How many hydrogen atoms are contained in 1.50 g of glucose (C 6H 12 06 )?
6.02 × 10 22
Calculating Hydrogen Atoms in Glucose: A Step-by-Step Guide
This question asks us to find the total number of hydrogen atoms present in a specific mass of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)). To solve this, we need to use the concepts of molar mass, moles, Avogadro's number, and the chemical formula of glucose.
Understanding Glucose and Molar Mass
Glucose has the chemical formula \(\text{C}_6\text{H}_{12}\text{O}_6\). This formula tells us that each molecule of glucose contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
The molar mass of a compound is the mass of one mole of that compound. We can calculate the molar mass of glucose by adding the atomic masses of all the atoms in its formula. Using approximate atomic masses:
- Carbon (C): approximately 12.0 g/mol
- Hydrogen (H): approximately 1.0 g/mol
- Oxygen (O): approximately 16.0 g/mol
Molar mass of \(\text{C}_6\text{H}_{12}\text{O}_6\):
\(\text{Molar Mass} = (6 \times \text{Molar Mass of C}) + (12 \times \text{Molar Mass of H}) + (6 \times \text{Molar Mass of O})\)
\(\text{Molar Mass} = (6 \times 12.0 \, \text{g/mol}) + (12 \times 1.0 \, \text{g/mol}) + (6 \times 16.0 \, \text{g/mol})\)
\(\text{Molar Mass} = 72.0 \, \text{g/mol} + 12.0 \, \text{g/mol} + 96.0 \, \text{g/mol}\)
\(\text{Molar Mass} = 180.0 \, \text{g/mol}\)
Calculating Moles of Glucose
We are given a mass of 1.50 g of glucose. We can convert this mass to moles using the molar mass we just calculated:
\(\text{Moles of Glucose} = \frac{\text{Mass of Glucose}}{\text{Molar Mass of Glucose}}\)
\(\text{Moles of Glucose} = \frac{1.50 \, \text{g}}{180.0 \, \text{g/mol}}\)
\(\text{Moles of Glucose} \approx 0.008333 \, \text{mol}\)
Or, as a fraction: \(\text{Moles of Glucose} = \frac{1.5}{180} = \frac{15}{1800} = \frac{1}{120} \, \text{mol}\)
Calculating Molecules of Glucose
One mole of any substance contains Avogadro's number of particles, which is approximately \(6.02 \times 10^{23}\) particles per mole. In this case, the particles are glucose molecules.
\(\text{Number of Glucose Molecules} = \text{Moles of Glucose} \times \text{Avogadro's Number}\)
\(\text{Number of Glucose Molecules} = \frac{1}{120} \, \text{mol} \times 6.02 \times 10^{23} \, \text{molecules/mol}\)
\(\text{Number of Glucose Molecules} = \frac{6.02 \times 10^{23}}{120} \, \text{molecules}\)
Calculating Total Hydrogen Atoms
From the formula \(\text{C}_6\text{H}_{12}\text{O}_6\), we know that each glucose molecule contains 12 hydrogen atoms. To find the total number of hydrogen atoms, we multiply the number of glucose molecules by 12.
\(\text{Total Hydrogen Atoms} = \text{Number of Glucose Molecules} \times \text{Number of H atoms per molecule}\)
\(\text{Total Hydrogen Atoms} = \left(\frac{6.02 \times 10^{23}}{120}\right) \times 12\)
\(\text{Total Hydrogen Atoms} = \frac{6.02 \times 10^{23} \times 12}{120}\)
\(\text{Total Hydrogen Atoms} = \frac{6.02 \times 10^{23}}{10}\)
\(\text{Total Hydrogen Atoms} = 0.602 \times 10^{23}\)
\(\text{Total Hydrogen Atoms} = 6.02 \times 10^{22}\)
Therefore, 1.50 g of glucose contains approximately \(6.02 \times 10^{22}\) hydrogen atoms.
| Step | Calculation | Result |
|---|---|---|
| Molar Mass of Glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) | \(6(12.0) + 12(1.0) + 6(16.0)\) g/mol | 180.0 g/mol |
| Moles of Glucose in 1.50 g | \(\frac{1.50 \, \text{g}}{180.0 \, \text{g/mol}}\) | \(\frac{1}{120}\) mol |
| Number of Glucose Molecules | \(\frac{1}{120} \, \text{mol} \times 6.02 \times 10^{23}\) molecules/mol | \(\frac{6.02 \times 10^{23}}{120}\) molecules |
| Total Hydrogen Atoms | \(\frac{6.02 \times 10^{23}}{120} \times 12\) atoms | \(6.02 \times 10^{22}\) atoms |
Revision Table: Key Concepts for Atom Calculations
| Concept | Definition/Use | Formula |
|---|---|---|
| Molar Mass | Mass of one mole of a substance | Sum of atomic masses (in g/mol) |
| Mole | Unit for amount of substance | \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\) |
| Avogadro's Number | Number of particles in one mole | \(6.022 \times 10^{23} \, \text{particles/mol (approx } 6.02 \times 10^{23})\) |
| Chemical Formula | Shows the number of atoms of each element in a molecule | e.g., \(\text{C}_6\text{H}_{12}\text{O}_6\) means 12 H atoms per molecule |
Additional Information: Mole Concept and Glucose
The mole concept is fundamental in chemistry as it allows us to relate macroscopic quantities (like mass) to microscopic quantities (like the number of atoms or molecules). One mole of any substance contains the same number of particles, making it a convenient unit for calculations involving chemical reactions.
Glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) is a simple sugar, an important carbohydrate, and a primary source of energy for living organisms. Calculations like the one performed here are essential in various fields, including biochemistry, nutrition, and chemical synthesis.
To accurately determine the number of specific atoms within a compound, always start by:
- Finding the molar mass of the compound.
- Converting the given mass to moles using the molar mass.
- Using Avogadro's number to find the total number of molecules.
- Using the chemical formula to find the number of the specific atom per molecule.
- Multiplying the number of molecules by the number of atoms per molecule to get the total number of atoms.
This systematic approach ensures correct calculations in stoichiometry problems.
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