Question

To form 100 g of `H_2O_2`, 5.927 g of H must react with 94.073 g of O. How many grams of O would be needed to form 104 g of `H_2O_2`?

Answer 1

Well, what is `5.927%` of `104*g`?

Explanation:

And I make this....

`5.927%xx104*g=6.164*g`

And let us just check this...and examine the empirical formula of the `104*g` mass...and for this we divide thru by the ATOMIC mass of each constituent element...

`"Moles of hydrogen"=(6.164*g)/(1.00794*g*mol^-1)=6.116*mol`

`"Moles of oxygen"=(104*g-6.164*g)/(16.00*g*mol^-1)=6.116*mol`

And thus the empirical formula is simply `HO`...why is this different from the chemical formula of hydrogen peroxide?