Molar Mass of Decanol (C10H22O)
Molar Mass of Decanol is often referenced when connecting molecular structure to quantitative conversion steps. Molar Mass of C₁₀H₂₂O is 158.28 g/mol, based on 3 element types, with C contributing the largest share.
For fast checks, use the calculate molar mass, verify element values in the periodic table with molar mass, or explore more molar mass page.
Molar Mass of Decanol is:
158.28 g/mol
Molar Mass of C₁₀H₂₂O equals 158.28 g/mol, so 158.28 grams is one mole.
Element Breakdown Table
| Element | Count | Atomic mass | Calculation | Contribution |
|---|---|---|---|---|
| Carbon (C) | 10 | 12.01 | 10 x 12.01 | 120.11 g/mol |
| Hydrogen (H) | 22 | 1.01 | 22 x 1.01 | 22.18 g/mol |
| Oxygen (O) | 1 | 16.00 | 1 x 16.00 | 16.00 g/mol |
Final molar mass 120.110 + 22.180 + 16.000 | 158.28 g/mol | |||
Computing Molar Mass of Decanol Step by Step
Molar Mass of Decanol: Step-by-Step Calculation
1. Identify Element Counts
Read C10H22O and list how many atoms of each element are present:
- 10 atoms of Carbon (C)
- 22 atoms of Hydrogen (H)
- 1 atom of Oxygen (O)
2. Determine Atomic Masses
Look up each element mass from the periodic table:
- Carbon (C) ~= 12.011 g/mol
- Hydrogen (H) ~= 1.008 g/mol
- Oxygen (O) ~= 15.999 g/mol
3. Multiply Atomic Mass by Quantity
Multiply atom count by atomic mass for each element:
- Carbon (C): 10 x 12.011 = 120.110 g/mol
- Hydrogen (H): 22 x 1.008 = 22.180 g/mol
- Oxygen (O): 1 x 15.999 = 16.000 g/mol
4. Sum Total Molar Mass
Add all contributions to get the final molar mass in g/mol.
Molar Mass = (10 x 12.011 + 22 x 1.008 + 1 x 15.999)
Molar Mass = 120.110 + 22.180 + 16.000
Molar Mass = 158.280 g/mol
Final rounded value shown on this page: 158.28 g/mol.
Visual Calculation Chart
| Element | Count | Mass | Count x mass | Contribution |
|---|---|---|---|---|
| Carbon (C) | 10 | 12.011 | 10 x 12.011 | = 120.110 |
| Hydrogen (H) | 22 | 1.008 | 22 x 1.008 | = 22.180 |
| Oxygen (O) | 1 | 15.999 | 1 x 15.999 | = 16.000 |
Final molar mass 120.110 + 22.180 + 16.000 | = 158.280 | |||
Easy Way to Remember
Easy way to remember Molar Mass of Decanol
- Molar Mass of Decanol: count atoms, multiply masses, and add totals.
- Write each element in a table so you do not miss subscripts.
- Keep 2-3 decimals during steps, then round only at the end.
Sample Reactions
| Type | Reaction |
|---|---|
| Combustion | C₁₀H₂₂O + ₁₅O₂ → ₁₀CO₂ + ₁₁H₂O |
| Sodium reaction | ₂C₁₀H₂₂O + ₂Na → ₂C₁₀H₂₁ONa + H₂ |
Do You Know?
– Decanol contains 3 element types: C, H, O.
– C contributes the largest share of this compound's total molar mass.
– In C₁₀H₂₂O, H appears with the highest atom count.
– Its molar mass is 158.28 g/mol, which is used directly in gram-to-mole conversions.
– A common reaction for Decanol is combustion (C₁₀H₂₂O + ₁₅O₂ → ₁₀CO₂ + ₁₁H₂O).
Why This Compound Matters
Decanol matters in organic chemistry, especially for fuel, solvent, or carbon-chain analysis.
Its formula pattern helps students practice molecular composition and yield calculation methods.
Similar calculations can be compared with Methanol (CH3OH) and Ethanol (C2H6O).
Where This Is Used
- Competitive exams and school chemistry tests.
- Lab work when preparing measured solutions.
- Real-world manufacturing and quality checks.
Whenever you need lab-ready totals in those settings, molar mass stays handy for cross-checking.
Common Mistakes When Calculating This
- Skipping subscripts in C₁₀H₂₂O and miscounting atoms.
- Rounding atomic masses too early before finishing all multiplication steps.
- Mixing up C element contribution with total molar mass.
- Reporting a value without units; final answer should be in g/mol for Decanol.
Quick Revision
Formula: C10H22O
Molar Mass: 158.28 g/mol
Key takeaway: count atoms accurately, multiply by atomic masses, and sum only at the end.
Formula Explanation
C10H22O contains Carbon (C) (10), Hydrogen (H) (22), Oxygen (O) (1). Add each element contribution to get total molar mass.
FAQ
Conclusion
Molar Mass of Decanol and Molar Mass of C₁₀H₂₂O are now easy to revise with this structured page. You can use this method in exams, lab reports, and daily chemistry practice.