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03/14 Chemical Composition and the Mole

A progressive guide to chemical formulas, nomenclature, mole calculations, molar mass, percent composition, and empirical and molecular formulas.

Reading Chemical Formulas

Chemical formulas translate particle-level composition into symbolic form. Element symbols identify the elements, subscripts show how many atoms of the preceding element are present, and a missing subscript means one.

For example, H2O\mathrm{H_2O} contains two hydrogen atoms and one oxygen atom in each molecule. A coefficient multiplies the entire formula, so 3H2O3\mathrm{H_2O} represents three water molecules, containing six hydrogen atoms and three oxygen atoms. Parentheses multiply a group: Ca(NO3)2\mathrm{Ca(NO_3)_2} contains one calcium atom, two nitrogen atoms, and six oxygen atoms per .

A gives the actual atoms in one molecule. An ionic compound is represented by a , the simplest electrically neutral ratio of its ions. An gives the simplest whole-number ratio of elements. Thus, glucose has C6H12O6\mathrm{C_6H_{12}O_6} and CH2O\mathrm{CH_2O}, while NaCl\mathrm{NaCl} represents the ionic ratio of sodium ions to chloride ions.

Takeaway: Read subscripts as composition within one particle and coefficients as the number of particles represented.

Formulas and Chemical

Ionic compounds must have an overall charge of zero. Write the cation first and the anion second, then choose subscripts that make the total positive and negative charge equal.

For aluminum oxide, the ions are Al3+\mathrm{Al^{3+}} and O2−\mathrm{O^{2-}}. The least common multiple of the charge magnitudes is 66, so two aluminum ions and three oxide ions are needed:

Al2O3\mathrm{Al_2O_3}

Do not carry ion charges into the final neutral formula, and reduce subscripts to the lowest whole-number ratio. When more than one polyatomic ion is required, use parentheses, as in Ca(NO3)2\mathrm{Ca(NO_3)_2}.

names substances systematically. For ionic compounds, name the cation first and the anion second. A monatomic anion usually ends in “-ide,” and a metal that forms more than one charge uses a Roman numeral: FeCl2\mathrm{FeCl_2} is iron(II) chloride, whereas FeCl3\mathrm{FeCl_3} is iron(III) chloride.

For binary molecular compounds, use numerical prefixes to show atom counts. Examples include carbon monoxide, CO\mathrm{CO}; carbon dioxide, CO2\mathrm{CO_2}; and dinitrogen pentoxide, N2O5\mathrm{N_2O_5}. Prefixes describe shared-electron molecules and are not used to determine ionic charges.

For introductory acid , binary acids use hydro- plus the element root and “-ic acid,” as in hydrochloric acid, HCl\mathrm{HCl}. Oxyacids from “-ate” ions end in “-ic acid,” while those from “-ite” ions end in “-ous acid.”

Takeaway: Identify the compound type before applying naming rules: charge balance for ionic compounds, prefixes for binary molecular compounds, and characteristic endings for acids.

The and Particle Counting

The connects microscopic particles with laboratory-scale quantities. One contains number of specified entities:

1 mol=6.02214076×1023 entities1\ \mathrm{mol}=6.02214076\times10^{23}\ \text{entities}

The entities must be identified. One of CO2\mathrm{CO_2} contains 6.02214076×10236.02214076\times10^{23} carbon dioxide molecules, while one of carbon contains the same number of carbon atoms.

Use conversion factors so that units cancel. For 0.2500.250 mol of water:

0.250 mol H2O(6.022×1023 molecules H2O1 mol H2O)=1.51×1023 molecules H2O0.250\ \mathrm{mol\ H_2O}\left(\frac{6.022\times10^{23}\ \mathrm{molecules\ H_2O}}{1\ \mathrm{mol\ H_2O}}\right)=1.51\times10^{23}\ \mathrm{molecules\ H_2O}

Each water molecule contains two hydrogen atoms, so:

1.51×1023 molecules H2O(2 H atoms1 H2O molecule)=3.01×1023 H atoms1.51\times10^{23}\ \mathrm{molecules\ H_2O}\left(\frac{2\ \mathrm{H\ atoms}}{1\ \mathrm{H_2O\ molecule}}\right)=3.01\times10^{23}\ \mathrm{H\ atoms}

Takeaway: Particle-count problems require both the conversion and the particle ratio supplied by the .

and Mass– Conversions

is the mass of one of a substance in g/mol\mathrm{g/mol}. Its numerical value matches the formula mass in atomic mass units per or the molecular mass in atomic mass units per molecule.

Add the atomic masses represented by the formula. For water:

M(H2O)=2(1.008)+16.00=18.016 g/mol≈18.02 g/molM(\mathrm{H_2O})=2(1.008)+16.00=18.016\ \mathrm{g/mol}\approx18.02\ \mathrm{g/mol}

The three related quantities are amount in moles nn, mass in grams mm, and MM:

n=mM,m=nM,M=mnn=\frac{m}{M},\qquad m=nM,\qquad M=\frac{m}{n}

For 36.0436.04 g of water:

n=36.04 g H2O(1 mol H2O18.02 g H2O)=2.000 mol H2On=36.04\ \mathrm{g\ H_2O}\left(\frac{1\ \mathrm{mol\ H_2O}}{18.02\ \mathrm{g\ H_2O}}\right)=2.000\ \mathrm{mol\ H_2O}

Ionic compounds use formula mass rather than molecular mass, but the mass calculation follows the same addition of atomic masses. For calcium chloride:

M(CaCl2)=40.08+2(35.45)=110.98 g/molM(\mathrm{CaCl_2})=40.08+2(35.45)=110.98\ \mathrm{g/mol}

Takeaway: Choose the equation that isolates the requested quantity, then track units through the calculation.

and Empirical Formulas

measures the mass contribution of each element. For an element XX:

%X=mass of X in one mole of compoundmolar mass of compound×100%\%X=\frac{\text{mass of }X\text{ in one mole of compound}}{\text{molar mass of compound}}\times100\%

For water, one contains 2.0162.016 g of hydrogen and 16.0016.00 g of oxygen, with total 18.0218.02 g/mol:

%H=2.01618.02×100%=11.19%\%\mathrm{H}=\frac{2.016}{18.02}\times100\%=11.19\%
%O=16.0018.02×100%=88.81%\%\mathrm{O}=\frac{16.00}{18.02}\times100\%=88.81\%

The values add to 100.00%100.00\%. Small discrepancies in other problems may result from rounding.

To determine an from percentages, treat the percentages as grams in an assumed 100.0100.0-g sample. Convert each mass to moles using n=m/Mn=m/M, divide every value by the smallest value, and convert the resulting ratios to the smallest whole numbers. For a compound containing 40.0%40.0\% carbon, 6.71%6.71\% hydrogen, and 53.29%53.29\% oxygen, the approximate amounts are 3.333.33, 6.666.66, and 3.333.33 mol. Dividing by 3.333.33 gives the ratio 1:2:11:2:1, so the is CH2O\mathrm{CH_2O}.

If a ratio is close to a fraction such as 1.51.5, 1.331.33, or 1.251.25, multiply all ratios by 22, 33, or 44, respectively, before writing the formula.

Takeaway: Convert mass information to moles before reducing a composition to an element ratio.

From Empirical to Molecular Formulas

A is a whole-number multiple of the . When the compound’s is known, first calculate the empirical-formula mass and then find the multiplier:

n=molar massempirical-formula massn=\frac{\text{molar mass}}{\text{empirical-formula mass}}

Suppose the is CH2O\mathrm{CH_2O} and the compound’s is 180.16180.16 g/mol. The empirical-formula mass is:

M(CH2O)=12.01+2(1.008)+16.00=30.03 g/molM(\mathrm{CH_2O})=12.01+2(1.008)+16.00=30.03\ \mathrm{g/mol}

The multiplier is:

n=180.1630.03≈6n=\frac{180.16}{30.03}\approx6

Multiply every empirical-formula subscript by 66:

(CH2O)6=C6H12O6(\mathrm{CH_2O})_6=\boxed{\mathrm{C_6H_{12}O_6}}

The result is consistent because the multiplier is a small whole number. The gives only the simplest ratio, while the gives the actual number of atoms in each molecule.

For any composition calculation, write the formula first, distinguish molecules from formula units, use dimensional analysis, retain extra digits during intermediate steps, and round only at the end. Check that ionic formulas are neutral, empirical subscripts are smallest whole numbers, and elemental percentages sum to approximately 100%100\%.

Final takeaway: Composition problems move through a reliable chain: formula interpretation, mass or conversion, ratio simplification, and a final check of charge, units, and significant figures.