Conservation of mass and balanced chemical equations — AQA GCSE Chemistry
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The law of conservation of mass states that no atoms are created or destroyed during a chemical reaction, meaning the total mass of the products always equals the total mass of the reactants. Balanced symbol equations reflect this fundamental principle by ensuring that equal numbers of each type of atom appear on both sides of the equation.
Conservation of mass and balanced chemical equations explained
Conservation of mass and balanced chemical equations covers 3 specification points, including The law of conservation of mass states that no… and This means that chemical reactions can be….
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The law of conservation of mass states that no atoms are lost or made during a chemical reaction so the mass of the products equals the mass of the reactants.
During a chemical reaction, bonds break and new bonds form, but the atoms themselves are rearranged rather than created or destroyed. Because the total number of each type of atom stays the same, the total mass of the products equals the total mass of the reactants. This is the law of conservation of mass. For example, when magnesium reacts with oxygen, the white magnesium oxide produced has a greater mass than the magnesium alone because oxygen atoms from the air have joined the magnesium; the mass of magnesium plus the mass of oxygen that reacted equals the mass of magnesium oxide formed. In a closed system, the total mass before and after a reaction is unchanged. If a reaction appears to lose or gain mass, it is usually because a gas has escaped to the surroundings or a gas from the air has been incorporated into the product.
This means that chemical reactions can be represented by symbol equations which are balanced in terms of the numbers of atoms of each element involved on both sides of the equation.
Because atoms are conserved, a symbol equation must show the same number of atoms of each element on both sides. You balance an equation by placing whole-number multipliers, called coefficients, in front of chemical formulae; you never change the formulae themselves, because that would change the substances involved. For example, hydrogen burning in oxygen is written as 2H₂ + O₂ → 2H₂O. Counting shows four hydrogen atoms and two oxygen atoms on each side. To balance an equation, start with the formula containing the most elements, adjust one element at a time using coefficients, and finish by checking every element. State symbols may be added to show solid, liquid, gas or aqueous conditions. A balanced equation is the quantitative link between the reaction and the masses of substances involved.
Students should understand the use of the multipliers in equations in normal script before a formula and in subscript within a formula.
Chemical formulae use two different multiplier positions with distinct meanings. A subscript, written small and low after a symbol, multiplies only the atom immediately before it within that formula unit: H₂O contains two hydrogen atoms and one oxygen atom, and Ca(OH)₂ contains one calcium, two oxygen and two hydrogen atoms because the bracket groups the OH unit. A coefficient in normal script before a formula multiplies the whole formula unit: 2H₂O means two water molecules, giving four hydrogen atoms and two oxygen atoms in total. When counting atoms in an equation, apply subscripts first, then coefficients. For example, 2Ca(NO₃)₂ contains two calcium atoms, four nitrogen atoms and twelve oxygen atoms. This distinction underpins balancing equations and all mole and mass calculations.
Your focus
- State the law of conservation of mass in terms of atoms and mass.
- Explain an observed change in mass during a reaction by referring to gases entering or leaving the system.
- Apply the law of conservation of mass to predict that the mass of products equals the mass of reactants in a closed system.
Show all 9 objectives
- Write a balanced symbol equation with the same number of atoms of each element on both sides.
- Explain why coefficients, not subscripts, are used to balance an equation.
- Check a completed equation by counting the atoms of every element on each side.
- Distinguish between a coefficient in normal script and a subscript within a formula.
- Count the atoms of each element in a formula containing coefficients, subscripts and brackets.
- Apply correct multiplier interpretation when balancing chemical equations.
Which command word is this?
Give a detailed account of the causes or reasons behind an observation (e.g., state why mass decreased by identifying the gas produced and noting that it escaped the unsealed container).
Conservation of mass and balanced chemical equations exam tips
Topic Overview
Conservation of mass and balanced chemical equations form the central foundation of quantitative chemistry in AQA GCSE Chemistry. This topic explains why the total mass of products strictly equals the total mass of reactants during any chemical transformation, as atoms are simply rearranged rather than created or destroyed.
Mastering this area requires students to represent chemical reactions accurately through balanced symbol equations and state symbols. It also equips learners to interpret real-world experimental anomalies, such as observed increases or decreases in mass in open systems where gases are absorbed or released.
Key Concepts
- →Law of Conservation of Mass: No atoms are created or destroyed in a chemical reaction; mass of reactants equals mass of products.
- →Balancing Equations: Coefficients are placed in front of chemical formulas so that the number of atoms of each element is identical on both sides.
- →Apparent Mass Changes: Open systems allow gases to escape (causing a recorded mass drop) or atmospheric gases to enter (causing a recorded mass gain).
- →State Symbols: Physical states must be identified using (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solutions.
Marking Points
- State that atoms are rearranged in a chemical reaction and are neither created nor destroyed.
- Explain that the total mass of the products equals the total mass of the reactants because the number of each type of atom is unchanged.
- Apply the law to a practical situation, such as a reaction in a closed vessel where the mass reading stays constant.
- Explain an apparent mass change by identifying a gas that has left the reaction mixture or a gas from the air that has joined it.
- Use the law to support the idea that a symbol equation must have the same number of each type of atom on both sides.
- Write correct chemical formulae for all reactants and products before attempting to balance.
- Use whole-number coefficients in front of formulae so that the number of atoms of each element is the same on both sides.
- Check each element in turn after balancing and adjust coefficients if any count is unequal.
- Explain that coefficients multiply the whole formula, so 2H₂O contains four hydrogen atoms and two oxygen atoms.
- Recognise that changing a subscript alters the substance and is not a valid way to balance an equation.
- A subscript applies only to the atom or bracketed group immediately before it, not to earlier elements in the formula.
- A coefficient in normal script before a formula multiplies every atom in that formula unit.
- When counting atoms, evaluate subscripts within the formula before applying any coefficient.
- Brackets with a subscript multiply every element inside the bracket, as in Ca(OH)₂.
- Correct interpretation of multipliers is needed to balance equations and to calculate relative formula masses.
- In 2Ca(NO₃)₂ the coefficient doubles the whole formula, giving 2 Ca, 4 N and 12 O atoms.
Examiner Tips
- 💡Name the gas that escapes or is gained when explaining an apparent change in mass, for example carbon dioxide escaping or oxygen being gained from the air.
- 💡Use the phrase closed system when describing a situation where the measured mass stays constant.
- 💡Link your explanation back to atoms being rearranged, because this is the reason the masses are equal.
- 💡Balance one element at a time and tick it off once its count matches on both sides.
- 💡Do a final count of every element, including any you balanced early, because changing one coefficient can affect another element.
- 💡Write the simplest whole-number ratio of coefficients in your final answer.
- 💡Underline or circle each coefficient and subscript before counting atoms so you do not confuse their roles.
- 💡Write out an atom tally for each side of an equation; this makes balancing errors visible and easy to correct.
- 💡Check bracketed groups separately, multiplying each element inside by the subscript outside the bracket.
- 💡If a count seems odd, re-read the formula from left to right, applying subscripts before coefficients.
- 💡Perform an atom audit before and after writing coefficients: list each element and count its atoms across both sides to ensure equality.
- 💡When explaining mass changes, explicitly identify the reaction vessel as an 'open system' or 'closed system' to secure AQA marking criteria.
- 💡Check for diatomic molecules (H2, N2, O2, F2, Cl2, Br2, I2) when writing balanced equations from word descriptions.
Common Mistakes
- Thinking that mass can disappear when a gas is produced; the correction is that the gas has moved into the surroundings, so the mass of the reaction mixture falls but the total mass of all substances is unchanged.
- Believing that a reaction can create new atoms; the correction is that atoms are rearranged into new substances, not made.
- Assuming that mass always stays the same in an open container; the correction is that a gas may escape or be taken in, so the measured mass can change even though total mass is conserved.
- Changing a subscript in a formula to make the counts match; the correction is to change only the coefficient in front of the formula.
- Forgetting to multiply every atom in a formula by its coefficient; the correction is to count each element after applying the coefficient to the whole formula.
- Leaving the equation with fractional coefficients; the correction is to multiply all coefficients by the same factor to obtain the simplest whole-number ratio.
- Treating a subscript as multiplying the whole formula, so reading H₂O as four hydrogen atoms; the correction is that the subscript multiplies only the preceding symbol, giving two hydrogen atoms.
- Applying a coefficient to only the first element of a formula, so reading 2CO₂ as two carbon and two oxygen atoms; the correction is that the coefficient multiplies every atom, giving two carbon and four oxygen atoms.
- Ignoring brackets when counting, so reading Ca(OH)₂ as one oxygen and one hydrogen; the correction is that the subscript after the bracket multiplies each element inside, giving two oxygen and two hydrogen atoms.
- Believing that mass is destroyed when an effervescent reaction bubbles away, rather than recognizing that gas has simply escaped into the room.
- Modifying formula subscripts to balance equations (e.g., writing H2 + O2 -> H2O2 instead of 2H2 + O2 -> 2H2O), which creates an entirely different chemical compound.
- Assuming that mass increases in oxidation reactions because mass was created, rather than accounting for oxygen gas combining from the air.
Revision Plan
- 1Day 1-2: Review chemical formulas and practice balancing 15-20 progressively harder symbol equations without changing subscripts.
- 2Day 3-4: Study open versus closed systems, focusing on experiments involving thermal decomposition and combustion of metals.
- 3Day 5: Practice exam-style quantitative questions calculating missing reactant or product masses using conservation laws.
- 4Day 6-7: Complete timed past paper questions covering AQA Paper 1 Quantitative Chemistry and review against mark schemes.
Exam Question Types
- 📋Explain the change in mass: Short-answer questions (2-3 marks) asking why the mass increased (e.g., heating magnesium) or decreased (e.g., acid on marble chips).
- 📋Balancing equations: 1-2 mark questions requiring balancing numbers in front of partially completed chemical equations.
- 📋Mass calculation from conservation: Calculating the exact mass of a reactant or product using given masses of other participants.
Command Word Expectations (AQA)
Give a detailed account of the causes or reasons behind an observation (e.g., state why mass decreased by identifying the gas produced and noting that it escaped the unsealed container).
Use the provided numerical data to reach a mathematical answer, showing working clearly and stating units where required.
Insert whole-number coefficients in front of formulas so that the number of atoms of each element is equal on both sides of the arrow.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A student heats 10.0 g of calcium carbonate (CaCO3) in an open crucible until it completely decomposes into calcium oxide (CaO) and carbon dioxide (CO2). At the end of the experiment, 5.6 g of calcium oxide remains in the crucible. Explain why the mass appeared to decrease, and calculate the mass of carbon dioxide released.
- 1.Step 1: State the relevant conservation law. The law of conservation of mass dictates that Total Mass of Reactants = Total Mass of Products.
- 2.Step 2: Explain the apparent decrease in mass. Calcium carbonate decomposes according to the equation: CaCO3(s) -> CaO(s) + CO2(g). Because the crucible is an open system, the carbon dioxide gas escapes into the air, reducing the balance reading.
- 3.Step 3: Calculate the mass of the escaping gas using the conservation equation: Mass of CaCO3 = Mass of CaO + Mass of CO2. Substitute the known values: 10.0 g = 5.6 g + Mass of CO2.
- 4.Step 4: Solve for the unknown mass: Mass of CO2 = 10.0 g - 5.6 g = 4.4 g.
Question: Balance the following chemical equation and include the correct state symbols: Al(s) + Fe2O3(s) -> Al2O3(s) + Fe(s).
- 1.Step 1: Count the atoms of each element on the reactant side: Al = 1, Fe = 2, O = 3.
- 2.Step 2: Count the atoms of each element on the product side: Al = 2, Fe = 1, O = 3.
- 3.Step 3: Balance aluminium by placing a coefficient of 2 in front of Al on the reactant side: 2Al(s) + Fe2O3(s) -> Al2O3(s) + Fe(s).
- 4.Step 4: Balance iron by placing a coefficient of 2 in front of Fe on the product side: 2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(s).
- 5.Step 5: Verify all atoms: Reactants: 2 Al, 2 Fe, 3 O. Products: 2 Al, 2 Fe, 3 O. Ensure state symbols are retained.