# If the heat of combustion for a specific compound is -1220 kJ/mol and its molar mass is 78.35 g/mol, how many grams of this compound must you burn to release 649.90 kJ of heat?

Jan 11, 2016

#### Answer:

$\text{41.7 g}$

#### Explanation:

Before doing anything else, make sure that you understand what a compound's heat of combustion tells you.

The heat of combustion, also referred to as the enthalpy change of combustion, $\Delta {H}_{\text{comb}}$, represents the heat given off when one mole of a substance undergoes complete combustion, usually under standard conditions.

In your case, you have

$\Delta {H}_{\text{comb" = -"1220 kJ/mol}}$

Two things to notice here

• the enthalpy change of combustion carries a negative sign - this confirms that you're dealing with heat given off
• the enthalpy change of combustion is given per mole of the compound

So, simply put, $\text{1220 kJ}$ are given off when one mole of your mystery compound undergoes complete combustion.

Now, to solve this problem, you can convert the enthalpy change of combustion from kilojoules per mole to kilojoules per gram, then calculate the mass needed to give off that much heat.

To do that, use the compound's molar mass, which is said to be equal to $\text{78.35 kJ/mol}$. This tells you that one mole of this compound has a mass of $\text{78.35 g}$.

This means that you have

$1220 \text{kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mol"))))/"78.35 g" = "15.571 kJ/g}$

So, when one gram undergoes complete combustion, $\text{15.571 kJ}$ of heat are given off, This means that you have

649.90 color(red)(cancel(color(black)("kJ"))) * "1 g"/(15.571color(red)(cancel(color(black)("kJ")))) = "41.739 g"

Rounded to three sig figs, the answer will be

$m = \textcolor{g r e e n}{\text{41.7 g}}$

So, you can conclude that

• when $\text{41.7 g}$ of this compound undergo complete combustion, $\text{649.90 kJ}$ of heat are being given off

• when $\text{41.7 g}$ of this compound undergo complete combustion, the enthalpy change of reaction is equal to $\Delta {H}_{\text{rxn" = -"649.90 kJ}}$

Remember, those two statements are equivalent.