Calculate the number of molecules of NBr3 in 25.0 g NBr3. Use a molar mass with at least as many significant figures as the data given.
How many moles do you have in 25.0 g NBr3? That is 25.0/molar mass = ?
How many molecules in that number of moles? Remember there are 6.02E23 molecules in a mole of anything Then multiply that by 3 since there are 3 atoms Br in each molecule of NBr3.
To calculate the number of molecules of NBr3 in 25.0 g of NBr3, you will need to follow these steps:
1. Determine the molar mass of NBr3.
- N has an atomic mass of 14.01 g/mol.
- Br has an atomic mass of 79.90 g/mol.
Therefore, the molar mass of NBr3 is:
(1 x 14.01 g/mol) + (3 x 79.90 g/mol) = 253.71 g/mol
2. Use the molar mass to convert grams of NBr3 to moles of NBr3.
Number of moles = Mass of substance (g) / Molar mass (g/mol)
Number of moles = 25.0 g / 253.71 g/mol = 0.0985 mol
3. Use Avogadro's number to convert moles of NBr3 to molecules of NBr3.
1 mole of any substance contains 6.022 × 10^23 molecules.
Number of molecules = Number of moles × (6.022 × 10^23 molecules/mol)
Number of molecules = 0.0985 mol × (6.022 × 10^23 molecules/mol) = 5.928 × 10^22 molecules
Therefore, there are approximately 5.928 × 10^22 molecules of NBr3 in 25.0 g of NBr3.
To calculate the number of molecules of NBr3 in 25.0 g of NBr3, we need to follow a few steps:
1. Determine the molar mass of NBr3:
- N: atomic mass = 14.01 g/mol
- Br: atomic mass = 79.90 g/mol
- NBr3 has one nitrogen atom and three bromine atoms, so the molar mass of NBr3 is:
(1 * N) + (3 * Br) = (1 * 14.01 g/mol) + (3 * 79.90 g/mol) = 252.71 g/mol
2. Calculate the number of moles of NBr3 using the given mass and molar mass:
- Moles = Mass / Molar mass
- Moles = 25.0 g / 252.71 g/mol
- Moles ≈ 0.099 mol (rounded to three significant figures)
3. Finally, calculate the number of molecules using Avogadro's number:
- Avogadro's number (NA) = 6.022 × 10^23 molecules/mol
- Number of molecules = Moles * NA
= 0.099 mol * (6.022 × 10^23 molecules/mol)
≈ 5.95 × 10^22 molecules
Therefore, there are approximately 5.95 × 10^22 molecules of NBr3 in 25.0 g NBr3.