Question

Question #f2ea8

Answer 1

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 atom of nitrogen, `1 xx "N"`
• three atoms of 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, `A`, tells you the number of protons, which is given by the atomic number, `Z`, 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` proton inside its nucleus

`"For N: " Z = 7 ->` a nitrogen atom contains `7` protons inside 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 `0` neutrons and for every atom of nitrogen you get `7` 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.