# 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

(a) how many

(b) how many

(c) how many protons are present

##### 1 Answer

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*, *how many moles* you have in that

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 **one mole** of hydrazine will have a mass of

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**, **one** molecule of hydrazine. As you can tell from its chemical formula, this molecule will contain

,two atomsof nitrogen#2 xx "N"# ,four atomsof hydrogen#4 xx "H"#

So, if you get **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 **every atom** of nitrogen contains **protons**.

Hydrogen has an atomic number equal to **every atom** of hydrogen contains **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.