Answer:
To calculate the number of moles of copper(II) bromate (Cu(BrO3)2) in 119.4 grams of the compound, you can use the formula:
moles = mass (grams) / molar mass (g/mol)
First, you'll need to find the molar mass of copper(II) bromate. To do this, you sum the atomic masses of all the elements in one mole of the compound:
- Copper (Cu) has a molar mass of approximately 63.55 g/mol.
- Bromine (Br) has a molar mass of approximately 79.90 g/mol.
- Oxygen (O) has a molar mass of approximately 16.00 g/mol.
Copper(II) bromate has two bromine atoms and six oxygen atoms in its chemical formula.
Molar mass of Cu(BrO3)2 = (1 * Cu) + (2 * Br) + (6 * O)
Molar mass of Cu(BrO3)2 = (1 * 63.55 g/mol) + (2 * 79.90 g/mol) + (6 * 16.00 g/mol)
Now, calculate the molar mass:
Molar mass of Cu(BrO3)2 = 63.55 g/mol + 159.80 g/mol + 96.00 g/mol
Molar mass of Cu(BrO3)2 = 319.35 g/mol
Now, you can calculate the number of moles of copper(II) bromate in 119.4 grams:
moles = 119.4 g / 319.35 g/mol ≈ 0.374 moles
So, there are approximately 0.374 moles of copper(II) bromate in 119.4 grams of the compound.