Weight Laws and Stoichiometric Relationships
Antoine Lavoisier weighed everything. Before him, chemists described reactions in words ("the metal is calcined," "the air is spoiled"); he put them on a balance, before and after, and discovered something that seems obvious today but that changed chemistry back then: the total mass doesn't change by even a gram. Imagine you drop an effervescent tablet into a glass sitting on the kitchen scale: when the fizzing stops, why does it read less? From that scale and two other equally precise observations came the three laws you will use every time you balance an equation or calculate how much reactant you need.
The Law of Conservation of Mass
If you burn wood in a bonfire, the ashes weigh much less than the wood; on the other hand, if you heat steel wool in air, the oxidized residue weighs more than the starting metal. Where does the mass that seems to disappear go, or where does the mass that is gained come from?
The answer to the puzzle of the wood and the steel wool: in the bonfire, some of the products (CO₂ and water vapor) escape into the air; in the steel, the oxygen from the air adds to the metal. In both cases mass is conserved if you weigh everything.
We dissolve an effervescent tablet of $3\ \text{g}$ in a beaker of $100\ \text{g}$ of water. When the bubbling stops, the whole system weighs $102.5\ \text{g}$. Observe…
We dissolve an effervescent tablet of $3\ \text{g}$ in a beaker of $100\ \text{g}$ of water. When the bubbling stops, the whole system weighs $102.5\ \text{g}$. Observe the diagram of the process:
- a) What mass of gas has escaped into the air?
- b) Does this result contradict the law of conservation of mass? Reason your answer.
When a candle is weighed on a balance before and after being left to burn for a while, we find that the whole system has lost mass. The diagram shows what happens: a)…
When a candle is weighed on a balance before and after being left to burn for a while, we find that the whole system has lost mass. The diagram shows what happens:
- a) Does this mean that matter has been destroyed? Explain what has actually happened.
- b) How would you design the experiment to check that the total mass is conserved?
In a kiln, $10\ \text{g}$ of calcium carbonate decomposes due to heat, producing $5.6\ \text{g}$ of calcium oxide and $4.4\ \text{g}$ of carbon dioxide, as summarised in…
In a kiln, $10\ \text{g}$ of calcium carbonate decomposes due to heat, producing $5.6\ \text{g}$ of calcium oxide and $4.4\ \text{g}$ of carbon dioxide, as summarised in the diagram:
- a) Check that the law of conservation of mass is obeyed.
- b) If $25\ \text{g}$ of calcium carbonate decomposes, what mass of each product will be obtained? Which second law of chemical combination are you applying?
The Law of Definite Proportions
In a hermetically sealed crucible, a magnesium ribbon is heated with oxygen. Exactly $2.4\ \text{g}$ of magnesium reacts with $1.6\ \text{g}$ of oxygen, forming…
In a hermetically sealed crucible, a magnesium ribbon is heated with oxygen. Exactly $2.4\ \text{g}$ of magnesium reacts with $1.6\ \text{g}$ of oxygen, forming magnesium oxide, as summarised in the figure:
- a) What mass of magnesium oxide is obtained?
- b) Which law of chemical combination have you applied? State it.
The analysis of any sample of pure water shows that hydrogen and oxygen always combine in the ratio of $1\ \text{g}$ of hydrogen for every $8\ \text{g}$ of oxygen. a)…
The analysis of any sample of pure water shows that hydrogen and oxygen always combine in the ratio of $1\ \text{g}$ of hydrogen for every $8\ \text{g}$ of oxygen.
- a) What mass of oxygen combines with $5\ \text{g}$ of hydrogen?
- b) What mass of water is formed? Which laws of chemical combination have you applied in each part?
In a laboratory, two samples of copper(II) oxide from different sources are analysed. Sample A contains $3.97\ \text{g}$ of copper and $1.00\ \text{g}$ of oxygen; Sample…
In a laboratory, two samples of copper(II) oxide from different sources are analysed. Sample A contains $3.97\ \text{g}$ of copper and $1.00\ \text{g}$ of oxygen; Sample B contains $11.91\ \text{g}$ of copper and $3.00\ \text{g}$ of oxygen.
- a) Calculate the ratio $\frac{\text{masa de cobre}}{\text{masa de oxígeno}}$ in each sample.
- b) Could they be the same compound? Which law of chemical combination guarantees this?
A sample of $15.9\ \text{g}$ of copper(II) oxide contains $12.7\ \text{g}$ of copper and $3.2\ \text{g}$ of oxygen. a) What mass of oxygen will be combined with…
A sample of $15.9\ \text{g}$ of copper(II) oxide contains $12.7\ \text{g}$ of copper and $3.2\ \text{g}$ of oxygen.
- a) What mass of oxygen will be combined with $25.4\ \text{g}$ of copper in another sample of the same oxide?
- b) What will be the total mass of that second sample?
Carbon reacts with oxygen in the ratio of $12\ \text{g}$ of carbon for every $32\ \text{g}$ of oxygen, forming carbon dioxide. $6\ \text{g}$ of carbon is reacted with…
Carbon reacts with oxygen in the ratio of $12\ \text{g}$ of carbon for every $32\ \text{g}$ of oxygen, forming carbon dioxide. $6\ \text{g}$ of carbon is reacted with $20\ \text{g}$ of oxygen.
- a) What mass of oxygen is consumed and how much is left over?
- b) What mass of carbon dioxide is formed? Check that the total mass is conserved.
A company buys industrial copper(II) oxide and suspects that it arrives with impurities. Pure oxide combines copper and oxygen in the ratio of $63.5\ \text{g}$ of copper…
A company buys industrial copper(II) oxide and suspects that it arrives with impurities. Pure oxide combines copper and oxygen in the ratio of $63.5\ \text{g}$ of copper for every $16\ \text{g}$ of oxygen. Analysis of a $10.0\ \text{g}$ sample of the product received shows that it contains only $7.6\ \text{g}$ of copper.
- a) Calculate what mass of copper a sample of $10.0\ \text{g}$ of pure copper(II) oxide should contain.
- b) Is the product received pure? Justify your answer using the law of definite proportions.
- c) Assuming that the impurity does not contain copper, estimate the purity of the product as a percentage.
The Law of Multiple Proportions
Carbon and oxygen form two different gases: for a fixed $3\ \text{g}$ of carbon, one contains $4\ \text{g}$ of oxygen and the other $8\ \text{g}$. Why doesn't oxygen combine in just any amount, but in exact jumps?
Carbon forms two gaseous oxides. In one of them, $3\ \text{g}$ of carbon combines with $4\ \text{g}$ of oxygen; in the other, $3\ \text{g}$ of carbon combines with…
Carbon forms two gaseous oxides. In one of them, $3\ \text{g}$ of carbon combines with $4\ \text{g}$ of oxygen; in the other, $3\ \text{g}$ of carbon combines with $8\ \text{g}$ of oxygen, as shown in the figure:
- a) Calculate the ratio between the two masses of oxygen that combine with the same mass of carbon.
- b) Which law of chemical combination do these data illustrate, and who stated it?
Dalton's atomic model states that atoms are indivisible and that in a chemical reaction they are only rearranged, being neither created nor destroyed, as represented by…
Dalton's atomic model states that atoms are indivisible and that in a chemical reaction they are only rearranged, being neither created nor destroyed, as represented by the particle model in the figure:
- a) Explain, using this model, why mass is conserved in any chemical reaction.
- b) Which law of chemical combination directly explains this idea?
Two iron oxide minerals are analysed. For the first, it is found that $7.0\ \text{g}$ of iron is combined with $3.0\ \text{g}$ of oxygen; for the second, that…
Two iron oxide minerals are analysed. For the first, it is found that $7.0\ \text{g}$ of iron is combined with $3.0\ \text{g}$ of oxygen; for the second, that $7.0\ \text{g}$ of iron is combined with $2.0\ \text{g}$ of oxygen.
- a) Calculate the iron:oxygen ratio in each mineral.
- b) Could they be the same compound? Justify your answer using the law of definite proportions.
- c) Knowing that the possible oxides are $\text{FeO}$ and $\text{Fe}_2\text{O}_3$ (atomic masses: $\text{Fe} = 56\ \text{u}$; $\text{O} = 16\ \text{u}$), identify each mineral.
Nitrogen forms several oxides. Fixing the mass of nitrogen at $14\ \text{g}$, the masses of combined oxygen are $16\ \text{g}$ in nitrogen monoxide, $32\ \text{g}$ in…
Nitrogen forms several oxides. Fixing the mass of nitrogen at $14\ \text{g}$, the masses of combined oxygen are $16\ \text{g}$ in nitrogen monoxide, $32\ \text{g}$ in nitrogen dioxide, and $40\ \text{g}$ in dinitrogen pentoxide.
- a) Express the ratio $16 : 32 : 40$ as a ratio of simple integers.
- b) State the law of chemical combination that these data predict and explain its meaning.
Sulphur forms two oxides with formulas $\text{SO}2$ and $\text{SO}3$ (atomic masses: $\text{S} = 32\ \text{u}$; $\text{O} = 16\ \text{u}$). a) Calculate what mass of…
Sulphur forms two oxides with formulas $\text{SO}_2$ and $\text{SO}_3$ (atomic masses: $\text{S} = 32\ \text{u}$; $\text{O} = 16\ \text{u}$).
- a) Calculate what mass of oxygen combines with $6.4\ \text{g}$ of sulphur in each of the two oxides.
- b) Check that these two masses of oxygen are in a ratio of simple integers. Which law is confirmed?
Chlorine forms four oxides: $\text{Cl}2\text{O}$, $\text{Cl}2\text{O}3$, $\text{Cl}2\text{O}5$, and $\text{Cl}2\text{O}7$ (atomic masses: $\text{Cl} = 35.5\ \text{u}$;…
Chlorine forms four oxides: $\text{Cl}_2\text{O}$, $\text{Cl}_2\text{O}_3$, $\text{Cl}_2\text{O}_5$, and $\text{Cl}_2\text{O}_7$ (atomic masses: $\text{Cl} = 35.5\ \text{u}$; $\text{O} = 16\ \text{u}$).
- a) Calculate the mass of oxygen that combines with $71\ \text{g}$ of chlorine in each of the four oxides.
- b) Check that these masses are in the ratio of simple integers $1 : 3 : 5 : 7$.
- c) Predict what mass of oxygen would combine with $71\ \text{g}$ of chlorine in a hypothetical oxide $\text{Cl}_2\text{O}_9$ and state in your own words the law you have used.
In a laboratory for the analysis of pollutant gases, two samples of nitrogen oxides are studied. Sample A contains $0.70\ \text{g}$ of nitrogen and $0.80\ \text{g}$ of…
In a laboratory for the analysis of pollutant gases, two samples of nitrogen oxides are studied. Sample A contains $0.70\ \text{g}$ of nitrogen and $0.80\ \text{g}$ of oxygen; Sample B contains $0.70\ \text{g}$ of nitrogen and $1.60\ \text{g}$ of oxygen (atomic masses: $\text{N} = 14\ \text{u}$; $\text{O} = 16\ \text{u}$).
- a) Check that the masses of oxygen combined with the same mass of nitrogen are in a ratio of simple integers. Which law is satisfied?
- b) Calculate the moles of each element in each sample and deduce the atom ratio $\text{N} : \text{O}$ in each oxide.
- c) Assign each sample its formula from among $\text{NO}$, $\text{NO}_2$, $\text{N}_2\text{O}$, and $\text{N}_2\text{O}_5$.
From Weight Laws to Stoichiometry
Together, the three laws explain why a balanced equation is not just a convention: it is a statement about mass proportions that can be checked with a balance and lets you predict how much reactant is needed or how much product is obtained, in any amount.
The synthesis of water shows that $4\ \text{g}$ of hydrogen reacts with $32\ \text{g}$ of oxygen to yield $36\ \text{g}$ of water, as represented by the particle model…
The synthesis of water shows that $4\ \text{g}$ of hydrogen reacts with $32\ \text{g}$ of oxygen to yield $36\ \text{g}$ of water, as represented by the particle model in the figure:
- a) If we start with $8\ \text{g}$ of hydrogen, what mass of oxygen is needed?
- b) What mass of water will form? State which laws you have used.
In water, the ratio by mass is $1\ \text{g}$ of hydrogen for every $8\ \text{g}$ of oxygen, and each molecule contains $2$ hydrogen atoms for every oxygen atom. a)…
In water, the ratio by mass is $1\ \text{g}$ of hydrogen for every $8\ \text{g}$ of oxygen, and each molecule contains $2$ hydrogen atoms for every oxygen atom.
- a) Deduce how many times heavier an oxygen atom is than a hydrogen atom.
- b) Explain why this deduction is only possible if it is known how many atoms of each element make up the molecule, and what would happen if the formula of water were $\text{HO}$.
The industrial synthesis of ammonia is summarised as follows: $28\ \text{g}$ of nitrogen reacts with $6\ \text{g}$ of hydrogen to yield $34\ \text{g}$ of ammonia. a)…
The industrial synthesis of ammonia is summarised as follows: $28\ \text{g}$ of nitrogen reacts with $6\ \text{g}$ of hydrogen to yield $34\ \text{g}$ of ammonia.
- a) What mass of hydrogen is needed to react with $56\ \text{g}$ of nitrogen?
- b) What mass of ammonia will then be produced?
- c) State which two laws of chemical combination you have applied and in which part.
Lavoisier obtained oxygen by heating mercury oxide, a red powder that decomposes into liquid mercury and gaseous oxygen. On heating $21.7\ \text{g}$ of oxide,…
Lavoisier obtained oxygen by heating mercury oxide, a red powder that decomposes into liquid mercury and gaseous oxygen. On heating $21.7\ \text{g}$ of oxide, $20.1\ \text{g}$ of mercury and $1.6\ \text{g}$ of oxygen are collected.
- a) Verify numerically that the law of conservation of mass is satisfied.
- b) If $43.4\ \text{g}$ of the same oxide were decomposed, what mass of mercury and oxygen would be obtained?
- c) Which law of chemical combination guarantees that the mercury:oxygen ratio is always the same in this compound?
In a laboratory practical, $2.4\ \text{g}$ of magnesium ribbon is burned in an open crucible. Knowing that magnesium combines with oxygen in the ratio of $2.4\ \text{g}$…
In a laboratory practical, $2.4\ \text{g}$ of magnesium ribbon is burned in an open crucible. Knowing that magnesium combines with oxygen in the ratio of $2.4\ \text{g}$ of magnesium for every $1.6\ \text{g}$ of oxygen, the oxide formed should weigh $4.0\ \text{g}$, but when weighing the ash, only $3.6\ \text{g}$ is collected.
- a) Explain why the ash weighs more than the initial ribbon, even though the metal has "disappeared".
- b) Reason why less oxide than expected is collected when the crucible is open.
- c) Calculate what fraction of the product has been lost, express it as a percentage.
Dalton knew two experimental facts: in water, $1\ \text{g}$ of hydrogen combines with $8\ \text{g}$ of oxygen; in hydrogen peroxide, $1\ \text{g}$ of hydrogen combines…
Dalton knew two experimental facts: in water, $1\ \text{g}$ of hydrogen combines with $8\ \text{g}$ of oxygen; in hydrogen peroxide, $1\ \text{g}$ of hydrogen combines with $16\ \text{g}$ of oxygen.
- a) Which law of chemical combination do these two facts jointly illustrate? Verify this using the ratio $8 : 16$.
- b) Dalton assumed that water was $\text{HO}$. With that hypothesis, what formula would correspond to hydrogen peroxide and what relative atomic mass would oxygen have?
- c) Today we know that water is $\text{H}_2\text{O}$. Deduce the formula of hydrogen peroxide and the correct relative atomic mass of oxygen, and explain why atomic masses depend on knowing the formulas.
$10\ \text{g}$ of hydrogen and $64\ \text{g}$ of oxygen are introduced into a closed reactor to produce water. The stoichiometric ratio by mass is $1\ \text{g}$ of…
$10\ \text{g}$ of hydrogen and $64\ \text{g}$ of oxygen are introduced into a closed reactor to produce water. The stoichiometric ratio by mass is $1\ \text{g}$ of hydrogen for every $8\ \text{g}$ of oxygen.
- a) Determine which reactant is used up first and justify your answer with a calculation.
- b) Calculate the mass of water formed and the mass of the remaining reactant.
- c) Check that the total mass inside the reactor before and after the reaction is the same. Which law requires this?
In a lime kiln, $200\ \text{g}$ of limestone (pure calcium carbonate) is heated to obtain $112\ \text{g}$ of quicklime (calcium oxide) and $88\ \text{g}$ of carbon…
In a lime kiln, $200\ \text{g}$ of limestone (pure calcium carbonate) is heated to obtain $112\ \text{g}$ of quicklime (calcium oxide) and $88\ \text{g}$ of carbon dioxide.
- a) Check that mass is conserved in the process.
- b) Express the stoichiometric mass ratio calcium carbonate : calcium oxide : carbon dioxide using integers.
- c) Predict what mass of quicklime and carbon dioxide $500\ \text{g}$ of limestone will produce, indicating which two laws of chemical combination make this prediction possible.
Go back to the beginning: in the glass with the effervescent tablet, the scale reads less because the CO₂ produced escapes into the air. The mass wasn't destroyed: if you weigh everything, in a closed container, the total is the same.