Question

In 6.72 g of `"N"_2"H"_4`, how many...?

(a) how many `"N"_2"H"_4` molecules are present?
(b) how many `"N"` atoms are present?
(c) how many protons are present

Answer 1

Here's what I got.

Explanation:

Your tool of choice for this problem will be Avogadro's number, which tells you how many molecules of a substance you get in one mole of that substance.

`color(blue)(|bar(ul(color(white)(a/a)"one mole" = 6.022 * 10^(23)"molecules"color(white)(a/a)|))) ->` Avogadro's number

So, you know that one mole of hydrazine, `"N"_2"H"_4`, contains `6.022 * 10^(23)` molecules of hydrazine. However, you don't know how many moles you have in that `"6.72-g"` sample.

To figure that out, use hydrazine's molar mass, which tells you the mass of one mole of hydrazine molecules. Since hydrazine has a molar mass of `"32.045 g mol"^(-1)`, you can say that one mole of hydrazine will have a mass of `"32.045 g"`.

This means that your sample contains

`6.72color(red)(cancel(color(black)("g"))) * overbrace(("1 mole N"_2"H"_4)/(32.045color(red)(cancel(color(black)("g")))))^(color(brown)("molar mass of N"_2"H"_4)) = "0.2097 moles N"_2"H"_4`

Now that you how many moles you have, use Avogadro's number to find how many molecules you have

`0.2097color(red)(cancel(color(black)("moles"))) * overbrace((6.022 * 10^(23)"molecules")/(1color(red)(cancel(color(black)("mole")))))^(color(purple)("Avogadro's number")) = color(green)(|bar(ul(color(white)(a/a)1.26 * 10^(23)"molecules"color(white)(a/a)|)))`

Now, for part (b) you must find the number of nitrogen atoms, `"N"`, present in the sample. To do that, focus on one molecule of hydrazine. As you can tell from its chemical formula, this molecule will contain

• two atoms of nitrogen, `2 xx "N"`
• four atoms of hydrogen, `4 xx "H"`

So, if you get `2` atoms of nitrogen for every molecule of hydrazine, you can say that the total number of nitrogen atoms will be

`1.26 * 10^(23)color(red)(cancel(color(black)("molecules N"_2"H"_4))) * "2 atoms of N"/(1color(red)(cancel(color(black)("molecule N"_2"H"_4)))) = color(green)(|bar(ul(color(white)(a/a)2.52 * 10^(23)"atoms of N"color(white)(a/a)|)))`

Finally, for part (c) you must find the total number of protons present in the sample. As you know, the number of protons present in the nucleus of a given atom is equal to that atom's atomic number.

A quick look in the periodic table will reveal that nitrogen has an atomic number equal to `7`, which means that every atom of nitrogen contains `7` protons.

Hydrogen has an atomic number equal to `1`, which means that every atom of hydrogen contains `1` proton in its nucleus.

Now, you already know how many atoms of nitrogen you have in the sample. Use the same approach to find the number of atoms of hydrogen

`1.26 * 10^(23)color(red)(cancel(color(black)("molecules N"_2"H"_4))) * "4 atoms of H"/(1color(red)(cancel(color(black)("molecule N"_2"H"_4)))) = 5.04 * 10^(23)"atoms of H"`

This means that the total number of protons present in the sample will be

`"no. of protons" = overbrace(7 xx 2.52 * 10^(23))^(color(blue)("coming from nitrogen")) + overbrace(1 xx 5.04 * 10^(23))^(color(red)("coming from hydrogen"))`

`"no. of protons" = color(green)(|bar(ul(color(white)(a/a)2.27 * 10^(24)"protons"color(white)(a/a)|)))`

The answers are rounded to three sig figs, the number of sig figs you have for the mass of hydrazine.