How moles of oxygen are there in 3.37 grams of of sodium dichromate (Na2Cr2O7)?

Question

(a) How moles of oxygen are there in 3.37 grams of of sodium dichromate (Na2Cr2O7)?

______moles
(b) How many oxygen atoms are present?
 
______atoms

Answer 1

PART (A)

 

Step 1: Determine the molar mass of sodium dichromate:

 

MM (Na2Cr2O7)

= 2 x MW(Na) + 2 x MW(Cr) + 7 x MW(O)

= 2 x 22.990 g/mol + 2 x 51.996 g/mol + 7 x 15.998 g/mol

= 261.958 g/mol

 

Step 2: Use stoichiometry:

 

3.37 g Na₂Cr₂O₇ x (1 mol Na₂Cr₂O₇ / 261.958 g Na₂Cr₂O₇) x (7 mol O / 1 mol Na₂Cr₂O₇)

 

Step 3: Determine the answer for (a) via calculation the above.

 

# moles of Oxygen = ???

 

 

PART (B)

 

Take the number of moles of oxygen in part (a). Multiply the number of moles of oxygen by Avogadro's number.

 

??? moles of Oxygen x (6.022 x 10²³ oxygen atoms / 1 mol Oxygen) = ??????