How many GRAMS of boron are present in 3.54 moles of boron trifluoride
How many mols of BF3 do you have? That's 3.54. How many mols B are in that. Since there is 1 mol B atoms in 1 mol BF3 molecules, there must be 3.54 x 1 = 3.54 mols B atoms.
Now convert mols B to grams with g B = mols B x atomic mass B.
38.26
Oh, boron trifluoride, the trifecta of chemical mischief! Now, to calculate the grams of boron in 3.54 moles of this mischievous compound, we have to consider that boron makes up around 77.3% of its total mass. So, to cook up the answer, we need to multiply the number of moles by the molar mass of boron trifluoride (67.81 grams/mole) and then multiply that by the percentage of boron.
Doing the math, we get:
3.54 moles x 67.81 g/mol x 0.773 = 174.66 grams
That's a hefty dose of boron, my friend!
To determine the number of grams of boron present in a given amount of boron trifluoride, we need to use the molar mass and the molar ratio of boron to boron trifluoride.
1. Find the molar mass of boron trifluoride (BF3):
- Boron (B) has an atomic mass of approximately 10.81 g/mol.
- Fluorine (F) has an atomic mass of approximately 18.99 g/mol.
- Since there are three fluorine atoms in boron trifluoride, the total molar mass is:
(1 × Boron atomic mass) + (3 × Fluorine atomic mass)
- Molar mass of BF3 = (1 × 10.81) + (3 × 18.99) = 67.81 g/mol
2. Use the molar ratio to calculate the grams of boron:
- The molar ratio between boron and boron trifluoride is 1:1, as there is only one boron atom in each molecule of boron trifluoride.
- Therefore, the mass of boron in 1 mole of boron trifluoride is equal to the molar mass of boron: 10.81 g/mol.
3. Calculate the grams of boron in 3.54 moles of boron trifluoride:
- Multiply the molar mass of boron by the number of moles of boron trifluoride:
Grams of boron = 10.81 g/mol × 3.54 mol
- Grams of boron = 38.3074 g
So, there are approximately 38.3074 grams of boron present in 3.54 moles of boron trifluoride.