CH4N.

Step 3: Determine the Molecular Formula

Now we compare the empirical formula's molar mass to the given molar mass of dimethylhydrazine to determine the molecular formula. The molar mass of CH4N is \((12.01 + 4 \times 1.008 + 14.01) \mathrm{~g/mol} = 30.05 \mathrm{~g/mol}\). The given molar mass of dimethylhydrazine is \(60.10 \mathrm{~g/mol}\). Therefore, we multiply the empirical formula by two to get the molecular formula: - C - 1 \(\times\) 2 = 2 - H - 4 \(\times\) 2 = 8 - N - 1 \(\times\) 2 = 2 The molecular formula of dimethylhydrazine is C2H8N2.

Step 4: Check the Final Formulas

We have the simplest formula (empirical formula) as CH4N and the molecular formula as C2H8N2. Both formulas match the given information and have the correct molar mass.

Key Concepts

Empirical Formula

The empirical formula is essentially the simplest integer ratio of elements in a compound. To determine the empirical formula, you start by calculating the mass of each element present in a compound. Then, you convert these masses to moles since moles provide a common unit of measurement that directly reflects the number of atoms, due to Avogadro's number.
Converting grams to moles involves dividing by the atomic or molecular weight of the element, determined by the periodic table. Once you have the moles of each element, you calculate the smallest mole ratio by dividing the moles of each element by the smallest number of moles calculated.

Understanding Ratios

For instance, if we are given a compound consisting of carbon, hydrogen, and nitrogen, and after calculations, we find the ratio of their moles to be close to whole numbers, we simply round them. In our example, ratios of 1.01, 4.03, and 1 for carbon, hydrogen, and nitrogen respectively are rounded to 1:4:1 to give us the empirical formula CH4N. This represents the simplest ratio of atoms in the compound, laying the foundation for finding the molecular formula.

Combustion Analysis

Combustion analysis is a standard laboratory method commonly used to determine the empirical formula of a hydrocarbon or carbon-hydrogen-oxygen compound. During complete combustion in excess oxygen, all carbon in the compound is converted to carbon dioxide (CO2), and hydrogen to water (H2O). Nitrogen in the compound can be found as different products, and is often determined by difference after carbon and hydrogen have been accounted for.
By knowing the molar mass of CO2 and H2O, and assuming complete conversion, the amount of carbon and hydrogen in the original compound can be calculated. This is done by multiplying the mass of CO2 and H2O produced by the combustion reaction with the respective fraction of carbon or hydrogen mass in the molecules' molar mass.

Practical Application

In the given exercise, the amount of carbon and hydrogen in dimethylhydrazine is derived from the mass of CO2 and H2O formed. Nitrogen content, not accounted for in these products directly, is found by the difference in mass before and after combustion. This process highlights how combustion analysis directly enables the determination of a compound's empirical formula.

Molar Mass Calculation

The molar mass of a compound is the mass of one mole of its molecules, typically expressed in grams per mole (g/mol). Once the empirical formula is known, the molar mass can be calculated by adding the molar masses of all the atoms in the empirical formula.
To find the molecular formula, you must compare the molar mass of the empirical formula to the known molar mass of the compound. This tells you how many empirical formula units make up the actual molecule. For instance, if the empirical formula's molar mass is half of the compound's molar mass, you multiply the subscripts in the empirical formula by two to get the molecular formula.

Relevance in Molecular Formula

In our example, the empirical formula CH4N has a molar mass of 30.05 g/mol, whereas dimethylhydrazine has a molar mass of 60.10 g/mol. Since 60.10 g/mol is twice the molar mass of the empirical formula, it indicates that the molecular formula of dimethylhydrazine is C2H8N2, affirming that there are two CH4N units per molecule of dimethylhydrazine. This step is crucial for transforming the simplicity of the empirical formula into the specificity of the molecular formula, which represents the actual number of atoms in a molecule.