# 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?

Jun 14, 2018

Well, what is 5.927% of $104 \cdot g$?

#### 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 $104 \cdot g$ mass...and for this we divide thru by the ATOMIC mass of each constituent element...

$\text{Moles of hydrogen} = \frac{6.164 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 6.116 \cdot m o l$

$\text{Moles of oxygen} = \frac{104 \cdot g - 6.164 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 6.116 \cdot m o l$

And thus the empirical formula is simply $H O$...why is this different from the chemical formula of hydrogen peroxide?