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Lesson 12 · Moles to heat
Two machines, one chain
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Two machines, one chain

No balance reads moles. Real problems hand you GRAMS, so the mole machine from chapter one runs first, n=mMn = \dfrac{m}{M} — and only then does q=n ΔHq = n\,\Delta H collect its heat. Grams → moles → kilojoules: two small formulas, one chain, and the mole count is the coupling between them. The chain runs backwards too — a heat budget fixes the moles via n=qΔHn = \dfrac{q}{\Delta H}, and m=n Mm = n\,M dresses them in grams for the balance.

One trap gets a name here: the sign slip. An exothermic ΔH\Delta H is negative, so q computes negative — heat leaving, exactly as it should. But when the ask says “heat released”, the answer is the MAGNITUDE: the word already spent the minus sign. Reporting −500 kJ released says the reaction absorbed it — the sign, spoken twice, flips the story.

mMn

n=mMn = \frac{m}{M}

  • nn= Amount of substance (amount of substance)
  • mm= Mass (mass)
  • MM= Molar mass (molar mass)
Moles from Mass (n = m/M) solver →
qΔHn

q=nΔHq = n \Delta H

  • qq= Heat released or absorbed (energy)
  • nn= Amount of substance (amount of substance)
  • ΔH\Delta H= Molar enthalpy change (molar energy)
Heat of Reaction solver →