# If a sample contains 1.228*10^23 molecules of "H"_2"SO"_4 then what is the mass of an individual "H"^(+) and "SO"_4^(2-) ion?

## I don't get how I can use ${N}_{A}$ to solve this. Guess i should revise the basics...

Jul 28, 2016

Here's what I got.

#### Explanation:

The interesting thing to note here is that you don't need to know how many molecules of sulfuric acid you have in your sample, all you need to know here is the acid's percent composition.

Sulfuric acid has a molar mass of ${\text{98.0785 g mol}}^{- 1}$, which means that one mole of sulfuric acid has a mass of $\text{98.0785 g}$. You know that one mole of sulfuric acid contains

• two moles of atoms of hydrogen, $2 \times \text{H}$
• one mole of atoms of sulfur, $1 \times \text{S}$
• four moles of atoms of oxygen, $4 \times \text{O}$

Use the molar masses of these elements to figure out the percent composition of hydrogen in one mole of sulfuric acid

"% H" = (2 xx 1.00794 color(red)(cancel(color(black)("g"))))/(2 xx color(red)(cancel(color(black)("g"))) + 1 xx 32.065 color(red)(cancel(color(black)("g"))) + 4 xx 15.994 color(red)(cancel(color(black)("g")))) xx 100 = 2.055%

This tells you that for every $\text{100 g}$ of sulfuric acid, you get $\text{2.005 g}$ of hydrogen. This is equivalent to saying that for every $\text{100 g}$ of sulfuric acid, you get

$\text{100 g " - " 2.055 g " = " 97.945 g}$

of sulfur and oxygen. Since the mass of an ion is approximately equal to the mass of the parent atom, you can say that $\text{100 g}$ of sulfuric acid can produce

$\textcolor{w h i t e}{a} {\text{2.055 g " -> color(white)(a)"H}}^{+}$ cations

${\text{97.945 g " -> "SO}}_{4}^{2 -}$ anions

Now, you know the molar mass of sulfuric acid, so use the percent composition to figure out how many grams of hydrogen cations and sulfate anions you can get from one mole of acid

98.0785 color(red)(cancel(color(black)("g H"_2"SO"_4))) * "2.055 g H"^(+)/(100color(red)(cancel(color(black)("g H"_2"SO"_4)))) = "2.0155 g H"^(+)

98.0785 color(red)(cancel(color(black)("g H"_2"SO"_4))) * "97.945 g SO"_4^(2-)/(100color(red)(cancel(color(black)("g H"_2"SO"_4)))) = "96.063 g SO"_4^(2-)

Since one mole of sulfuric acid contains $6.022 \cdot {10}^{23}$ molecules of acid, as given by Avogadro's number, ${N}_{\text{A}}$, you can say that the mass of a single hydrogen cation will be

2.0155color(white)(a)"g"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) * (1color(red)(cancel(color(black)("mole H"_2"SO"_4))))/(6.022 * 10^(23)color(red)(cancel(color(black)("molec H"_2"SO"_4)))) * (1color(red)(cancel(color(black)("molec H"_2"SO"_4))))/("2 H"^(+)"cations") =

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{1.673 \cdot {10}^{- 24} \text{g / H"^(+)"cation}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

SIDE NOTE As you know, a hydrogen cation is simply a proton. The mass of a proton is listed as

${m}_{\text{proton" ~~ 1.672 622 * 10^(-24)"g}}$

which makes this an excellent result!

Similarly, the mass of a single sulfate anion will be

96.063color(white)(a)"g"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) * (1color(red)(cancel(color(black)("mole H"_2"SO"_4))))/(6.022 * 10^(23)color(red)(cancel(color(black)("molec H"_2"SO"_4)))) * (1color(red)(cancel(color(black)("molec H"_2"SO"_4))))/("1 SO"_4^(2-)"anion") =

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{1.595 \cdot {10}^{- 22} \text{g / SO"_4^(2-)"anion}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

SIDE NOTE The only possibility I can see for using the number of molecules of sulfuric acid would also require the mass of the sample.

If the problem provides you the mass of $1.228 \cdot {10}^{23}$ molecules of sulfuric acid, then you can use this information to find the acid's molar mass.

From that point on, the answer would look pretty much the same. The answers would most certainly come out the same.