Stoichiometric Calculations
The gas tank at your home runs empty and someone asks: "How much oxygen does the stove use to burn all that gas?" The balanced equation only says that $1$ molecule of fuel needs $2$ of oxygen. To find out how many kilograms of air are used, or how much CO₂ goes out into the atmosphere, you need a bridge between particles and the scale: the mole. At the end of the page we'll come back to this question.
Molar mass: from weight in grams to moles
If you weigh $12\ \text{g}$ of carbon and $12\ \text{g}$ of magnesium, both scales read the same. But a magnesium atom has twice the mass of a carbon atom, so the magnesium pan holds half as many atoms. How can you count particles just by weighing?
The glucose used by your muscles when playing sport has the formula $\text{C}6\text{H}{12}\text{O}6$. Calculate its molar mass and explain in a sentence what this value…
The glucose used by your muscles when playing sport has the formula $\text{C}_6\text{H}_{12}\text{O}_6$. Calculate its molar mass and explain in a sentence what this value means (atomic masses: $\text{H} = 1\ \text{u}$; $\text{C} = 12\ \text{u}$; $\text{O} = 16\ \text{u}$).
A glass contains $36\ \text{g}$ of water. Knowing that the molar mass of water is $18\ \text{g/mol}$, calculate how many moles of water are in the glass.
A glass contains $36\ \text{g}$ of water. Knowing that the molar mass of water is $18\ \text{g/mol}$, calculate how many moles of water are in the glass.
A flask contains $0.5\ \text{mol}$ of carbon dioxide. Knowing that the molar mass of $\text{CO}2$ is $44\ \text{g/mol}$, calculate the mass of gas contained in the flask.
A flask contains $0.5\ \text{mol}$ of carbon dioxide. Knowing that the molar mass of $\text{CO}_2$ is $44\ \text{g/mol}$, calculate the mass of gas contained in the flask.
A balanced equation is read in moles
A balanced equation tells you how many particles react with each other. Since a mole is simply a large, fixed package of particles, the same coefficients also apply to moles.
$$\text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O}$$It reads: $1$ mol of methane reacts with $2$ mol of oxygen and produces $1$ mol of CO₂ and $2$ mol of water. If you double the moles of methane, all the others double too.
The balanced equation for the combustion of methane, $\text{CH}4 + 2\ \text{O}2 \rightarrow \text{CO}2 + 2\ \text{H}2\text{O}$, can also be read in moles. a) How many…
The balanced equation for the combustion of methane, $\text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O}$, can also be read in moles.
- a) How many moles of oxygen are needed to burn $3\ \text{mol}$ of methane?
- b) How many moles of carbon dioxide and how many of water are formed then?
The equation for the formation of water is $2\ \text{H}2 + \text{O}2 \rightarrow 2\ \text{H}2\text{O}$. If $4\ \text{mol}$ of hydrogen is completely burned: a) How many…
The equation for the formation of water is $2\ \text{H}_2 + \text{O}_2 \rightarrow 2\ \text{H}_2\text{O}$. If $4\ \text{mol}$ of hydrogen is completely burned:
- a) How many moles of oxygen are consumed?
- b) How many moles of water are formed?
In the synthesis of ammonia, $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$, $6\ \text{mol}$ of hydrogen reacts with sufficient nitrogen. a) How many moles of…
In the synthesis of ammonia, $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$, $6\ \text{mol}$ of hydrogen reacts with sufficient nitrogen.
- a) How many moles of nitrogen are consumed?
- b) How many moles of ammonia are obtained?
The general method: mass → mole → mole → mass
In the lab and in industry, you don't measure moles, you weigh grams. That's why almost every problem follows four steps.
In a barbecue, $12\ \text{g}$ of carbon burns according to the equation $\text{C} + \text{O}2 \rightarrow \text{CO}2$. Calculate, by converting through moles, what mass…
In a barbecue, $12\ \text{g}$ of carbon burns according to the equation $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$. Calculate, by converting through moles, what mass of carbon dioxide is formed (atomic masses: $\text{C} = 12\ \text{u}$; $\text{O} = 16\ \text{u}$).
A home boiler burns $16\ \text{g}$ of methane according to the equation $\text{CH}4 + 2\ \text{O}2 \rightarrow \text{CO}2 + 2\ \text{H}2\text{O}$. Calculate, by…
A home boiler burns $16\ \text{g}$ of methane according to the equation $\text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O}$. Calculate, by converting through moles:
- a) The mass of oxygen it consumes.
- b) The mass of carbon dioxide it emits.
(molar masses: $\text{CH}_4 = 16\ \text{g/mol}$; $\text{O}_2 = 32\ \text{g/mol}$; $\text{CO}_2 = 44\ \text{g/mol}$).
An electrolysis apparatus decomposes $90\ \text{g}$ of water according to the equation $2\ \text{H}2\text{O} \rightarrow 2\ \text{H}2 + \text{O}2$. Calculate, by…
An electrolysis apparatus decomposes $90\ \text{g}$ of water according to the equation $2\ \text{H}_2\text{O} \rightarrow 2\ \text{H}_2 + \text{O}_2$. Calculate, by converting through moles:
- a) The mass of hydrogen obtained.
- b) The mass of oxygen released.
(molar masses: $\text{H}_2\text{O} = 18\ \text{g/mol}$; $\text{H}_2 = 2\ \text{g/mol}$; $\text{O}_2 = 32\ \text{g/mol}$).
More practice with the same method, using different equations:
A steel part containing $56\ \text{g}$ of iron oxidises completely according to the equation $4\ \text{Fe} + 3\ \text{O}2 \rightarrow 2\ \text{Fe}2\text{O}3$. Calculate,…
A steel part containing $56\ \text{g}$ of iron oxidises completely according to the equation $4\ \text{Fe} + 3\ \text{O}_2 \rightarrow 2\ \text{Fe}_2\text{O}_3$. Calculate, by converting through moles, the mass of iron(III) oxide formed (molar masses: $\text{Fe} = 56\ \text{g/mol}$; $\text{Fe}_2\text{O}_3 = 160\ \text{g/mol}$).
A reactor aims to produce $34\ \text{g}$ of ammonia according to the equation $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$. Calculate, by converting through…
A reactor aims to produce $34\ \text{g}$ of ammonia according to the equation $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$. Calculate, by converting through moles, the minimum mass of hydrogen that must be fed in (molar masses: $\text{H}_2 = 2\ \text{g/mol}$; $\text{NH}_3 = 17\ \text{g/mol}$).
Compare the oxygen required by two gaseous fuels. On the one hand, the combustion of methane: $\text{CH}4 + 2\ \text{O}2 \rightarrow \text{CO}2 + 2\ \text{H}2\text{O}$;…
Compare the oxygen required by two gaseous fuels. On the one hand, the combustion of methane: $\text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O}$; on the other, that of propane: $\text{C}_3\text{H}_8 + 5\ \text{O}_2 \rightarrow 3\ \text{CO}_2 + 4\ \text{H}_2\text{O}$.
- a) Calculate the mass of oxygen required to burn $16\ \text{g}$ of methane.
- b) Calculate the mass of oxygen required to burn $44\ \text{g}$ of propane.
- c) Express both results in grams of oxygen per gram of fuel and reason, looking at the formulas of the two gases, why the one with a higher proportion of hydrogen in its molecule consumes more oxygen per gram.
(molar masses: $\text{CH}_4 = 16\ \text{g/mol}$; $\text{C}_3\text{H}_8 = 44\ \text{g/mol}$; $\text{O}_2 = 32\ \text{g/mol}$).
Impure raw materials: purity
Real raw materials are almost never pure. The limestone that is heated to make lime, or iron ore, contains impurities that don't react. Only the pure part goes into the calculation.
In a lime kiln, $100\ \text{g}$ of calcium carbonate decomposes according to the equation $\text{CaCO}3 \rightarrow \text{CaO} + \text{CO}2$. Calculate, by converting…
In a lime kiln, $100\ \text{g}$ of calcium carbonate decomposes according to the equation $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$. Calculate, by converting through moles, the mass of quicklime ($\text{CaO}$) obtained (molar masses: $\text{CaCO}_3 = 100\ \text{g/mol}$; $\text{CaO} = 56\ \text{g/mol}$).
A quarry supplies a lime factory with limestone containing $80\ \%$ calcium carbonate (the remainder is unreactive impurities). $250\ \text{g}$ of this rock is loaded…
A quarry supplies a lime factory with limestone containing $80\ \%$ calcium carbonate (the remainder is unreactive impurities). $250\ \text{g}$ of this rock is loaded into the kiln: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$. The bar shows the composition of the rock:
- a) What mass of calcium carbonate actually enters the kiln?
- b) Calculate, by converting through moles, the mass of quicklime ($\text{CaO}$) obtained.
(molar masses: $\text{CaCO}_3 = 100\ \text{g/mol}$; $\text{CaO} = 56\ \text{g/mol}$).
A steelworks receives iron ore with $80\ \%$ iron(III) oxide (the remainder is inert gangue). In the blast furnace, the oxide reacts according to…
A steelworks receives iron ore with $80\ \%$ iron(III) oxide (the remainder is inert gangue). In the blast furnace, the oxide reacts according to $\text{Fe}_2\text{O}_3 + 3\ \text{CO} \rightarrow 2\ \text{Fe} + 3\ \text{CO}_2$. $200\ \text{g}$ of ore is loaded:
- a) What mass of $\text{Fe}_2\text{O}_3$ does that charge contain?
- b) Calculate, by converting through moles, the mass of iron obtained.
(molar masses: $\text{Fe}_2\text{O}_3 = 160\ \text{g/mol}$; $\text{Fe} = 56\ \text{g/mol}$).
The limiting reactant
When you make tacos, if you have 30 tortillas and 10 steaks, you only have enough for 10 tacos even though there are tortillas left over. The same thing happens in reactions.
In a container, $10\ \text{g}$ of hydrogen is mixed with $64\ \text{g}$ of oxygen and a spark is ignited: $2\ \text{H}2 + \text{O}2 \rightarrow 2\ \text{H}2\text{O}$.…
In a container, $10\ \text{g}$ of hydrogen is mixed with $64\ \text{g}$ of oxygen and a spark is ignited: $2\ \text{H}_2 + \text{O}_2 \rightarrow 2\ \text{H}_2\text{O}$. The diagram summarises the experiment data:
- a) Calculate the moles of each reactant and deduce which one is in excess.
- b) What mass of water is formed?
- c) What mass of excess reactant remains unreacted?
(molar masses: $\text{H}_2 = 2\ \text{g/mol}$; $\text{O}_2 = 32\ \text{g/mol}$; $\text{H}_2\text{O} = 18\ \text{g/mol}$).
$14\ \text{g}$ of nitrogen and $6\ \text{g}$ of hydrogen are introduced into a reactor to produce ammonia: $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$. a)…
$14\ \text{g}$ of nitrogen and $6\ \text{g}$ of hydrogen are introduced into a reactor to produce ammonia: $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$.
- a) Calculate the moles of each reactant and deduce which one is used up first.
- b) What mass of ammonia is obtained?
- c) What mass of the other reactant is left over?
(molar masses: $\text{N}_2 = 28\ \text{g/mol}$; $\text{H}_2 = 2\ \text{g/mol}$; $\text{NH}_3 = 17\ \text{g/mol}$).
In a furnace, $32\ \text{g}$ of methane is burned with $96\ \text{g}$ of oxygen: $\text{CH}4 + 2\ \text{O}2 \rightarrow \text{CO}2 + 2\ \text{H}2\text{O}$. a) Calculate…
In a furnace, $32\ \text{g}$ of methane is burned with $96\ \text{g}$ of oxygen: $\text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O}$.
- a) Calculate the moles of each reactant and deduce which one is the limiting reactant.
- b) What mass of carbon dioxide is emitted?
- c) What mass of oxygen is left over?
(molar masses: $\text{CH}_4 = 16\ \text{g/mol}$; $\text{O}_2 = 32\ \text{g/mol}$; $\text{CO}_2 = 44\ \text{g/mol}$).
Percent yield
The earlier calculations give the theoretical amount: the most you would get if everything reacted perfectly. In practice you always get less.
When calcining $100\ \text{g}$ of pure calcium carbonate, $\text{CaCO}3 \rightarrow \text{CaO} + \text{CO}2$, theory states that $56\ \text{g}$ of quicklime should be…
When calcining $100\ \text{g}$ of pure calcium carbonate, $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$, theory states that $56\ \text{g}$ of quicklime should be obtained. However, a real kiln has only produced $42\ \text{g}$, as compared by the bars in the figure.
- a) Calculate the percentage yield of the kiln.
- b) Propose a reason explaining why the kiln does not reach $100\ \%$.
According to equation $\text{N}2 + 3\ \text{H}2 \rightarrow 2\ \text{NH}3$, $28\ \text{g}$ of nitrogen should yield $34\ \text{g}$ of ammonia. In a real reactor, only…
According to equation $\text{N}_2 + 3\ \text{H}_2 \rightarrow 2\ \text{NH}_3$, $28\ \text{g}$ of nitrogen should yield $34\ \text{g}$ of ammonia. In a real reactor, only $20.4\ \text{g}$ has been collected.
- a) Calculate the percentage yield of the reaction.
- b) Explain two reasons why a factory might not achieve the theoretical yield of $100\ \%$.
Back to the beginning: when $1\ \text{kg}$ of methane ($M = 16\ \text{g/mol}$) burns, there are $62.5\ \text{mol}$; the $1:2$ ratio requires $125\ \text{mol}$ of O₂, or $125 \times 32 = 4{,}000\ \text{g}$. The stove needs $4\ \text{kg}$ of oxygen for every kilogram of gas, and it releases $62.5 \times 44 = 2.75\ \text{kg}$ of CO₂.