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Chemical reactions · The fundamental laws of chemistry

Quantitative Calculations in Everyday Life

A fertilizer label, a lab analysis, or the spec sheet for an iron ore almost never gives the chemical formula: they give percentages. "NPK 15-15-15," "ore with 80% purity," "85.7% carbon." How do you go from those percentages to real kilograms, and from there to the formula of the compound? That is what you will learn here.

The Percent Composition of a Compound

If you break down $100\ \text{g}$ of pure water, you always get $11.1\ \text{g}$ of hydrogen and $88.9\ \text{g}$ of oxygen, no matter which faucet or bottle the water comes from. Can you predict that proportion from the formula $\text{H}_2\text{O}$ alone? Yes.

You use the mass of the whole formula as the "total" and the mass each element contributes as the "part." It is a proportion: part divided by total, times one hundred.

Segmented bar: the molar mass of water, 18 g/mol, is split between hydrogen (11.1 %) and oxygen (88.9 %). H: 11.1 % O: 88.9 % H₂O: 18 g/mol split between H and O
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Calculate the percentage composition of water, $\text{H}2\text{O}$ (atomic masses: $\text{H} = 1\ \text{u}$; $\text{O} = 16\ \text{u}$). Check that the two percentages…

Calculate the percentage composition of water, $\text{H}_2\text{O}$ (atomic masses: $\text{H} = 1\ \text{u}$; $\text{O} = 16\ \text{u}$). Check that the two percentages add up to $100\ \%$.

Solve
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Carbon dioxide emitted by cars is the main greenhouse gas. Calculate the mass percentage of carbon and oxygen in $\text{CO}2$ (atomic masses: $\text{C} = 12\ \text{u}$;…

Carbon dioxide emitted by cars is the main greenhouse gas. Calculate the mass percentage of carbon and oxygen in $\text{CO}_2$ (atomic masses: $\text{C} = 12\ \text{u}$; $\text{O} = 16\ \text{u}$).

Solve
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Table salt is sodium chloride, $\text{NaCl}$. Calculate the mass percentage of sodium and chlorine (atomic masses: $\text{Na} = 23\ \text{u}$;…

Table salt is sodium chloride, $\text{NaCl}$. Calculate the mass percentage of sodium and chlorine (atomic masses: $\text{Na} = 23\ \text{u}$; $\text{Cl} = 35.5\ \text{u}$). Nutritionists warn about the sodium in salt: explain, using your results, why $5\ \text{g}$ of salt is not $5\ \text{g}$ of sodium.

Solve
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Urea, $\text{CO}(\text{NH}2)2$, is the most widely used nitrogenous fertiliser in the world (atomic masses: $\text{C} = 12\ \text{u}$; $\text{O} = 16\ \text{u}$;…

Urea, $\text{CO}(\text{NH}_2)_2$, is the most widely used nitrogenous fertiliser in the world (atomic masses: $\text{C} = 12\ \text{u}$; $\text{O} = 16\ \text{u}$; $\text{N} = 14\ \text{u}$; $\text{H} = 1\ \text{u}$).

  1. a) Calculate its molar mass and its mass percentage of nitrogen.
  2. b) How many kilograms of nitrogen does a bag of $25\ \text{kg}$ of urea provide?
Solve

Using Percent Composition with Real Masses

With the percentage of an element you can calculate how much of that element is in any sample: $m_{\text{element}} = m_{\text{sample}} \times \dfrac{\%}{100}$.

This is the same thing farmers do with fertilizers. The label "NPK 15-15-15" says the bag contains $15\%$ nitrogen, $15\%$ phosphorus(V) oxide, and $15\%$ potassium oxide. In a $20\ \text{kg}$ bag, each nutrient provides $20 \times 0.15 = 3\ \text{kg}$.

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A bag of fertiliser from an agricultural warehouse has the label "NPK $15$-$15$-$15$" printed on it, indicating $15\ \%$ by mass of nitrogen, $15\ \%$ of phosphorus(V)…

A bag of fertiliser from an agricultural warehouse has the label "NPK $15$-$15$-$15$" printed on it, indicating $15\ \%$ by mass of nitrogen, $15\ \%$ of phosphorus(V) oxide, and $15\ \%$ of potassium oxide.

  1. a) Explain what the three numbers on the label mean.
  2. b) If the bag contains $20\ \text{kg}$ of fertiliser, how many kilograms of each nutrient does it provide?
Solve
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A tank contains $90\ \text{g}$ of water. Knowing that the percentage composition of water is $11.1\ \%$ of hydrogen and $88.9\ \%$ of oxygen, as summarised by the bar:…

A tank contains $90\ \text{g}$ of water. Knowing that the percentage composition of water is $11.1\ \%$ of hydrogen and $88.9\ \%$ of oxygen, as summarised by the bar:

90 g of water H: 11.1 % → ? g O: 88.9 % → ? g
  1. a) How many grams of hydrogen and how many of oxygen does the tank contain?
  2. b) Check that the two masses you calculated add up to the total mass of the water.
Solve
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A farmer buys NPK fertiliser $15$-$15$-$15$ in $40\ \text{kg}$ bags. a) How many kilograms of nitrogen does each bag contain? b) If their plot needs an input of…

A farmer buys NPK fertiliser $15$-$15$-$15$ in $40\ \text{kg}$ bags.

  1. a) How many kilograms of nitrogen does each bag contain?
  2. b) If their plot needs an input of $12\ \text{kg}$ of nitrogen, how many bags should they apply? How many kilograms of phosphorus(V) oxide and potassium oxide will they be adding in the process?
Solve
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Sodium nitrate, $\text{NaNO}3$, known as "Chilean nitrate", was one of the first commercial nitrogenous fertilisers, of mineral origin (atomic masses:…

Sodium nitrate, $\text{NaNO}_3$, known as "Chilean nitrate", was one of the first commercial nitrogenous fertilisers, of mineral origin (atomic masses: $\text{Na} = 23\ \text{u}$; $\text{N} = 14\ \text{u}$; $\text{O} = 16\ \text{u}$).

  1. a) Calculate its percentage by mass of nitrogen.
  2. b) How many kilograms of nitrogen does a $40\ \text{kg}$ bag provide?
Solve

Hydrates: Water Inside the Crystal

Some salts trap water molecules in their crystal. Copper(II) sulfate pentahydrate, $\text{CuSO}_4 \cdot 5\ \text{H}_2\text{O}$ (the blue salt), has $5$ water molecules for every unit of $\text{CuSO}_4$. Its molar mass already includes that water: $249.5\ \text{g/mol}$.

The percentage of water is calculated like any other: $\dfrac{5 \times 18}{249.5} \times 100 = 36.1\%$. When you heat the salt until it is dehydrated, that percentage of the mass escapes as vapor.

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Copper(II) sulphate pentahydrate, $\text{CuSO}4 \times 5\ \text{H}2\text{O}$, is the blue salt used in crystallisation practicals. Its molar mass is…

Copper(II) sulphate pentahydrate, $\text{CuSO}_4 \times 5\ \text{H}_2\text{O}$, is the blue salt used in crystallisation practicals. Its molar mass is $249.5\ \text{g/mol}$ and that of water is $18\ \text{g/mol}$. The figure shows what "pentahydrate" means:

copper(II) sulphate 5 H₂O of crystallisation
  1. a) Calculate the percentage by mass of water of crystallisation in the hydrated salt.
  2. b) If $50\ \text{g}$ of the blue salt is heated until completely dehydrated, what mass of water is given off?
Solve

Purity of a Raw Material

In mining and industry, almost nothing is $100\%$ pure. An iron ore that arrives at the steel mill carries useful mineral mixed with worthless rock (gangue).

Bar of 50 kg of ore: 80 % (40 kg) is iron(III) oxide and 20 % (10 kg) is impurities. Ore: 50 kg (100 %) 80 % Fe₂O₃ 20 % 40 kg of Fe₂O₃ (useful) 10 kg of impurities
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An iron ore arrives at the steelworks with a purity of $80\ \%$ in iron(III) oxide. a) How many kilograms of iron(III) oxide are in $50\ \text{kg}$ of ore? b) What does…

An iron ore arrives at the steelworks with a purity of $80\ \%$ in iron(III) oxide.

  1. a) How many kilograms of iron(III) oxide are in $50\ \text{kg}$ of ore?
  2. b) What does the remaining $20\ \%$ refer to?
Solve
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A mine supplies an iron ore with a purity of $80\ \%$ iron(III) oxide, $\text{Fe}2\text{O}3$ (atomic masses: $\text{Fe} = 56\ \text{u}$; $\text{O} = 16\ \text{u}$). a)…

A mine supplies an iron ore with a purity of $80\ \%$ iron(III) oxide, $\text{Fe}_2\text{O}_3$ (atomic masses: $\text{Fe} = 56\ \text{u}$; $\text{O} = 16\ \text{u}$).

  1. a) Calculate the percentage by mass of iron in the pure oxide.
  2. b) How many kilograms of iron can be obtained, at most, from one metric ton ($1\,000\ \text{kg}$) of this ore?
Solve
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A steelworks can buy two iron ores. Ore A is haematite with a purity of $80\ \%$ in $\text{Fe}2\text{O}3$; Ore B is magnetite with a purity of $75\ \%$ in…

A steelworks can buy two iron ores. Ore A is haematite with a purity of $80\ \%$ in $\text{Fe}_2\text{O}_3$; Ore B is magnetite with a purity of $75\ \%$ in $\text{Fe}_3\text{O}_4$ (atomic masses: $\text{Fe} = 56\ \text{u}$; $\text{O} = 16\ \text{u}$).

  1. a) Calculate the percentage of iron in each of the two pure oxides.
  2. b) Calculate how many kilograms of iron are obtained from one metric ton of each ore.
  3. c) If both ores cost the same per metric ton, which one is worth buying? Justify the decision with the previous calculations.
Solve
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An agricultural technician needs to apply $46\ \text{kg}$ of nitrogen to a field and is deciding between two fertilisers: urea containing $46\ \%$ nitrogen, in…

An agricultural technician needs to apply $46\ \text{kg}$ of nitrogen to a field and is deciding between two fertilisers: urea containing $46\ \%$ nitrogen, in $25\ \text{kg}$ bags at $\$500$ per bag, or ammonium nitrate containing $34\ \%$ nitrogen, in $25\ \text{kg}$ bags at $\$400$ per bag.

  1. a) Calculate how many kilograms of each fertiliser are needed to provide the $46\ \text{kg}$ of nitrogen.
  2. b) Taking into account that only whole bags can be bought, calculate how many bags of each type must be bought.
  3. c) Which of the two options is cheaper? Justify this by calculating the total cost of each option.
Solve

From Composition to the Empirical Formula

A lab analyzes a pure substance and finds, in $100\ \text{mg}$, $40.0\ \text{mg}$ of carbon, $6.7\ \text{mg}$ of hydrogen, and $53.3\ \text{mg}$ of oxygen. How do you deduce the formula of the compound?

Three steps: convert percentages to grams in 100 g, convert grams to moles, and divide by the smallest number of moles. % → g take 100 g of sample g → mol ÷ atomic mass of each ÷ smallest by the smallest number of moles
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Glucose has the molecular formula $\text{C}6\text{H}{12}\text{O}6$. a) What is its empirical formula? Explain what information each of the two formulas gives. b) Give an…

Glucose has the molecular formula $\text{C}_6\text{H}_{12}\text{O}_6$.

  1. a) What is its empirical formula? Explain what information each of the two formulas gives.
  2. b) Give an example of a compound whose empirical and molecular formulas coincide and explain why this occurs.
Solve
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The analysis of a laboratory moisturiser shows a percentage composition of $40.0\ \%$ of carbon, $6.7\ \%$ of hydrogen, and $53.3\ \%$ of oxygen (atomic masses:…

The analysis of a laboratory moisturiser shows a percentage composition of $40.0\ \%$ of carbon, $6.7\ \%$ of hydrogen, and $53.3\ \%$ of oxygen (atomic masses: $\text{C} = 12\ \text{u}$; $\text{H} = 1\ \text{u}$; $\text{O} = 16\ \text{u}$).

  1. a) Calculate the moles of each element present in $100\ \text{g}$ of the compound.
  2. b) Deduce its empirical formula.
Solve
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The analysis of a reddish rock from a quarry shows that the iron oxide it contains has a percentage composition of $70.0\ \%$ iron and $30.0\ \%$ oxygen (atomic masses:…

The analysis of a reddish rock from a quarry shows that the iron oxide it contains has a percentage composition of $70.0\ \%$ iron and $30.0\ \%$ oxygen (atomic masses: $\text{Fe} = 56\ \text{u}$; $\text{O} = 16\ \text{u}$).

  1. a) Determine the empirical formula of the oxide from the percentages.
  2. b) Calculate the percentage of iron in the three known oxides, $\text{FeO}$, $\text{Fe}_2\text{O}_3$ and $\text{Fe}_3\text{O}_4$, and check that only one matches the analysis.
  3. c) Why can percentage analysis not distinguish between pure $\text{Fe}_2\text{O}_3$ and a mixture of oxides that gives the same overall proportion? Comment on what other test you would perform.
Solve
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A sample of $4.0\ \text{g}$ of a highly corrosive cleaning product contains $2.3\ \text{g}$ sodium, $1.6\ \text{g}$ oxygen and $0.1\ \text{g}$ hydrogen (atomic masses:…

A sample of $4.0\ \text{g}$ of a highly corrosive cleaning product contains $2.3\ \text{g}$ sodium, $1.6\ \text{g}$ oxygen and $0.1\ \text{g}$ hydrogen (atomic masses: $\text{Na} = 23\ \text{u}$; $\text{O} = 16\ \text{u}$; $\text{H} = 1\ \text{u}$).

  1. a) Check that the three partial masses add up to the total mass of the sample and interpret this check using the law of conservation of mass.
  2. b) Determine the empirical formula of the compound.
  3. c) Its measured molar mass is $40\ \text{g/mol}$: do the empirical and molecular formulas coincide in this case? What substance is it?
Solve

From the Empirical Formula to the Molecular Formula

Three CH₂ units of 14 g/mol each form the C₃H₆ molecule of 42 g/mol, with k = 3. CH₂ CH₂ CH₂ 14 g/mol 14 g/mol 14 g/mol k = 42 / 14 = 3: C₃H₆ (42 g/mol)

When you are not given the molar mass, sometimes you can get it from the density of a gas at standard conditions: $M = \rho \times 22.4\ \text{L/mol}$ (only valid at $0\ °\text{C}$ and $1\ \text{atm}$).

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A gaseous hydrocarbon has a percentage composition of $85.7\ \%$ carbon and $14.3\ \%$ hydrogen, and its molar mass is $42\ \text{g/mol}$ (atomic masses:…

A gaseous hydrocarbon has a percentage composition of $85.7\ \%$ carbon and $14.3\ \%$ hydrogen, and its molar mass is $42\ \text{g/mol}$ (atomic masses: $\text{C} = 12\ \text{u}$; $\text{H} = 1\ \text{u}$).

  1. a) Determine its empirical formula.
  2. b) Determine its molecular formula. What additional piece of data was needed compared to the previous part?
Solve
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A liquid hydrocarbon found in gasoline has a percentage composition of $85.7\ \%$ carbon and $14.3\ \%$ hydrogen, and its molar mass is $70\ \text{g/mol}$ (atomic…

A liquid hydrocarbon found in gasoline has a percentage composition of $85.7\ \%$ carbon and $14.3\ \%$ hydrogen, and its molar mass is $70\ \text{g/mol}$ (atomic masses: $\text{C} = 12\ \text{u}$; $\text{H} = 1\ \text{u}$).

  1. a) Determine its empirical formula and the molar mass of that empirical formula.
  2. b) Determine its molecular formula and check that it is consistent with the given molar mass.
  3. c) Explain why percentage analysis alone can never distinguish between two compounds with the same empirical formula.
Solve
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A gas sold in cylinders for cooking has a density of $1.96\ \text{g/L}$ at standard temperature and pressure (molar volume under these conditions: $22.4\ \text{L/mol}$)…

A gas sold in cylinders for cooking has a density of $1.96\ \text{g/L}$ at standard temperature and pressure (molar volume under these conditions: $22.4\ \text{L/mol}$) and a percentage composition of $81.8\ \%$ carbon and $18.2\ \%$ hydrogen (atomic masses: $\text{C} = 12\ \text{u}$; $\text{H} = 1\ \text{u}$).

  1. a) Calculate its molar mass from the density and the molar volume.
  2. b) Determine its empirical formula.
  3. c) Determine its molecular formula and identify which gas it is.
Solve

Formulas from a Combustion Analysis

If you burn a hydrocarbon, all the carbon ends up in the $\text{CO}_2$ and all the hydrogen in the $\text{H}_2\text{O}$. By weighing those products you recover the masses of C and H in the sample:

  • $m_\text{C} = m_{\text{CO}_2} \times \dfrac{12}{44}$

  • $m_\text{H} = m_{\text{H}_2\text{O}} \times \dfrac{2}{18}$

If $m_\text{C} + m_\text{H}$ equals the mass burned, the compound contains only C and H (conservation of mass). Then you convert to moles and follow the procedure for the empirical formula.

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In a combustion analysis, $4.4\ \text{g}$ of a hydrocarbon is completely burned and $13.2\ \text{g}$ of carbon dioxide and $7.2\ \text{g}$ of water are collected (atomic…

In a combustion analysis, $4.4\ \text{g}$ of a hydrocarbon is completely burned and $13.2\ \text{g}$ of carbon dioxide and $7.2\ \text{g}$ of water are collected (atomic masses: $\text{C} = 12\ \text{u}$; $\text{H} = 1\ \text{u}$; $\text{O} = 16\ \text{u}$; molar masses: $\text{CO}_2 = 44\ \text{g/mol}$; $\text{H}_2\text{O} = 18\ \text{g/mol}$). The diagram summarises the analysis:

hydrocarbon 4.4 g + O₂ combustion CO₂ 13.2 g H₂O 7.2 g
  1. a) Calculate the mass of carbon contained in the $\text{CO}_2$ collected and the mass of hydrogen contained in the water collected. Check that they add up to $4.4\ \text{g}$: which stoichiometric law justifies this check?
  2. b) Determine the empirical formula of the hydrocarbon.
  3. c) Knowing that the $4.4\ \text{g}$ burned was $0.1\ \text{mol}$ of substance, determine its molecular formula.
Solve

Back to the beginning: in a $25\ \text{kg}$ bag of NPK 15-15-15, $15\%$ nitrogen means $25 \times 0.15 = 3.75\ \text{kg}$ of nitrogen: you translate the percentage on the label into real mass. And with the composition from an analysis, plus the molar mass, you arrive at the formula.

Summary

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