# Question #86743

##### 1 Answer

#### Answer:

Here's why that is the case.

#### Explanation:

A compound's **percent composition** essentially tells you how many grams of each constituent element you get **per** **of compound**.

In this case, morphine is said to have a percent composition of nitrogen of **every**

Now, in order to show why this is the case, you must use the chemical formula of morphine, which is

So, each molecule of morphine contains **one atom** of nitrogen, **every mole** of morphine will contain **one mole** of nitrogen.

You can thus use the **molar mass** of morphine and the **molar mass** of nitrogen to find morphine's percent composition of nitrogen.

So, use the molar masses of carbon, oxygen, hydrogen, and nitrogen to find the molar mass of morphine. Since **one mole** of morphine contains

#17# molesof carbon,#17 xx "C"# #19# molesof hydrogen,#19 xx "H"# #3# molesof oxygen,#3 xx "O"# #1# moleof nitrogen,#1 xx "N"#

you can say that each element will contribute

#17 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "204.19 g C"#

#19 color(red)(cancel(color(black)("moles H"))) * "1.00794 g"/(1color(red)(cancel(color(black)("mole H")))) = "19.15 g H"#

#3 color(red)(cancel(color(black)("moles O"))) * "15.9994 g"/(1color(red)(cancel(color(black)("mole O")))) = "48.0 g O"#

#1 color(red)(cancel(color(black)("mole N"))) * "14.007 g"/(1color(red)(cancel(color(black)("mole N")))) = "14.01 g N"#

to the molar mass of the compound. The mass of **one mole** of morphine will thus be

#M_"M morphine" = "204.19 g" + "19.15 g" + "48.0 g" + "14.01 g"#

#M_"M morphine" = "285.35 g"#

Now all you have to do is use the fact that **one mole** of morphine has a mass of **one mole** of nitrogen, which has a mass of

#"% N" = (14.007 color(red)(cancel(color(black)("g"))))/(285.35color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(4.91%)color(white)(a/a)|)))#