Problem 30 - Chemistry Matter and Change

Question

Plants and animals depend on glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) as an energy source. Calculate the number of moles of each element in 1.25 \(\mathrm{mol} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) .

Step-by-Step Solution

Answer

In 1.25 moles of glucose (C6H12O6), there are 7.5 moles of Carbon atoms (C), 15 moles of Hydrogen atoms (H), and 7.5 moles of Oxygen atoms (O).

Step 1: Identify the elements and their stoichiometry in glucose

In a glucose molecule (C6H12O6), there are 6 Carbon (C) atoms, 12 Hydrogen (H) atoms, and 6 Oxygen (O) atoms.

Step 2: Calculate the moles of Carbon atoms

Since there are 6 Carbon atoms in a glucose molecule, the number of moles of Carbon atoms in 1.25 moles of glucose can be found by multiplying 1.25 moles of glucose by 6: C moles = 1.25 moles of glucose * 6 C atoms/glucose = 7.5 moles of C.

Step 3: Calculate the moles of Hydrogen atoms

There are 12 Hydrogen atoms in a glucose molecule. To find the number of moles of Hydrogen atoms in 1.25 moles of glucose, multiply 1.25 moles of glucose by 12: H moles = 1.25 moles of glucose * 12 H atoms/glucose = 15 moles of H.

Step 4: Calculate the moles of Oxygen atoms

There are 6 Oxygen atoms in a glucose molecule. To find the number of moles of Oxygen atoms in 1.25 moles of glucose, multiply 1.25 moles of glucose by 6: O moles = 1.25 moles of glucose * 6 O atoms/glucose = 7.5 moles of O.

Step 5: Summarize the results

In 1.25 moles of glucose (C6H12O6), there are: - 7.5 moles of Carbon atoms (C) - 15 moles of Hydrogen atoms (H) - 7.5 moles of Oxygen atoms (O)

Key Concepts

Moles Calculation

Moles are a fundamental concept in chemistry that helps us quantify the amount of a substance. A mole is a unit that measures the number of particles, such as atoms or molecules. Essentially, one mole equals Avogadro's number, approximately \(6.022 \times 10^{23}\) particles.
When calculating moles from a compound, it can become crucial when dealing with a balanced chemical equation. By understanding how many moles of each element are in a compound, like glucose, it helps us understand more about the chemical reactions involved and their proportions.
For example, if you have 1.25 moles of glucose, and you want to determine how many moles of a specific atom, such as Carbon, are present, you'll multiply the number of moles of glucose by the number of that atom present in a glucose molecule. As seen in the original exercise, this would be \(1.25 \times 6 = 7.5\) moles of Carbon atoms since there are 6 Carbon atoms in a glucose molecule.

Glucose Molecule

Glucose is a simple sugar that plays a vital role in biology, as it is a primary source of energy for cells in both plants and animals. When we talk about a glucose molecule, we refer to its chemical formula \( ext{C}_6 ext{H}_{12} ext{O}_6\).
Each glucose molecule contains six Carbon atoms, twelve Hydrogen atoms, and six Oxygen atoms. This specific arrangement of atoms creates a highly important carbohydrate that participates in multiple biological processes. The structure helps in the extraction of energy during cellular respiration.
In calculating the number of moles of each atom in a specific amount of glucose, like 1.25 moles, you multiply the mole amount by the number of each type of atom. Therefore, for glucose, you will have 7.5 moles of Carbon atoms, 15 moles of Hydrogen atoms, and 7.5 moles of Oxygen atoms, as showcased in the original step-by-step solution.

Chemical Formula of Glucose

The chemical formula of glucose, \( ext{C}_6 ext{H}_{12} ext{O}_6\), tells us a great deal about its composition. The formula represents the types and quantities of atoms present in one molecule of glucose.
- **C** indicates Carbon, and the subscript **6** signifies that there are six Carbon atoms.- **H** stands for Hydrogen, and the subscript **12** means there are twelve Hydrogen atoms.- **O** represents Oxygen, and the subscript **6** shows that six Oxygen atoms are present.
Understanding a chemical formula is fundamental to chemistry because it provides insight into how molecules can react and what amounts are needed in chemical reactions. By recognizing the stoichiometry of glucose, one can perform calculations such as mole ratios which are key to predicting the outcomes of chemical reactions. This forms the basis on which we proceeded to calculate moles of individual atoms in the given glucose amount in the solution provided.