High School

What mass of MnBr₃ will contain [tex]6.00 \times 10^{22}[/tex] formula units?

Answer :

To find the mass of MnBr3 with 6.00 x 10^22 formula units, divide the formula units by Avogadro's number to get moles, and multiply by MnBr3's molar mass, resulting in a mass of about 29.37 grams.

Firstly, calculate the amount (in moles) of MnBr3 by dividing the number of formula units by Avogadro's number:

To find the mass of MnBr3 that has 6.00 x 1022 formula units, we need to use Avogadro's number (6.022 x 1023 formula units/mole) to relate formula units to moles and then use the molar mass of MnBr3 to convert moles to grams.

Number of moles = (6.00 x 1022 formula units) / (6.022 x 1023 formula units/mole)

= 0.0996 moles

Next, determine the molar mass of MnBr3 by adding the atomic masses of Mn and three Br atoms: Molar mass of MnBr3 = (Mn) + 3(Br). Since the atomic mass of Mn is 54.94 g/mol and Br is 79.904 g/mol, we get:

Molar mass of MnBr3 = 54.94 g/mol + 3(79.904 g/mol)

= 294.652 g/mol

Finally, calculate the mass of MnBr3 by multiplying the moles by its molar mass:

Mass of MnBr3 = 0.0996 moles x 294.652 g/mol

= 29.37 grams