The mole, composition and measurement
A balance does not count atoms, it measures grams. A chemical reaction knows nothing about grams, only about how many particles collide. The mole is the bridge between what you measure in the lab (mass) and what really happens at atomic scale (numbers of particles).
The mole and the Avogadro constant
Weigh out $12.01\text{ g}$ of carbon and $4.00\text{ g}$ of helium. The two masses are not equal, and the two elements are very different, yet each is one mole, so both samples contain the same enormous number of atoms: $6.022 \times 10^{23}$ of them. How can we define a unit that lets us count microscopic particles by weighing them?
The molar mass of an element in $\text{g mol}^{-1}$ has the same number as its relative atomic mass $A_r$. For a compound, add up the $A_r$ values of all the atoms in the formula to get the relative molecular mass $M_r$ (it has no units); the molar mass has the same number, with units $\text{g mol}^{-1}$. Checking that your units cancel (unit analysis) is the best way to catch a slip.
A sample of $44.10\text{ g}$ of propane gas ($\mathrm{C_3H_8}$) is placed in a closed container. a) Calculate the amount of substance of propane contained in the…
A sample of $44.10\text{ g}$ of propane gas ($\mathrm{C_3H_8}$) is placed in a closed container.
- a) Calculate the amount of substance of propane contained in the container and determine the total number of propane molecules present.
- b) Calculate the number of moles of hydrogen atoms and the absolute number of hydrogen atoms present in the sample.
Data: Atomic molar masses: $\mathrm{C} = 12.01\text{ g}\times\text{mol}^{-1}$; $\mathrm{H} = 1,008\text{ g}\times\text{mol}^{-1}$. Avogadro constant: $N_{\mathrm{A}} = 6,022\times 10^{23}\text{ mol}^{-1}$.
Molar mass with brackets and water of crystallisation
Many salts crystallise with water trapped inside the crystal. This is called water of crystallisation, and the salt is a hydrate. In a formula such as $\mathrm{CuSO_4\cdot 5H_2O}$, the dot means "with $5$ water molecules for every formula unit of $\mathrm{CuSO_4}$". When you work out the molar mass, you include the water.
Brackets work the same way: in $\mathrm{Ca(OH)_2}$ the subscript $2$ multiplies everything inside, so there are $2$ O atoms and $2$ H atoms.
Iron(II) sulfate usually crystallises as the heptahydrate, $\mathrm{FeSO_4\times 7H_2O}$. In a gravimetry practical, the starting point is a pure crystal of this salt…
Iron(II) sulfate usually crystallises as the heptahydrate, $\mathrm{FeSO_4\times 7H_2O}$. In a gravimetry practical, the starting point is a pure crystal of this salt with a mass of $6,950\text{ g}$.
- a) Calculate the mass fraction and the percentage by mass of water of crystallisation present in the compound.
- b) Determine the total amount of substance of oxygen atoms present in the $6,950\text{ g}$ of salt, specifying how many moles of these atoms come from the water of hydration molecules and how many come from the sulfate anion.
Data: Atomic molar masses: $\mathrm{Fe} = 55.85\text{ g}\times\text{mol}^{-1}$; $\mathrm{S} = 32.06\text{ g}\times\text{mol}^{-1}$; $\mathrm{O} = 16.00\text{ g}\times\text{mol}^{-1}$; $\mathrm{H} = 1,008\text{ g}\times\text{mol}^{-1}$.
Relative atomic mass from isotopes
Most elements in nature are mixtures of isotopes with different masses. Isotopes are atoms of the same element (same number of protons) with different numbers of neutrons. The relative atomic mass $A_r$ is the weighted mean mass of the atoms of an element, compared with $\tfrac{1}{12}$ of the mass of a carbon-12 atom. The value in the periodic table is this weighted mean.
The same equation can be used backwards. If $A_r$ is known and there are two isotopes, call the fractional abundances $x$ and $1 - x$.
Natural gallium consists exclusively of two stable isotopes: $^{69}\mathrm{Ga}$, with isotopic mass $68,926\text{ u}$ and a natural abundance of $60.11\,\%$, and…
Natural gallium consists exclusively of two stable isotopes: $^{69}\mathrm{Ga}$, with isotopic mass $68,926\text{ u}$ and a natural abundance of $60.11\,\%$, and $^{71}\mathrm{Ga}$, with isotopic mass $70,925\text{ u}$ and an abundance of $39.89\,\%$.
- a) Calculate the weighted mean relative atomic mass of gallium, giving the result to four significant figures.
- b) Briefly explain why the atomic mass given in the periodic table for chemical elements rarely coincides with an exact integer.
Natural boron has a mean atomic mass of $10,811\text{ u}$ and is composed only of two stable isotopes: $^{10}\mathrm{B}$, with isotopic mass $10.0129\text{ u}$, and…
Natural boron has a mean atomic mass of $10,811\text{ u}$ and is composed only of two stable isotopes: $^{10}\mathrm{B}$, with isotopic mass $10.0129\text{ u}$, and $^{11}\mathrm{B}$, with isotopic mass $11.0093\text{ u}$.
- a) Set up the corresponding system of equations and determine, to four significant figures, the percentage natural abundance of each of the two isotopes.
- b) In a piece of pure elemental boron of $2.7028\text{ g}$, calculate the mass due exclusively to the isotope $^{10}\mathrm{B}$ and the total number of neutrons contributed jointly by all the $^{10}\mathrm{B}$ atoms contained in that piece.
Data: Atomic number of boron: $Z = 5$. Avogadro constant: $N_{\mathrm{A}} = 6,022\times 10^{23}\text{ mol}^{-1}$.
Significant figures and uncertainty in a measurement
So $2.3376\text{ g}$ has $5$ significant figures, and is written $(2.3376 \pm 0.0001)\ \text{g}$. Zeros between digits count, leading zeros do not ($0.0766$ has $3$), and trailing zeros after a decimal point do count ($4.4750$ has $5$).
Two ideas are easy to confuse. Accuracy is how close a measurement is to the true value; precision is how close repeated measurements are to each other. A random error makes repeated readings scatter either side of the true value; averaging several readings reduces it. A systematic error (for example a balance that has not been zeroed) shifts every reading the same way, so repeating does not remove it. Taring the balance and calibrating instruments helps guard against it.
Always read the balance carefully before using numbers from a diagram: write down both readings with all their decimal places, including zeros.
In a laboratory test, an analytical balance with an absolute uncertainty of $\pm 0.0001\text{ g}$ is used. The watch glass is first tared and a portion of sodium…
In a laboratory test, an analytical balance with an absolute uncertainty of $\pm 0.0001\text{ g}$ is used. The watch glass is first tared and a portion of sodium chloride ($\mathrm{NaCl}$) is weighed, with a reading of $2.3376\text{ g}$ recorded on the display.
- a) State the number of significant figures of this mass measurement and express the weighing result together with its associated absolute uncertainty.
- b) Calculate the amount of substance (in moles) of $\mathrm{NaCl}$ weighed, giving the result to the correct number of significant figures.
Data: Atomic molar masses: $\mathrm{Na} = 22.99\text{ g}\times\text{mol}^{-1}$; $\mathrm{Cl} = 35.45\text{ g}\times\text{mol}^{-1}$.
In the laboratory, a pure anhydrous sample of copper(II) sulfate, $\mathrm{CuSO_4}$, is weighed. To avoid direct contact with the pan of the single-pan digital balance,…
In the laboratory, a pure anhydrous sample of copper(II) sulfate, $\mathrm{CuSO_4}$, is weighed. To avoid direct contact with the pan of the single-pan digital balance, a clean watch glass is used. First the empty watch glass is read, and then the solid crystalline salt is placed on it, giving the readings shown in the following instrument diagram:
Using the experimental measurements shown:
- a) Calculate the net mass of $\mathrm{CuSO_4}$ and the amount of substance weighed, rounding to the correct number of significant figures.
- b) Determine the number of formula units of $\mathrm{CuSO_4}$ weighed and the total number of oxygen atoms present in that portion.
Data: Atomic molar masses: $\mathrm{Cu} = 63.55\text{ g}\times\text{mol}^{-1}$; $\mathrm{S} = 32.06\text{ g}\times\text{mol}^{-1}$; $\mathrm{O} = 16.00\text{ g}\times\text{mol}^{-1}$. Avogadro constant: $N_{\mathrm{A}} = 6,022\times 10^{23}\text{ mol}^{-1}$. Reading of the empty glass: $15,320\text{ g}$; reading with sample: $19,795\text{ g}$.
Empirical and molecular formulae
Formaldehyde, a toxic preservative, and glucose, a basic fuel for our cells, both contain $40.0$% C, $6.7$% H and $53.3$% O by mass. How can two substances with such different properties have the same proportion of elements?
Formaldehyde is $\mathrm{CH_2O}$ and glucose is $\mathrm{C_6H_{12}O_6}$: same empirical formula, different $k$ (1 and 6).
Quantitative elemental analysis of a pure sample of an open-chain carboxylic acid shows a percentage composition by mass of $40.00\,\%$ carbon, $6.71\,\%$ hydrogen and…
Quantitative elemental analysis of a pure sample of an open-chain carboxylic acid shows a percentage composition by mass of $40.00\,\%$ carbon, $6.71\,\%$ hydrogen and $53.29\,\%$ oxygen. By a complementary physical method, its molar mass is found to be $90.08\text{ g}\times\text{mol}^{-1}$.
- a) Determine, with justification, the empirical formula of this organic compound.
- b) Deduce its molecular formula and propose a compatible semi-structural formula containing one carboxyl group and one hydroxyl group.
Data: Atomic molar masses: $\mathrm{C} = 12.01\text{ g}\times\text{mol}^{-1}$; $\mathrm{H} = 1,008\text{ g}\times\text{mol}^{-1}$; $\mathrm{O} = 16.00\text{ g}\times\text{mol}^{-1}$.
Formulae from combustion data, gas measurements and mass loss
Sometimes there is no percentage composition, only experimental masses. Three tools turn these into moles.
Combustion analysis. When a compound of C, H (and N or O) burns completely, all its carbon ends up in $\mathrm{CO_2}$ and all its hydrogen in $\mathrm{H_2O}$. So $n(\mathrm{C}) = n(\mathrm{CO_2})$ and $n(\mathrm{H}) = 2\,n(\mathrm{H_2O})$. Nitrogen or oxygen can be found by difference: subtract the masses of C and H from the sample mass.
A molar mass from a gas. For an ideal gas, $PV = nRT$ (it is in the Formulae and Tables booklet). Use SI units: $P$ in Pa ($1\text{ atm} = 101\,325\text{ Pa}$), $V$ in $\text{m}^3$ ($1\text{ mL} = 10^{-6}\text{ m}^3$), $T$ in K ($T = \theta + 273.15$) and $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$. Then $n = PV/RT$ and $M = m/n$.
Mass loss on heating. If a solid releases a gas on heating, the mass lost is the mass of that gas (when nothing else leaves or enters).
A gaseous volatile organic compound consists only of carbon, hydrogen and nitrogen. Complete combustion of a sample of $1,452\text{ g}$ of this substance produces…
A gaseous volatile organic compound consists only of carbon, hydrogen and nitrogen. Complete combustion of a sample of $1,452\text{ g}$ of this substance produces $2,201\text{ g}$ of $\mathrm{CO_2}$ and $1,351\text{ g}$ of $\mathrm{H_2O}$, with all the nitrogen being converted into gaseous diatomic nitrogen ($\mathrm{N_2}$).
Separately, a sample of $0.5900\text{ g}$ of the same compound is placed in a closed flask of $311.0\text{ mL}$ at a constant temperature of $100.0\,^\circ\mathrm{C}$, reaching an equilibrium pressure of $1,000\text{ atm}$. Assuming ideal gas behaviour:
- a) Determine the mass of each element in the combusted sample and deduce the empirical formula.
- b) Calculate the experimental molar mass of the gas and determine its definitive molecular formula.
Data: $R = 0.08206\text{ atm}\times\text{L}\times\text{mol}^{-1}\times\text{K}^{-1}$. Atomic molar masses: $\mathrm{C} = 12.01\text{ g}\times\text{mol}^{-1}$; $\mathrm{H} = 1,008\text{ g}\times\text{mol}^{-1}$; $\mathrm{N} = 14.01\text{ g}\times\text{mol}^{-1}$; $\mathrm{O} = 16.00\text{ g}\times\text{mol}^{-1}$.
A solid sample of $5.0000\text{ g}$ consisting exclusively of a homogeneous mixture of sodium chloride ($\mathrm{NaCl}$) and calcium carbonate ($\mathrm{CaCO_3}$) is…
A solid sample of $5.0000\text{ g}$ consisting exclusively of a homogeneous mixture of sodium chloride ($\mathrm{NaCl}$) and calcium carbonate ($\mathrm{CaCO_3}$) is subjected to prolonged calcination in an open crucible. At the working temperature, the sodium chloride remains unchanged, whereas the calcium carbonate decomposes completely on heating, releasing carbon dioxide gas according to the reaction: $\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}$.
After cooling the crucible in a desiccator containing silica gel to avoid absorption of atmospheric moisture, the dry solid residue has a final mass of $3.8998\text{ g}$.
- a) Calculate the mass of $\mathrm{CaCO_3}$ and of $\mathrm{NaCl}$ present in the initial mixture, as well as the percentage by mass of each component.
- b) Express the composition of the original mixture as mole fractions and explain, with reasoning, the sign of the error that would be made in the percentage of calcium carbonate if the crucible had been left to cool in the open air without a desiccator.
Data: Atomic molar masses: $\mathrm{Ca} = 40.08\text{ g}\times\text{mol}^{-1}$; $\mathrm{C} = 12.01\text{ g}\times\text{mol}^{-1}$; $\mathrm{O} = 16.00\text{ g}\times\text{mol}^{-1}$; $\mathrm{Na} = 22.99\text{ g}\times\text{mol}^{-1}$; $\mathrm{Cl} = 35.45\text{ g}\times\text{mol}^{-1}$.
Summary
Back to the start: the balance gives grams, the molar mass converts them to moles, and $N_{\mathrm{A}}$ turns moles into counted particles.