Question

(3) 3. How many grams of oxygen are present in a sample containing 5.75 x 1021 formula units of manganese (II) phosphate ? (6) 4. Determine the percent composition (by mass) for each element in ammonium dichromate, (NH4)2Cr207.

Answer 1

Molecular weight of Mn3(PO4)2 is 355 g/mol.

1mole Mn3(PO4)2 = 355g

6.022*1023 formula unit = 1mole

Thus, 355g Mn3(PO4)2 have 6.022*1023 formula unit of Mn3(PO4)2

6.022*1023 formula unit : 355g

1 formula unit : 355/6.022*1023 g

5.75*1021 : 355/6.022*1023 * 5.75*1021 g

Thus, 3.389 g Mn3(PO4)2 is present in 5.75*1021 formula unit.

Now,

355 g Mn3(PO4)2 contains 16*8g oxygen.

355 g: 128g oxygen

1 g : 0.36 g oxygen

3.389 g : 1.22g oxygen

Thus, 1.22 g oxygen is present in 5.75*1021 formula units of Mn3(PO4)2 .