Problem 101 - Chemistry The Central Science In Si Units

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 metabolism of 1.0 kg of tristearin produces about 1.111 kg of water.

Step 1: Write the Balanced Chemical Equation

The chemical reaction for the metabolism of tristearin (\(\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6\)) with oxygen is written as:\[\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6 + 81\mathrm{O}_2 \rightarrow 57 \mathrm{CO}_2 + 55 \mathrm{H}_2\mathrm{O}\]This equation is balanced for atoms of carbon, hydrogen, and oxygen.

Step 2: Determine the Molar Mass of Tristearin

Calculate the molar mass of tristearin, \(\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6\), by adding the atomic masses of all the atoms in the formula.\[\text{C: } 57 \times 12.01 = 684.57 \text{ g/mol}\]\[\text{H: } 110 \times 1.008 = 110.88 \text{ g/mol}\]\[\text{O: } 6 \times 16.00 = 96.00 \text{ g/mol}\]\[\text{Total molar mass = 891.45 \text{ g/mol}}\]

Step 3: Calculate the Moles of Tristearin

Using the molar mass calculated above, convert 1.0 kg of tristearin into moles: \[\text{Moles of } \mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6 = \frac{1000 \text{ g}}{891.45 \text{ g/mol}} \approx 1.122 \text{ mol}\]

Step 4: Use Stoichiometry to Find Moles of Water Produced

From the balanced equation, 1 mole of \(\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6\) produces 55 moles of \(\mathrm{H}_2\mathrm{O}\). Thus, the moles of \(\mathrm{H}_2\mathrm{O}\) are: \[1.122 \text{ mol} \times 55 \text{ moles } \mathrm{H}_2 \mathrm{O}/ \text{mole } \mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6 = 61.71 \text{ mol } \mathrm{H}_2 \mathrm{O}\]

Step 5: Calculate the Mass of Water Produced

Convert the moles of water to grams using the molar mass of water (18.02 g/mol): \[\text{Mass of } \mathrm{H}_2 \mathrm{O} = 61.71 \text{ mol} \times 18.02 \text{ g/mol} \approx 1111.44 \text{ g}\]Hence, the weight of water produced is approximately 1111 grams or 1.111 kg.

Key Concepts

Balanced Chemical Equation

Creating a balanced chemical equation is crucial for understanding how different substances interact in a reaction. In the example of the metabolism of tristearin, the reaction is described by the equation:\[\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6 + 81\mathrm{O}_2 \rightarrow 57 \mathrm{CO}_2 + 55 \mathrm{H}_2\mathrm{O}\]This equation shows the transformation of tristearin and oxygen into carbon dioxide and water. Each element on the reactant side has the same number of atoms as on the product side. This is essential to satisfy the law of conservation of mass, indicating that matter cannot be created or destroyed.

• Carbon atoms: 57 on both sides
• Hydrogen atoms: 110 on both sides
• Oxygen atoms: 183 on both sides of the reaction

A balanced equation is achieved by adjusting the coefficients in front of each compound, aligning with stoichiometry principles. This provides a solid foundation to proceed with calculating product quantities.

Molar Mass Calculation

The molar mass is a useful tool for converting grams of a substance to moles, which is critical for comparing reactant and product quantities in a chemical reaction. For tristearin, we compute its molar mass by summing the mass contributions of each element:

• Carbon (C): 57 atoms, each \(12.01 \, \text{g/mol}\)
• Hydrogen (H): 110 atoms, each \(1.008 \, \text{g/mol}\)
• Oxygen (O): 6 atoms, each \(16.00 \, \text{g/mol}\)

Calculating gives a total molar mass of 891.45 g/mol. This value tells us how much one mole of tristearin weighs, and is necessary for converting the 1 kg of tristearin into moles. Understanding this conversion is imperative for moving ahead with stoichiometric calculations in reactions.

Stoichiometric Coefficients

Stoichiometric coefficients are the factors that allow us to translate moles of reactants to moles of products in a chemical reaction. These coefficients are assigned to the balanced chemical equation:In the example equation:\[1 \text{ mole of } \mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_6\] combines with \[81 \text{ moles of } \mathrm{O}_2\] to yield \[57 \text{ moles of } \mathrm{CO}_2\] and \[55 \text{ moles of } \mathrm{H}_2\mathrm{O}\].The coefficients (1, 81, 57, 55) indicate the proportion of molecules involved. In practical applications, this means if you start with 1.122 moles of tristearin, it will produce \[1.122 \times 55 = 61.71\] moles of water. These coefficients grant us the ability to predict the amounts of products formed, allowing chemists to prepare for reactions on a larger scale.

Energy Metabolism of Fats

In biology, the metabolism of fats such as tristearin involves breaking down these molecules for energy production and water generation. Tristearin, a common animal fat, converts into carbon dioxide and water when reacted with oxygen via a process known as beta-oxidation.

• Fats contain a high number of hydrogen chains, ideal for producing energy.
• They also produce water as a byproduct, critical for organisms in arid environments like camels.

In camels, this water production is a survival adaptation, as their fat reserves yield the water needed when liquid water is scarce. The metabolism of fats, therefore, supports energy production and water synthesis, with calculations showing about 1.111 kg of water produced from 1 kg of tristearin metabolism. This showcases the dual role of fats in providing both caloric energy and hydration.