lesson
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If you ignite 12 g of magnesium ribbon in pure oxygen, you don't end up with 12 g of white powder — you get 20 g.
Because chemical reactions rearrange individual atoms rather than bulk mass, we cannot convert directly from grams of reactant to grams of product.
How do chemists bridge the gap between counting invisible atoms and weighing real substances on a balance?
The Three-Step Mole Bridge
The mole (n) is our counting unit: one mole contains exactly 6.022×1023 particles, and its mass in grams equals the substance's molar mass (M) in g/mol.
In 1792, German chemist Jeremias Richter discovered that chemicals react in strict, whole-number proportions, coining the term stoichiometry to describe measuring these exact chemical ratios.
📊Create an interactive/animated flow chart diagram showing the 3-step 'Mole Bridge' for stoichiometric mass calculations. Left card: 'Mass of Reactant A (g)' with a scale icon. Arrow pointing down labeled '÷ Molar Mass A (g/mol)'. Middle-left card: 'Moles of Reactant A (mol)'. Horizontal arrow bridging across labeled '× (Coefficient B / Coefficient A) [Mole Ratio from Balanced Equation]'. Middle-right card: 'Moles of Product B (mol)'. Arrow pointing up labeled '× Molar Mass B (g/mol)'. Right card: 'Mass of Product B (g)' with a beaker/crystal icon. Use a clean modern palette with light slate background (#f8fafc), navy text (#1e2945), sky blue accent (#0284c7), amber highlight (#d97706) for the bridge ratio, rounded 12px borders, responsive to 350px width.
Every mass-to-mass stoichiometry problem follows the exact same three-step route: grams to moles, mole ratio conversion, and moles back to grams.
What does this look like in a real laboratory reaction?
Worked Example: Burning Magnesium
Consider the balanced equation for burning magnesium in oxygen: 2Mg+O2→2MgO
Let's calculate the theoretical mass of magnesium oxide (MgO) produced when 12.15 g of Mg burns in excess oxygen. Relative atomic masses: Ar(Mg)=24.3, Ar(O)=16.0.
📊Create a step-by-step visual solution card showing the 3 calculations for 12.15 g Mg: Step 1 (Molar mass Mg = 24.3 g/mol): n(Mg) = 12.15 g / 24.3 g/mol = 0.50 mol. Step 2 (Mole Ratio): 2 mol Mg : 2 mol MgO -> 1:1 ratio, so n(MgO) = 0.50 mol. Step 3 (Molar mass MgO = 24.3 + 16.0 = 40.3 g/mol): m(MgO) = 0.50 mol × 40.3 g/mol = 20.15 g. Use distinct colored step badges (Step 1 blue, Step 2 orange, Step 3 green), clear typography, math notation display, responsive container 350px width.