Question

How many oxygen atoms are in 3.00 grams of sodium dichromate, Na2Cr2O7?

Answer 1

The solution is as follows: Convert 3.00 g Na2Cr2O7 into mol. To do this, we must know the molar mass. Molar mass of Na2Cr2O7 is 261.97 g/mol Mol Na2Cr2O7 = (3.00 g Na2Cr2O7)( mol / 261.97 g) = 0.01145 mol In 1 mol of Na2Cr2O7 there are 7 mol of O (oxygen). Therefore: Mol O = 0.01145 mol Na2Cr2O7 (7 mol O/ 1 mol Na2Cr2O7) = 0.08016 mol O In 1 mol of any substance, there are 6.02 x 10^23 atoms Atoms of O = 0.08016 mol O ( 6.02 x 10^23 atoms / mol O) = 4.8257 x 10^22 atoms

Answer 2

Answer : The number of oxygen atoms are, $4.8\times 10^{22}$

Explanation : Given,

Mass of $Na_2Cr_2O_7$ = 3 g

Molar mass of $Na_2Cr_2O_7$ = 261.97 g/mole

First we have to calculate the moles of $N_2O_3$

$\text{Moles of }Na_2Cr_2O_7=\frac{\text{Mass of }Na_2Cr_2O_7}{\text{Molar mass of }Na_2Cr_2O_7}=\frac{3g}{261.97g/mole}=0.0114mole$

As we know that $Na_2Cr_2O_7$ contains 2 number of sodium atoms, 2 number of chromium atoms and 7 number of oxygen atoms.

Now we have to calculate the number of atoms of oxygen.

As, 1 mole of $Na_2Cr_2O_7$ contains $7\times 6.022\times 10^{23}$ number of oxygen atoms

As, 0.0114 mole of $Na_2Cr_2O_7$ contains $0.0114\times 7\times 6.022\times 10^{23}=4.8\times 10^{22}$ number of oxygen atoms

Therefore, the number of oxygen atoms are, $4.8\times 10^{22}$