Molar Mass of Bromoheptane (C7H15Br)
Molar Mass of Bromoheptane is often referenced when connecting molecular structure to quantitative conversion steps. Molar Mass of C₇H₁₅Br is 179.10 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 Bromoheptane is:
179.10 g/mol
Molar Mass of C₇H₁₅Br equals 179.10 g/mol, so 179.10 grams is one mole.
Element Breakdown Table
| Element | Count | Atomic mass | Calculation | Contribution |
|---|---|---|---|---|
| Carbon (C) | 7 | 12.01 | 7 x 12.01 | 84.08 g/mol |
| Hydrogen (H) | 15 | 1.01 | 15 x 1.01 | 15.12 g/mol |
| Bromine (Br) | 1 | 79.90 | 1 x 79.90 | 79.90 g/mol |
Final molar mass 84.080 + 15.120 + 79.900 | 179.10 g/mol | |||
Computing Molar Mass of Bromoheptane Step by Step
Molar Mass of Bromoheptane: Step-by-Step Calculation
1. Identify Element Counts
Read C7H15Br and list how many atoms of each element are present:
- 7 atoms of Carbon (C)
- 15 atoms of Hydrogen (H)
- 1 atom of Bromine (Br)
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
- Bromine (Br) ~= 79.904 g/mol
3. Multiply Atomic Mass by Quantity
Multiply atom count by atomic mass for each element:
- Carbon (C): 7 x 12.011 = 84.080 g/mol
- Hydrogen (H): 15 x 1.008 = 15.120 g/mol
- Bromine (Br): 1 x 79.904 = 79.900 g/mol
4. Sum Total Molar Mass
Add all contributions to get the final molar mass in g/mol.
Molar Mass = (7 x 12.011 + 15 x 1.008 + 1 x 79.904)
Molar Mass = 84.080 + 15.120 + 79.900
Molar Mass = 179.100 g/mol
Final rounded value shown on this page: 179.10 g/mol.
Visual Calculation Chart
| Element | Count | Mass | Count x mass | Contribution |
|---|---|---|---|---|
| Carbon (C) | 7 | 12.011 | 7 x 12.011 | = 84.080 |
| Hydrogen (H) | 15 | 1.008 | 15 x 1.008 | = 15.120 |
| Bromine (Br) | 1 | 79.904 | 1 x 79.904 | = 79.900 |
Final molar mass 84.080 + 15.120 + 79.900 | = 179.100 | |||
Easy Way to Remember
Easy way to remember Molar Mass of Bromoheptane
- Molar Mass of Bromoheptane: 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 |
|---|---|
| Synthesis | ₂H₂ + O₂ → ₂H₂O |
| Combustion | CH₄ + ₂O₂ → CO₂ + ₂H₂O |
Do You Know?
– Bromoheptane contains 3 element types: C, H, Br.
– C contributes the largest share of this compound's total molar mass.
– In C₇H₁₅Br, H appears with the highest atom count.
– Its molar mass is 179.10 g/mol, which is used directly in gram-to-mole conversions.
– A common reaction for Bromoheptane is synthesis (₂H₂ + O₂ → ₂H₂O).
Why This Compound Matters
Bromoheptane 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 Bromomethane (C1H3Br) and Bromoethane (C2H5Br).
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 C₇H₁₅Br 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 Bromoheptane.
Quick Revision
Formula: C7H15Br
Molar Mass: 179.10 g/mol
Key takeaway: count atoms accurately, multiply by atomic masses, and sum only at the end.
Formula Explanation
C7H15Br contains Carbon (C) (7), Hydrogen (H) (15), Bromine (Br) (1). Add each element contribution to get total molar mass.
FAQ
Conclusion
Molar Mass of Bromoheptane and Molar Mass of C₇H₁₅Br are now easy to revise with this structured page. You can use this method in exams, lab reports, and daily chemistry practice.