Deducing stoichiometry from masses of reactants and products

Predict, then check

Use the ammonia recipe: 1 N₂ + 3 H₂ → 2 NH₃. Now double the nitrogen.

If 2 moles of nitrogen react, how many moles of hydrogen are needed and how many moles of ammonia form?

The magnesium oxide practical

Step through it. Each stage has a reason.

1. 1Protect your eyesBurning magnesium gives off a bright light that can damage eyes, so safety precautions are needed.
2. 2Heat the magnesiumHeating magnesium in a crucible makes it react with oxygen to form magnesium oxide.
3. 3Lift the lid now and thenRaising the lid periodically lets enough oxygen in to oxidise all the magnesium.+
4. 4Heat until the mass is constantHeating continues until the mass stays constant.
5. 5Keep magnesium limitingMagnesium is the limiting reactant and oxygen is in excess, so every atom of magnesium reacts and the maximum amount of magnesium oxide forms.+
6. 6Find the massesThe masses of magnesium that reacted and magnesium oxide produced come from the measurements. Conservation of mass then gives the mass of oxygen that reacted.+

What you need to know

• The big numbers in front of formulae in a balanced symbol equation are coefficients, and they give the molar ratio.
• Read 1 N₂ + 3 H₂ → 2 NH₃ as 1 mole of nitrogen, 3 moles of hydrogen and 2 moles of ammonia.
• Subscripts, the small numbers inside a formula, show the ratio of atoms or ions within one substance.
• Have a goMaya says the 2 in O₂ means “two moles of oxygen”. She sounds very sure. What has she mixed up?

  She has read a subscript as if it were a coefficient.

  The small 2 in O₂ shows the ratio of atoms within one substance; only a big number in front would show moles.

• Change the amount of one substance and every other coefficient is multiplied or divided by the same factor.
• One mole is 6.02 × 10²³ particles, and moles = mass in grams ÷ relative mass.
• Have a goQuick one: how many moles are there in 36 g of a substance with a relative mass of 18?

  2 mol

  Moles = mass ÷ relative mass, so 36 ÷ 18 = 2. Flipping it to 18 ÷ 36 would wrongly give 0.5.

• Start the calculation grid with each formula, remembering that oxygen is O₂ because it is a diatomic element.
• The unitary ratio comes from dividing every value in the moles row by the smallest, so at least one value is 1.
• Molar ratios need whole numbers: round values very close to one, otherwise multiply all the values by the same number.
• Have a goA unitary ratio comes out as 1 : 2.04. Do you round it or multiply it?

  Round it, to 1 : 2.

  2.04 is very close to a whole number, so rounding is fine. You only multiply when a value isn't close enough to round.

• Use the molar ratio as the coefficients, then check the atoms balance. Product masses give the products' ratio too.
• In the magnesium oxide practical, magnesium is the limiting reactant and oxygen's mass comes from conservation of mass.

The big picture

The big numbers in a balanced equation are a recipe in moles, and you can work them out from masses measured in the lab. Turn every mass into moles, divide by the smallest value, tidy up to whole numbers, and that molar ratio becomes the coefficients.

⚠ Watch out

Treating subscripts and coefficients as the same kind of number. Subscripts sit inside a formula and fix what the substance is; coefficients sit in front and give the moles of each substance reacting. A second easy slip is writing oxygen as O in the grid when it should be O₂.

Memory hook

Mass → moles → ÷ smallest → whole numbers → coefficients. Big numbers in front are moles; small numbers inside are atoms.

Check yourself

Cover the page. A moles row reads 0.05, 0.075, 0.05. Find the unitary ratio, round or multiply, then write the molar ratio. (Check: 1 : 1.5 : 1, times 2, 2 : 3 : 2.)