Question #46fd3
1 Answer
Here's what I got.
Explanation:
An interesting way to approach this problem is to determine the oxide's molecular formula first, then backtrack and find its empirical formula.
So, you know that this oxide of phosphorus has a molar mass of
You also know that the oxide has a percent composition of
This means that every
#"43.66 g "-># phosphorus,#"P"# #"56.34 g " -># oxygen,#"O"#
Pick a
So, you can say that this sample will contain
#284 color(red)(cancel(color(black)("g oxide"))) * "43.66 g P"/(100color(red)(cancel(color(black)("g oxide")))) = "123.994 g P"#
#284color(red)(cancel(color(black)("g oxide"))) * "56.34 g O"/(100color(red)(cancel(color(black)("g oxide")))) = "160.006 g O"#
Use the molar masses of the two elements to convert the masses to moles
#"For P: " 123.994 color(red)(cancel(color(black)("g"))) * "1 mole P"/(30.9738color(red)(cancel(color(black)("g")))) = 4.003 ~~ "4 moles P"#
#"For O: " 160.006color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = 10.001 ~~ "10 moles O"#
Since this is how many moles of phosphorus and oxygen you get in one mole of the oxide, it follows that its molecular formula will be
#"molecular formula" = color(green)(|bar(ul(color(white)(a/a)"P"_4"O"_10color(white)(a/a)|))) -># phosphorus pentoxide
To find the empirical formula, all you have to do is rewrite the
In this case, you can say that
#4:10 " "<=>" " 2 : 5#
This means that the empirical formula will be
#"empirical formula" = color(green)(|bar(ul(color(white)(a/a)"P"_2"O"_5color(white)(a/a)|)))#