A #0.967g# mass of an oxide of phosphorus contains #0.422g# of phosphorus. What is its empirical formula?

Answer 1

We gots #P_2O_5#.....

We figure out the empirical formula.

"Moles of phosphorus" are equal to (0.422g)/(30.9737gmol^-1) = 0.01361mol.

#"Moles of oxygen"=(0.967*g-0.422*g)/(15.999*g*mol^-1)=0.0341*mol#

We divide thru by the SMALLEST MOLAR quantity to give an empirical formula of #P_((0.01361)/(0.01361))O_((0.0341)/(0.01361))#

#-=PO_(2.50)#....

We now double the trial formula to obtain the empirical formula, which is by definition the simplest WHOLE number ratio.

#P_2O_5#

The actual molecule is #P_4O_10# but we would need further data to assess this.

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Answer 2

Owen Ambrose

The empirical formula of the oxide of phosphorus is P₂O₅.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

A #0.967*g# mass of an oxide of phosphorus contains #0.422*g# of phosphorus. What is its empirical formula?

A #0.967g# mass of an oxide of phosphorus contains #0.422g# of phosphorus. What is its empirical formula?