Question #a1555

The idea here is that the difference between the mass of phosphorus that undergoes combustion and the mass of the oxide will represent the mass of oxygen that took part in the reaction #-># think the Law of mass conservation here.

This means that you have

#m_"oxide" = m_(P) + m_(O)#

#m_(O) = "3.55 g" - "1.55 g" = "2.00 g"#

So, you know that your #"3.55-g"# sample of phosphorus oxide contains #"1.55 g"# of phosphorus and #"2.00 g"# of oxygen.

Use the molar masses of the two elements to determine how many moles of each you get in the sample

#1.55 color(red)(cancel(color(black)("g"))) * "1 mole P"/(30.974 color(red)(cancel(color(black)("g")))) = "0.05004 moles P"#

#2.00color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.1250 moles O"#

To get the mole ratio that exists between the two elements in the oxide, divide both values by the smallest one

#"For P: " (0.05004color(red)(cancel(color(black)("moles"))))/(0.05004color(red)(cancel(color(black)("moles")))) = 1#

#"For O: " (0.1250color(red)(cancel(color(black)("moles"))))/(0.05004color(red)(cancel(color(black)("moles")))) = 2.498 ~~ 2.5#

Now, a compound's empirical formula tells you the smallest whole number ratio that exists between its constituent elements.

In your case, you get #1# mole of phosphorus for every #2.5# moles of oxygen. This is equivalent to saying that you get #2# moles of phosphorus for every #5# moles of oxygen.

Here #2:5# is the smallest whole number ratio that exists for #1:2.5#, which can be thought of as #1:5/2#.

The compound's empirical formula will thus be

#color(green)(|bar(ul(color(white)(a/a)"P"_2"O"_5color(white)(a/a)|)))#