The molar mass and molecular weight of Manganese(VII) selenide (Mn2Se7) is 662.5960 g/mol.
The composition of the Mn2Se7 formula is as follows:
| Element | Symbol | Atomic Weight | Atoms | Total Atomic Weight | Mass Percent |
|---|---|---|---|---|---|
| Selenium | Se | 78.96 g/mol | 7 | 552.7200 g/mol | 83.4173% |
| Manganese | Mn | 54.938 g/mol | 2 | 109.8760 g/mol | 16.5827% |
How to find the molar mass of Mn2Se7?
The molar mass and molecular weight of Mn2Se7 can be calculated in 4 steps.
Step I: Identify the different elemental atoms present in the Mn2Se7 compound.
The given compound is Mn2Se7. It comprises atoms from 2 different elements i.e., Selenium (Se) and Manganese (Mn).
Step II: Find the atomic weight of each element in the Mn2Se7.
Here is the list of atomic weights for all elements, let’s check the Selenium (Se) and Manganese (Mn) atom’s atomic weight.
• The atomic weight of Selenium (Se) is 78.96 g/mol.
• The atomic weight of Manganese (Mn) is 54.938 g/mol.
Step III: Determine the number of atoms of each element present in the Mn2Se7 compound.
As per the chemical formula, Mn2Se7, It is made up of 7 Selenium atoms and 2 Manganese atoms.
| Element | Number of Atoms |
|---|---|
| Se (Selenium) | 7 |
| Mn (Manganese) | 2 |
Step IV: Calculate the molar mass of the Mn2Se7 compound by applying the formula:
For Mn2Se7:
Substituting into the above formula, the values determined in steps II and III:
Molar mass of Mn2Se7 = [(78.96 x 7) + (54.938 x 2)]
Molar mass of Mn2Se7 = 552.719 + 109.876 = 662.5960 g/mol
Result: The molecular weight and molar mass of Manganese(VII) selenide (Mn2Se7) is 662.5960 g/mol.
FAQs
What is the mass percent composition of Selenium (Se) in Mn2Se7? |
To find the mass percent of Selenium in Mn2Se7, follow the steps given below:
∴ Mass of Selenium in Mn2Se7 = 78.96 x 7 = 552.7200 g/mol. Mass Percent Composition (%) Formula = (Mass of Element in the Compound/Molar Mass of the Compound) x 100 ∴ Mass percent of Selenium in Mn2Se7 = (Mass of Selenium in Mn2Se7/Molar mass of Mn2Se7) × 100% Result: Mn2Se7 contains 83.42 % of Selenium as per its chemical composition.
|
What is the mass percent composition of Manganese (Mn) in Mn2Se7? |
To find the mass percent of Manganese in Mn2Se7, follow the steps given below:
∴ Mass of Manganese in Mn2Se7 = 54.938 x 2 = 109.8760 g/mol. Mass Percent Composition (%) Formula = (Mass of Element in the Compound/Molar Mass of the Compound) x 100 ∴ Mass percent of Manganese in Mn2Se7 = (Mass of Manganese in Mn2Se7/Molar mass of Mn2Se7) × 100% Result: Mn2Se7 contains 16.58 % of Manganese as per its chemical composition.
|
What is the atomic percentage composition of elements in Mn2Se7? |
| Atomic percentage composition formula (%) = Number of atoms of a particular element in a compound/Total number of atoms in that compound
For Mn2Se7: The atomic percentage composition of elements in the compound is: Atomic percentage of Selenium (Se) in Mn2Se7: Atomic percentage of Manganese (Mn) in Mn2Se7:
|
How many grams of Selenium are present in 1 mole of Mn2Se7? |
| There are 7 Se-atoms in a Mn2Se7 molecule. • Atomic weight of a Se-atom = 78.96 g/mol ∴ Total mass of Selenium in 1 mole of Mn2Se7 = 7 x 78.96 = 552.72 g. Therefore there are 552.72 grams of Selenium in 1 mole of Mn2Se7. |
What is the mass of 9 mol of Mn2Se7? |
| The molecular mass of Mn2Se7 is 662.5960 g/mol. This means 662.5960 grams of Mn2Se7 are present per mole. Therefore, we can find the mass of Mn2Se7 in 9 moles as follows: ∴ Moles = Mass/Molar Mass ∴ Mass of Mn2Se7 = Moles x Molar mass ∴ Mass of Mn2Se7 = 9 x 662.5960 = 5963.36 g Thus, the mass of 9 mol of Mn2Se7 is 5963.36 g. |
If you have 211.4792 grams of Mn2Se7, how many moles do you have? |
| ∴ Moles = Mass/Molar mass ∴ Moles of Mn2Se7 = 211.4792/662.5960 = 0.3192 Thus, there are 0.3192 moles of Mn2Se7 in its 211.4792 grams. |
What is the mass in kg of 110.43 x 1025 molecules of Mn2Se7? |
| 1 mole of a substance contains Avogadro number of particles i.e., 6.02 x 1023. Therefore, the number of moles in 110.43 x 1025 molecules of Mn2Se7 are: ∴ Moles of Mn2Se7 = 110.43 x 1025/6.02 x 1023 = 1834.4
Now that we have its number of moles, so, we can use the molar mass of Mn2Se7 (662.5960 g/mol) to find its mass as shown below. ∴ Mass of Mn2Se7 = Moles x Molar mass
∴ Mass of Mn2Se7 = 1834.4 x 662.5960 = 1215485.77 g
Converting mass from grams (g) to kilograms (kg) gives us: ∴ Mass of Mn2Se7 = 1215485.77/1000 = 1215.49 kg
The mass of 110.43 x 1025 molecules of Mn2Se7 is 1215.49 kg.
|
What is the molar mass and molecular weight of Manganese(VII) selenide (Mn2Se7)? |
Mn2Se7 is composed of 7 Selenium (Se), and 2 Manganese (Mn) atoms.
To find the molecular mass of Mn2Se7, one multiplies the atomic weight of each element by its number of atoms in the molecule and then sums the results. ∴ For Mn2Se7, it’s (7 x 78.9600) + (2 x 54.9380).
Therefore, the molar mass of Mn2Se7 is 662.5960 g/mol.
|
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/