The molar mass and molecular weight of Ethanol (C2H5OH) is 46.0692 g/mol.
The composition of the C2H5OH formula is as follows:
| Element | Symbol | Atomic Weight | Atoms | Total Atomic Weight | Mass Percent |
|---|---|---|---|---|---|
| Hydrogen | H | 1.00797 g/mol | 6 | 6.0478 g/mol | 13.1277% |
| Oxygen | O | 15.9994 g/mol | 1 | 15.9994 g/mol | 34.7290% |
| Carbon | C | 12.011 g/mol | 2 | 24.0220 g/mol | 52.1433% |
How to find the molar mass of Ethanol (C2H5OH)?
The molar mass and molecular weight of Ethanol (C2H5OH) can be calculated in 4 steps.
Step I: Identify the different elemental atoms present in the C2H5OH compound.
The given compound is C2H5OH. It comprises atoms from 3 different elements i.e., Hydrogen (H), Oxygen (O) and Carbon (C).
Step II: Find the atomic weight of each element in the C2H5OH.
Here is the list of atomic weights for all elements, let’s check the Hydrogen (H), Oxygen (O) and Carbon (C) atom’s atomic weight.
• The atomic weight of Hydrogen (H) is 1.00797 g/mol.
• The atomic weight of Oxygen (O) is 15.9994 g/mol.
• The atomic weight of Carbon (C) is 12.011 g/mol.
Step III: Determine the number of atoms of each element present in the C2H5OH compound.
As per the chemical formula, C2H5OH, It is made up of 6 Hydrogen atoms, 1 Oxygen atom and 2 Carbon atoms.
| Element | Number of Atoms |
|---|---|
| H (Hydrogen) | 6 |
| O (Oxygen) | 1 |
| C (Carbon) | 2 |
Step IV: Calculate the molar mass of the Ethanol (C2H5OH) compound by applying the formula:
For Ethanol (C2H5OH):
Substituting into the above formula, the values determined in steps II and III:
Molar mass of C2H5OH = [(1.00797 x 6) + (15.9994 x 1) + (12.011 x 2)]
Molar mass of C2H5OH = 6.04782 + 15.9994 + 24.022 = 46.0692 g/mol
Result: The molecular weight and molar mass of Ethanol (C2H5OH) is 46.0692 g/mol.
FAQs
What is the mass percent composition of Hydrogen (H) in Ethanol (C2H5OH)? |
To find the mass percent of Hydrogen in C2H5OH, follow the steps given below:
∴ Mass of Hydrogen in C2H5OH = 1.00797 x 6 = 6.0478 g/mol. Mass Percent Composition (%) Formula = (Mass of Element in the Compound/Molar Mass of the Compound) x 100 ∴ Mass percent of Hydrogen in C2H5OH = (Mass of Hydrogen in C2H5OH/Molar mass of C2H5OH) × 100% Result: C2H5OH contains 13.13 % of Hydrogen as per its chemical composition.
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What is the mass percent composition of Oxygen (O) in Ethanol (C2H5OH)? |
To find the mass percent of Oxygen in C2H5OH, follow the steps given below:
∴ Mass of Oxygen in C2H5OH = 15.9994 x 1 = 15.9994 g/mol. Mass Percent Composition (%) Formula = (Mass of Element in the Compound/Molar Mass of the Compound) x 100 ∴ Mass percent of Oxygen in C2H5OH = (Mass of Oxygen in C2H5OH/Molar mass of C2H5OH) × 100% Result: C2H5OH contains 34.73 % of Oxygen as per its chemical composition.
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What is the mass percent composition of Carbon (C) in Ethanol (C2H5OH)? |
To find the mass percent of Carbon in C2H5OH, follow the steps given below:
∴ Mass of Carbon in C2H5OH = 12.011 x 2 = 24.0220 g/mol. Mass Percent Composition (%) Formula = (Mass of Element in the Compound/Molar Mass of the Compound) x 100 ∴ Mass percent of Carbon in C2H5OH = (Mass of Carbon in C2H5OH/Molar mass of C2H5OH) × 100% Result: C2H5OH contains 52.14 % of Carbon as per its chemical composition.
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What is the atomic percentage composition of elements in Ethanol (C2H5OH)? |
| Atomic percentage composition formula (%) = Number of atoms of a particular element in a compound/Total number of atoms in that compound
For C2H5OH: The atomic percentage composition of elements in the compound is: Atomic percentage of Hydrogen (H) in C2H5OH: Atomic percentage of Oxygen (O) in C2H5OH: Atomic percentage of Carbon (C) in C2H5OH:
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How many grams of Hydrogen are present in 1 mole of C2H5OH? |
| There are 6 H-atoms in a C2H5OH molecule. • Atomic weight of a H-atom = 1.00797 g/mol ∴ Total mass of Hydrogen in 1 mole of C2H5OH = 6 x 1.00797 = 6.05 g. Therefore there are 6.05 grams of Hydrogen in 1 mole of C2H5OH. |
What is the mass of 9 mol of C2H5OH? |
| The molecular mass of C2H5OH is 46.0692 g/mol. This means 46.0692 grams of C2H5OH are present per mole. Therefore, we can find the mass of C2H5OH in 9 moles as follows: ∴ Moles = Mass/Molar Mass ∴ Mass of C2H5OH = Moles x Molar mass ∴ Mass of C2H5OH = 9 x 46.0692 = 414.62 g Thus, the mass of 9 mol of C2H5OH is 414.62 g. |
If you have 10.2218 grams of C2H5OH, how many moles do you have? |
| ∴ Moles = Mass/Molar mass ∴ Moles of C2H5OH = 10.2218/46.0692 = 0.2219 Thus, there are 0.2219 moles of C2H5OH in its 10.2218 grams. |
What is the mass in kg of 7.68 x 1025 molecules of C2H5OH? |
| 1 mole of a substance contains Avogadro number of particles i.e., 6.02 x 1023. Therefore, the number of moles in 7.68 x 1025 molecules of C2H5OH are: ∴ Moles of C2H5OH = 7.68 x 1025/6.02 x 1023 = 127.5
Now that we have its number of moles, so, we can use the molar mass of C2H5OH (46.0692 g/mol) to find its mass as shown below. ∴ Mass of C2H5OH = Moles x Molar mass
∴ Mass of C2H5OH = 127.5 x 46.0692 = 5875.89 g
Converting mass from grams (g) to kilograms (kg) gives us: ∴ Mass of C2H5OH = 5875.89/1000 = 5.88 kg
The mass of 7.68 x 1025 molecules of C2H5OH is 5.88 kg.
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What is the molar mass and molecular weight of Ethanol (C2H5OH)? |
C2H5OH is composed of 6 Hydrogen (H), 1 Oxygen (O), and 2 Carbon (C) atoms.
To find the molecular mass of C2H5OH, one multiplies the atomic weight of each element by its number of atoms in the molecule and then sums the results. ∴ For C2H5OH, it’s (6 x 1.0080) + (1 x 15.9994) + (2 x 12.0110).
Therefore, the molar mass of Ethanol (C2H5OH) is 46.0692 g/mol.
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About the author
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/