# Why is it that in order to convert the change in enthalpy per mole of product into the change in enthalpy per gram of the same product, you need to divide the change in enthalpy by the molar mass? Can you provide an example, please?

Jun 20, 2017

Here's why we need to do that.

#### Explanation:

The idea here is that the molar mass of a compound essentially acts as a bridge between moles and grams.

More specifically, the molar mass of a substance tells you the mass of exactly $1$ mole of said substance.

$\text{molar mass" = "a number of grams"/"1 mole}$

This means that every time you're working with moles, you can convert to grams by multiplying by the molar mass of the compound.

Let' say, for example, that you're working with $2$ moles of water. Water has a molar mass of ${\text{18.015 g mol}}^{- 1}$, so you can say that

"molar mass H"_2"O" = "18.015 g"/("1 mole H"_2"O")

To convert the sample to grams, use the molar mass as a conversion factor

2 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "36.03 g"

Similarly, when you're working with grams, you can convert to moles by flipping the conversion factor

36.03 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "2 moles H"_2"O"

Now, the molar enthalpy change of a reaction tells you the enthalpy change that occurs when $1$ mole of a substance takes part in the reaction.

Let's take a combustion reaction as an example.

${\text{CH"_ (4(g)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O}}_{\left(l\right)}$

The enthalpy change of combustion is equal to

$\Delta {H}_{\text{comb" = -"882.0 kJ mol}}^{- 1}$

This tells you that when $1$ mole of methane undergoes combustion, $\text{882.0 kJ}$ of heat are being given off.

$\Delta {H}_{\text{comb" = (-"882.0 kJ")/"1 mole CH}} _ 4$

To convert this to kilojoules per gram, we must use the molar mass of methane, ${\text{16.04 g mol}}^{- 1}$ as a conversion factor.

${\text{molar mass CH"_4 = "16.04 g"/"1 mole CH}}_{4}$

However, notice that the enthalpy change of combustion has ${\text{1 mole CH}}_{4}$ on the denominator, so you must flip the molar mass conversion factor to get ${\text{1 mole CH}}_{4}$ on the numerator--this will ensure that when you multiply the two, the ${\text{1 mole CH}}_{4}$ cancels out.

$\text{1 mole CH"_4/"16.04 g}$

Now all you have to do is multiply the two

(-"882.0 kJ")/(1 color(red)(cancel(color(black)("mole CH"_4)))) * (1color(red)(cancel(color(black)("mole CH"_4))))/"16.04 g" = -"54.99 kJ g"^(-1)

Notice that we get the same result if we simply divide the enthalpy change by the molar mass

(-"882.0 kJ" color(red)(cancel(color(black)("mol"^(-1)))))/("16.04 g"color(red)(cancel(color(black)("mol"^(-1))))) = (-"882.0 kJ")/"16.04 g" = - "54.99 kJ g"^(-1)

So remember, think of the enthalpy change of reaction and of the molar mass as two conversion factors that have $\text{1 mole}$ of the substance in the denominator.