Problem 101 - Chemistry: The Central Science

Question

The fat stored in a camel's hump is a source of both energy and water. Calculate the mass of \(\mathrm{H}_{2} \mathrm{O}\) produced by the metabolism of \(1.0 \mathrm{~kg}\) of fat, assuming the fat consists entirely of tristearin \(\left(\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}\right)\), a typical animal fat, and assuming that during metabolism, tristearin reacts with \(\mathrm{O}_{2}\) to form only \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\).

Step-by-Step Solution

Answer

The mass of water produced by the metabolism of 1.0 kg of tristearin is approximately 1.162 kg.

Step 1: Write the Chemical Equation for Tristearin Metabolism

Write a balanced chemical equation for the reaction of tristearin (C57H110O6) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). The equation is: C57H110O6 + O2 -> CO2 + H2O We need to balance this equation. C57H110O6 + 79O2 -> 57CO2 + 55H2O Now, the equation is balanced.

Step 2: Calculate the Moles of Tristearin

To calculate the moles of tristearin, first, find its molar mass: \(M(C_{57}H_{110}O_{6}) = 57(12.01) + 110(1.01) + 6(16.00) = 852.12\ \textrm{g/mol}\) Given that there is 1.0 kg (1000 g) of tristearin, convert the mass to moles: \(\textrm{moles of tristearin} = \frac{1000\ \textrm{g}}{852.12\ \textrm{g/mol}} = 1.1732\ \textrm{mol}\)

Step 3: Apply Stoichiometry to Find Moles of H2O Produced

Use the balanced chemical equation to find the ratio between tristearin and water to determine the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) produced. From the balanced equation, C57H110O6 + 79O2 -> 57CO2 + 55H2O 1 mol of C57H110O6 produces 55 mol of H2O. Then, the moles of \(\mathrm{H}_{2} \mathrm{O}\) produced are: \(1.1732\ \textrm{mol tristearin} \times \frac{55\ \textrm{mol}\ H_{2}O}{1\ \textrm{mol tristearin}} = 64.526\ \textrm{mol}\ H_{2}O\)

Step 4: Determine the Mass of Water Formed

Now, convert moles of water to mass in grams: \(M(H_{2}O) = 2(1.01) + 1(16.00) = 18.02\ \textrm{g/mol}\) Total mass of \(\mathrm{H}_{2} \mathrm{O}\) produced = \(64.526\ \textrm{mol}\ H_{2}O \times 18.02\ \textrm{g/mol} = 1162.28\ \textrm{g}\) Therefore, the mass of water produced by the metabolism of 1.0 kg of tristearin is 1162.28 g, or approximately 1.162 kg.

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is essential in stoichiometry because it illustrates the reaction's context by showing the reactants and products involved in a chemical reaction. This balance is crucial to ensure the law of conservation of mass is satisfied — meaning the number of atoms for each element in the reactants equals the number in the products.

• In our exercise, we start with tristearin o(C_{57}H_{110}O_6) o. Combining it with oxygen o(O_2) o, we get carbon dioxide o(CO_2) o and water o(H_2O) o.
• Balanced equations are written by adjusting coefficients before the chemical formulas, which also determines the accurate mole ratio of reactants to products.
• The balanced equation for the combustion of tristearin is: oC_{57}H_{110}O_6 + 79O_2 -> 57CO_2 + 55H_2O o.

Understanding how to balance an equation helps in determining the precise amounts of reactants needed and products formed, which is a fundamental part of stoichiometry.

Moles Calculation

Calculating the number of moles is a critical step in chemical computations. It involves using the known mass of a substance and its molar mass to determine how much of the substance is used or produced in a reaction.
For our specific problem, we initially need to find out the molar mass of tristearin by considering the elements involved.

• The molar mass of oC_{57}H_{110}O_6o is calculated by summing the atomic masses of all the atoms present:
• Carbon (C): 57 atoms × 12.01 g/mol = 684.57 g/mol
• Hydrogen (H): 110 atoms × 1.01 g/mol = 111.1 g/mol
• Oxygen (O): 6 atoms × 16.00 g/mol = 96.00 g/mol
• Adding these gives the total molar mass: 852.12 g/mol.

Given a mass of 1.0 kg (1000 g) for tristearin, the number of moles is easily calculated by dividing the mass by the molar mass: o\( \frac{1000 \text{ g}}{852.12 \text{ g/mol}} = 1.1732 \text{ mol} \)o. Calculating moles enables us to apply stoichiometry directly to predict product formation, vital for further calculations in the process.

Molar Mass

The concept of molar mass links the mass of a substance to its moles, serving as a conversion factor in stoichiometric calculations. Molar mass is expressed in grams per mole, indicating how many grams one mole of that substance weighs.

• For molecules like oH_2O o (water), the process involves summing the masses of the constituent atoms.
• The molar mass is determined as follows: Hydrogen (H): 2 atoms × 1.01 g/mol = 2.02 g/mol and Oxygen (O): 1 atom × 16.00 g/mol = 16.00 g/mol.
• Thus, the total molar mass oM(H_2O) o equals 18.02 g/mol.

With molar mass, you convert between the mass of a substance and the moles of that substance, which is essential for stoichiometric calculations and understanding chemical reactions better.

Chemical Reactions

Chemical reactions are at the heart of stoichiometry. They represent the transformation of reactants into products through chemical changes. Understanding the nature of these reactions is crucial in comprehending the processes involved.

• The metabolism of tristearin in camels, as detailed in our exercise, involves its reaction with oO_2 o to form oCO_2 o and oH_2O o.
• This is an example of a combustion reaction — where a compound reacts with oxygen to produce water and carbon dioxide — a type of exothermic process releasing energy.
• Recognizing the type of reaction, like combustion, helps predict the products formed and calculate energy changes.

Appreciating chemical reactions helps us understand many natural and industrial processes, giving a deeper insight into how substances interact and change.