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How much heat is required to raise the temperature of $230\ \mathrm g$ $\ce{CH3OH(l)}$ from $22.0\ ^\circ\mathrm C$ to $30.0\ ^\circ\mathrm C$ and then vaporize it at $30.0\ ^\circ\mathrm C$? Use enthalpy of vaporization $\ce{CH3OH(l)} = 38.0\ \mathrm{kJ\ mol^{−1}}$ and a molar heat capacity of $\ce{CH3OH(l)}$ of $81.1\ \mathrm{J\ mol^{−1}\ K^{−1}}$.

I Used the equation $Q_\text{total} = Q_\text{heat} + Q_\text{vap}$. My answer was $88.9\ \mathrm{kJ}$.

However, it is wrong.

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  • $\begingroup$ Without seeing your work it's difficult to tell where you went wrong. Your basic method is correct though. Just to check, I get 7.18 moles of methanol. Just considering the enthalpy of vaporization, that's going to be much more than 89 kJ. $\endgroup$ – Jason Patterson Sep 17 '15 at 3:05
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The first thing to be sure of is that you're doing all the conversions correctly. Molar mass is the number of grams in a mol. It's found by adding the molar mass of all the elements in a compound times the number in the compound. For example the molar mass of $CH_4$ would be $$\textrm{Molar Mass of }C + 4\times\textrm{ Molar Mass of }H$$ Once you've found the correct molar mass it should have units of mass per mol - most commonly $\textrm{grams mol}^{-1}$. Thus if you have the mass of a substance and the mass required to have a mol, you can divide mass by molar mass to find moles. Then you simply: $$Q_{heat} = \Delta T\times \textrm{ number of moles } \times C_p$$ Where $C_p$ is heat capacity. To find the heat of vaporisation again you just multiply the per mol value by the number of moles: $$Q_{vap} = \textrm{ number of moles } \times \Delta_{vap}H^\circ$$ Where $\Delta_{vap}H^\circ$ is the heat of vaporisation. Be sure you got all these steps right and be aware of the difference between kJ and J! ($1\textrm{ kJ} = 1000\textrm{ J}$) Using this method I got a very different answer to yours. Feel free to post your own working for us to look at!

Note: Sometimes enthalpy of vaporisation values are actually the energy to vaporise from 298K (24.5C) but this can't be your problem because, as Jason Patterson said, even with just the enthalpy of vaporisation, no heating, your value is too low.

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