# Question #f2ea8

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Start by calculating how many atoms of hydrogen and how many atoms are present in your sample.

Use the **molar mass** of ammonia to convert the number of grams to *moles*

#1.7 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17.031color(red)(cancel(color(black)("g")))) = "0.0998 moles NH"_3#

Next, use **Avogadro's constant** to calculate how many *molecules* of ammonia are present in the sample

#0.0998 color(red)(cancel(color(black)("moles NH"_3))) * (6.022 * 10^(23)color(white)(.)"molecules NH"_3)/(1color(red)(cancel(color(black)("mole NH"_3))))#

# = 6.01 * 10^(22)# #"molecules NH"_3#

Now, every molecule of ammonia contains

,one atomof nitrogen#1 xx "N"# ,three atomsof hydrogen#3 xx "H"#

This means that your sample contains

#6.01 * 10^(22) color(red)(cancel(color(black)("molecules NH"_3))) * "1 atom N"/(1color(red)(cancel(color(black)("molecule NH"_3))))#

# = 6.01 * 10^(22)# #"atoms of N"#

#6.01 * 10^(22) color(red)(cancel(color(black)("molecules NH"_3))) * "3 atoms H"/(1color(red)(cancel(color(black)("molecule NH"_3))))#

# = 1.8 * 10^(23)# #"atoms of H"#

In order to determine how many **neutrons** are present in the sample, use the **molar masses** of the two elements.

#"For H: " M_M = "1.008 g mol"^(-1)#

#"For N: " M_M = "14.007 g mol"^(-1)#

If you **round** the molar mass of an element to the **nearest whole number**, you will get the **mass number** of the most common isotope of said element.

As you know, the mass number, **atomic number**, **and** neutrons present inside an atom's nucleus.

#color(blue)(ul(color(black)(A= Z + "no. of neutrons")))#

So, you will have

#"For H: " 1.008 ~~ 1#

#"For N: " 14.007 ~~ 14#

This means that you will have

#"For H: " A = 1#

#"For N: " A = 14#

Grab a Periodic Table and look for the **atomic numbers** of the two elements

#"For H: " Z = 1 -># a hydrogen atom contains#1# protoninside its nucleus

#"For N: " Z = 7 -># a nitrogen atom contains#7# protonsinside its nucleus

This means the most common isotope of hydrogen will have

#"no. of neutrons" = 1 -1 = 0#

The most common isotope of nitrogen will have

#"no. of protons" = 14 - 7 = 7#

You can thus *approximate* that for every atom of hydrogen you get **neutrons** and for every atom of nitrogen you get **neutrons**.

Therefore, the number of neutrons present in your sample will be equal to

#6.01 * 10^(22) color(red)(cancel(color(black)("atoms N"))) * "7 neutrons"/(1color(red)(cancel(color(black)("atom N")))) + 1.8 * 10^(23) color(red)(cancel(color(black)("atoms H"))) * "0 neutrons"/(1color(red)(cancel(color(black)("atom H"))))#

which gets you

#color(darkgreen)(ul(color(black)("no. of neutrons" ~~ 4.2 * 10^(23)color(white)(.)"neutrons")))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the mass of ammonia.