Photosynthesis: balance the symbol equation and quantities
Conserve carbon, hydrogen and oxygen in the Extended equation, then use its mole ratios under explicit idealised conditions.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Recall the Core photosynthesis word equation and conditions.
- Use molar masses, coefficient ratios and conservation of atoms.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many carbon atoms occur in one C₆H₁₂O₆ molecule?
What you will learn
- State and balance the required photosynthesis symbol equation with the light/chlorophyll conditions.
- Use the equation's quantity ratios without treating a theoretical calculation as a measured leaf yield.
Balance the glucose formula, keeping all atoms
The required symbol equation is 6CO₂+6H₂O→C₆H₁₂O₆+6O₂, using light energy in the presence of chlorophyll. One glucose molecule contains six carbon atoms and twelve hydrogen atoms, so six CO₂ and six H₂O supply those counts. Oxygen then totals eighteen on each side: twelve plus six on the left, and six plus twelve on the right.
Coefficients multiply complete formulas, whereas subscripts specify atoms within a substance. Do not replace C₆H₁₂O₆ with a different formula to make an unbalanced equation appear easier. Light provides energy and chlorophyll is a condition in the simplified net description; neither adds new carbon, hydrogen or oxygen atoms to the balance.
Use a net equation as a quantity model, with limits
The ratios are six moles CO₂ to one mole glucose to six moles O₂, with six moles water consumed in the stated net equation. With C=12, H=1 and O=16, glucose has molar mass 180 g/mol, CO₂ 44 g/mol, water 18 g/mol and O₂ 32 g/mol. The ratio can check a hypothetical theoretical calculation when the other reactant, light conditions and conversion assumptions are supplied.
A real plant's measured growth depends on more processes than this net equation, including respiration and formation of other substances. A theoretical glucose amount is not a guaranteed crop yield or an observed instantaneous photosynthesis rate. Use the equation for the stated accounting question and identify the assumptions before making a biological or environmental claim.
Sources for this lesson
Pause and explain
What is the CO₂:glucose mole ratio in the required equation?
Worked example
Under idealised complete net-equation conversion with sufficient water and suitable light/chlorophyll conditions, use 1.20 mol CO₂ to find glucose and O₂ masses. Use C=12, H=1, O=16.
Show the worked solution
- Glucose amount is 1.20/6=0.200 mol and molar mass is 180 g/mol.
- Its mass is 0.200×180=36.0 g.
- Oxygen amount is 1.20 mol, so its mass is 1.20×32=38.4 g; these are theoretical net-equation amounts.
Answer 36.0 g glucose and 38.4 g O₂ under the stated idealised conditions.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Write the balanced equation and independently count each element on both sides.
Give me a hint
Include oxygen from water as well as CO₂ and glucose.
Compare my reasoning
- Write 6CO₂+6H₂O→C₆H₁₂O₆+6O₂ with light and chlorophyll conditions.
- Each side has six carbon and twelve hydrogen atoms.
- Each side has eighteen oxygen atoms: 12+6 on the left and 6+12 on the right.
6CO₂+6H₂O→C₆H₁₂O₆+6O₂; C6/H12/O18 conserved, using light energy with chlorophyll.
Look for these in your work
- The equation and conditions are correct.
- Every oxygen contribution is counted.
For a hypothetical 0.60 mol CO₂ net-equation conversion with sufficient water and suitable conditions, find glucose and oxygen amounts and masses using C=12, H=1, O=16.
Give me a hint
Use the 6:1:6 CO₂:glucose:O₂ ratio before masses.
Compare my reasoning
- Glucose amount is 0.60/6=0.10 mol, mass 0.10×180=18.0 g.
- O₂ amount is 0.60 mol, mass 0.60×32=19.2 g.
- These are idealised theoretical amounts, not a measured crop yield.
0.10 mol/18.0 g glucose; 0.60 mol/19.2 g O₂, under the hypothetical complete-conversion conditions.
Look for these in your work
- Mole ratios precede mass conversion.
- The result keeps the stated theoretical limit.
For theoretical production of 9.0 g glucose, find CO₂ and water masses consumed and O₂ mass formed using the net equation and C=12, H=1, O=16. Check total mass and explain why light does not supply extra atoms.
Give me a hint
Nine grams is 0.050 mol glucose; multiply by six for the other quantities.
Compare my reasoning
- CO₂ and H₂O amounts are each 0.300 mol, giving 13.2 g CO₂ and 5.4 g H₂O.
- O₂ amount is 0.300 mol, mass 9.6 g. Reactant mass 13.2+5.4=18.6 g equals product mass 9.0+9.6=18.6 g.
- Light supplies energy rather than new elements; the quantities are theoretical net-equation accounting with suitable conditions.
13.2 g CO₂ and 5.4 g H₂O consumed; 9.6 g O₂ formed; 18.6 g total on each side. Light supplies energy, not extra atoms.
Look for these in your work
- All coefficient ratios and masses are consistent.
- Conservation and the energy/atom distinction are explicit.
Common mistakes
- Counting twelve oxygen atoms on the left while omitting those in water.
- Using a 1:1 CO₂:glucose mole ratio.
- Changing glucose subscripts to balance the equation.
- Reporting a theoretical glucose amount as a measured plant-growth result.
What can you explain now?
State the required balanced photosynthesis symbol equation.
Compare with the explanation
6CO₂+6H₂O→C₆H₁₂O₆+6O₂.
C6, H12 and O18 are conserved, with light energy and chlorophyll as stated conditions.
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Make it stick
Balance the symbol equation, count C/H/O, then calculate a new theoretical amount while keeping its conditions separate from observed plant growth.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.