Lesson 1.3 · Unit 1 · CED topics 1.3 and 1.4

Composition of pure substances and mixtures

How do you know what a substance is? Color, melting point and smell help, but the most direct answer is to find out what it is made of and in what proportions. A chemist who burns a sample, weighs the products and works out the masses of carbon and hydrogen in it is doing elemental analysis, and from those masses the formula follows. The same measurement can also tell you how pure a sample is. This lesson covers both: composition by mass of pure substances, and composition of mixtures.

Learning objectives · 1.3.A and 1.4.A

1.3.A Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance. 1.4.A Explain the quantitative relationship between the elemental composition by mass and the composition of substances in a mixture.

Required content · EK 1.3.A.1 to 1.3.A.3 and 1.4.A.1 to 1.4.A.2
  • Some pure substances are composed of individual molecules, while others consist of atoms or ions held together in fixed proportions as described by a formula unit.
  • According to the law of definite proportions, the ratio of the masses of the constituent elements in any pure sample of that compound is always the same.
  • The chemical formula that lists the lowest whole number ratio of atoms of the elements in a compound is the empirical formula.
  • Pure substances contain atoms, molecules or formula units of a single type. Mixtures contain two or more types, whose relative proportions can vary.
  • Elemental analysis can be used to determine the relative numbers of atoms in a substance and to determine its purity.

Pure substances: fixed proportions

A pure substance is made of one kind of particle throughout. An element is pure (every atom is the same element); a compound is pure (every molecule or formula unit is the same). Some pure compounds are built of separate molecules (water, carbon dioxide, glucose). Others, such as sodium chloride and other ionic solids, are lattices with no separate molecules, so their formula describes a formula unit: the simplest repeating ratio of ions.

The law of definite proportions says that a compound always contains the same elements in the same ratio by mass. Pure water from a lake, from a glacier or from a chemistry lab is always 8 parts oxygen to 1 part hydrogen by mass (more precisely 8.0 g of oxygen for each 1.0 g of hydrogen). A sample in which the ratio were different would be either not water or not pure.

Mass of hydrogen and oxygen in three samples of pure waterSample 1O 8 gH 1 gmass ratio O:H = 8:1Sample 2O 16 gH 2 gmass ratio O:H = 8:1Sample 3O 4 gH 0.5 gmass ratio O:H = 8:1
Figure 1.3a The law of definite proportions: whatever the size of the sample or where it came from, pure water is always 8 parts oxygen to 1 part hydrogen by mass. The ratio identifies the compound.

Why is the ratio by mass fixed? Because the ratio of atoms is fixed (2 H for each O) and each kind of atom has its own fixed mass. Two hydrogen atoms (2 × 1.008 amu) bonded to one oxygen atom (15.999 amu) give 16.0 : 2.0, which is 8 : 1.

Empirical formulas

The empirical formula gives the smallest whole number ratio of atoms in a compound. Glucose is C6H12O6 (the molecular formula, the actual atoms in one molecule), but its empirical formula is CH2O because 6 : 12 : 6 reduces to 1 : 2 : 1. For ionic compounds the formula unit is the empirical formula: NaCl, CaCl2, Al2O3.

Experiments deliver masses or percentages, not atoms. To go from data to formula you convert to moles, because moles are proportional to numbers of atoms.

1Assume 100 gpercent becomes grams2Grams to molesdivide by molar mass3Divide by the smallestgives a mole ratio4Whole numbersmultiply if neededExample: 40.0% C, 6.7% H, 53.3% O → 3.33 : 6.65 : 3.33 → 1 : 2 : 1 → CH2O
Figure 1.3b From mass percent to empirical formula. The example is a compound that is 40.0% C, 6.7% H and 53.3% O by mass, which has the empirical formula CH2O.
Worked example · percent composition to empirical formula

A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula, and its molecular formula if the molar mass is 180.16 g/mol.

Step 1. Assume 100.0 g, so the masses are 40.0 g C, 6.7 g H, 53.3 g O.

Step 2, moles. C: 40.0 / 12.011 = 3.330 mol. H: 6.7 / 1.008 = 6.65 mol. O: 53.3 / 15.999 = 3.331 mol.

Step 3, divide by the smallest (3.330). C: 1.00. H: 2.00. O: 1.00. The ratio is 1 : 2 : 1, so the empirical formula is CH2O.

Step 4, molecular formula. The empirical formula mass is 12.011 + 2(1.008) + 15.999 = 30.026 g/mol. \(\dfrac{180.16}{30.026} = 6.00\), so multiply every subscript by 6: C6H12O6.

Worked example · when the ratio is not whole

An alcohol is 52.14% C, 13.13% H and 34.73% O. In 100.0 g: C 4.341 mol, H 13.03 mol, O 2.171 mol. Dividing by 2.171 gives 2.00 : 6.00 : 1.00, so the formula is C2H6O. If a ratio comes out as 1 : 1.5, multiply all values by 2; if 1 : 1.33, multiply by 3. A value such as 1.98 or 2.03 is rounding error and is rounded to 2, but 1.5 is not.

Worked example · a hydrate from mass data

A 2.500 g sample of blue copper(II) sulfate crystals is heated until all the water is driven off, leaving 1.598 g of white CuSO4. Find the formula of the hydrate, CuSO4·xH2O.

Mass of water lost: 2.500 − 1.598 = 0.902 g, which is 0.902 / 18.015 = 0.0501 mol H2O. Moles CuSO4: 1.598 / 159.602 = 0.01001 mol. Ratio: 0.0501 / 0.01001 = 5.00. The formula is CuSO4·5H2O.

Mixtures and purity

A mixture contains two or more kinds of particles in proportions that can vary. Salt water can be weakly or strongly salty; air has varying amounts of water vapor. This variability is the key difference from a compound. In a homogeneous mixture (a solution) the parts are uniformly spread; in a heterogeneous mixture (sand and gravel) they are not.

Elemental analysis tests purity. Suppose a sample is supposed to be calcium carbonate, CaCO3. Pure CaCO3 is \(\dfrac{40.078}{100.086} \times 100 = 40.04\%\) calcium by mass. If an analysis finds less than that, the sample contains other material with less calcium (or none).

0102030405040.04%Pure CaCO338.04%Sample A34%Sample BMass percent Ca
Figure 1.3c Elemental analysis and purity. Pure CaCO3 is 40.04% calcium by mass. A sample with a lower percentage contains other material that has less calcium (or none), and the ratio of the percentages gives the purity.
Worked example · purity from elemental analysis

A limestone sample is analyzed and found to contain 38.04% calcium by mass. Assuming the only calcium compound is CaCO3, what is the percent purity?

Purity \(= \dfrac{38.04}{40.04} \times 100 = 95.0\%\). A second sample that is 34.0% calcium is \(\dfrac{34.0}{40.04} = 84.9\%\) CaCO3, with the remainder impurities such as sand.

Worked example · composition of a mixture

A 1.000 g mixture of sodium chloride and potassium nitrate is dissolved, and analysis finds 0.2200 g of chlorine. Only the NaCl contains chlorine. NaCl is \(\dfrac{35.45}{58.44} = 60.66\%\) chlorine, so the mass of NaCl is \(0.2200 / 0.6066 = 0.3627\) g, and the mixture is 36.27% NaCl and 63.73% potassium nitrate by mass.

The skill in both purity problems is to identify which component carries the element you measured, find that element’s mass fraction in the component, and divide.

Exam focus
  • Percent means grams if you assume 100 g. This one move starts nearly every empirical formula problem.
  • Always convert to moles before comparing. Comparing masses tells you nothing about atom ratios.
  • Do not round a non-whole ratio. 1.5 means multiply by 2. Only round values within about 0.1 of a whole number.
  • Empirical is not molecular. The molecular formula is a whole-number multiple of the empirical one, found from the molar mass.
  • A pure substance has a fixed composition; a mixture does not. Use this to decide between a compound and a mixture from data on several samples.
Practice
  1. Iron(III) oxide is Fe2O3. What is its mass percent iron?
    Show answer

    M = 2(55.845) + 3(15.999) = 159.687 g/mol. Iron: \(111.690 / 159.687 = 0.6994\), so 69.94% iron.

  2. Two samples of a pure compound of carbon and oxygen contain 3.00 g C with 8.00 g O, and 6.00 g C with 16.0 g O. Does this agree with the law of definite proportions?
    Show answer

    Yes. The mass ratio O : C is 8.00 / 3.00 = 2.67 in the first sample and 16.0 / 6.00 = 2.67 in the second. The ratio is the same for different sample sizes, which is what the law predicts for a pure compound.

  3. Analysis of a compound gives 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula.
    Show answer

    In 100 g: C = 85.7 / 12.011 = 7.14 mol; H = 14.3 / 1.008 = 14.19 mol. Dividing by 7.14: C 1.00, H 1.99. The empirical formula is CH2.

  4. The compound in question 3 has a molar mass of 56.1 g/mol. What is its molecular formula?
    Show answer

    The empirical formula mass of CH2 is 12.011 + 2(1.008) = 14.027. \(56.1 / 14.027 = 4.00\), so the molecular formula is C4H8.

  5. A sample labeled pure CaCO3 is found to contain 36.0% calcium. Is it pure? If not, estimate its purity assuming the only calcium compound present is CaCO3.
    Show answer

    Pure CaCO3 is 40.04% calcium, so the sample is not pure. Its purity is \(36.0 / 40.04 = 0.899\), about 89.9% CaCO3.

  6. A student claims that because seawater contains the same elements everywhere, it is a compound. Evaluate the claim.
    Show answer

    The claim is wrong. Seawater is a mixture: its salt content varies from place to place (and can be changed by evaporation), and its components can be separated physically. A compound has a fixed composition by mass, as stated in the law of definite proportions.

← 1.2 Mass spectra of elements1.4 Atomic structure and electron configuration →

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