Question

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

Answer 1

Answer : The number of oxygen atoms are,
`4.8* 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}=(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* 6.022* 10^(23)` number of oxygen atoms

As, 0.0114 mole of
`Na_2Cr_2O_7` contains
`0.0114* 7* 6.022* 10^(23)=4.8* 10^(22)` number of oxygen atoms

Therefore, the number of oxygen atoms are,
`4.8* 10^(22)`

Answer 2

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