AP Chem 1.3: Elemental Composition of Pure Substances

River water, glacier ice, and lab-made water

Take a glass of river water, chip a piece of ice off a glacier, and make some water in the lab by burning hydrogen in oxygen. Purify each one, decompose it, and measure the mass percent of hydrogen: all three give 11.19%. That’s no coincidence. Pure water always has exactly this composition.

Lesson 1.5.6 went from a formula to mass percents. This lesson runs the other way: from measured masses to a formula. That’s how chemists work out the composition of a new substance.

Two kinds of particles in pure substances

A pure substance is made of one kind of particle, but “particle” can mean two things:

Molecules, such as $\ce{H2O}$ , $\ce{CO2}$ , and $\ce{C6H12O6}$ . Each molecule has a definite number of each kind of atom, and the formula describes one molecule.
Formula units, such as NaCl and $\ce{CaCl2}$ . An ionic compound has no separate molecules; its ions are packed in a crystal in a fixed ratio, and the formula gives that ratio.

Either way, the ratio of atoms of each element is fixed, so the ratio of their masses is fixed too.

The law of definite proportions

A given compound always contains its elements in the same proportion by mass, wherever it comes from and however it was made. This is the law of definite proportions定比定律 (Proust, 1799), also called the law of constant composition.

Here are analyses of three samples of black copper oxide:

Sample	Mass of Cu (g)	Mass of O (g)	Mass ratio O : Cu
1	1.000	0.252	0.252
2	2.500	0.630	0.252
3	0.635	0.160	0.252

The samples differ in size, but the mass ratio is the same, as the law predicts. The formula CuO gives 16.00 ÷ 63.55 = 0.252, a perfect match.

A red copper oxide also exists, in which 1.000 g of Cu combines with only 0.126 g of O, half the ratio. It’s a different compound ( $\ce{Cu2O}$ ). A different mass ratio means a different pure substance.

To compare two samples with the law of definite proportions, you need the masses of both elements in each sample (or the sample’s total mass and the mass of one element). Knowing only that sample A has 1.0 g of Cu and sample B has 2.0 g of Cu proves nothing, since samples can be any size.

Percent composition

The mass percents of the elements in a compound are its percent composition百分组成. From the formula:

\text{mass percent of an element} = \frac{\text{number of atoms} \times \text{atomic mass}}{\text{molar mass}} \times 100\%

For water, $M(\ce{H2O}) = 18.02$ g/mol, so

\text{mass \% H} = \frac{2 \times 1.008}{18.02} \times 100\% = 11.19\%

and oxygen is 88.81%. The mass percents always add up to 100%.

Empirical formulas

The lowest whole-number ratio of atoms in a compound is its empirical formula经验式. Experiments measure masses, and four steps turn masses into an empirical formula:

Percent to mass. If you’re given mass percents, assume a 100 g sample, so each percent becomes that many grams.
Mass to moles. Divide each mass by the element’s atomic mass. The ratio of atoms equals the ratio of moles, not the ratio of masses.
Divide by the smallest. Divide every mole value by the smallest one.
Multiply until whole. If the results are close to whole numbers, round. If you see something like 1.5, 1.33, or 1.25, multiply everything by 2, 3, or 4 until all are whole.

A common memory aid: “Percent to mass, mass to mole, divide by small, multiply ‘til whole.”

Decimal part of a ratio	about .5	about .33 or .67	about .25 or .75
Multiply everything by	2	3	4

Molecular formulas

A molecular formula分子式 gives the actual number of each kind of atom in one molecule. It’s a whole-number multiple of the empirical formula:

\text{molecular formula} = (\text{empirical formula})_n

n = \frac{M(\text{molecule})}{M(\text{empirical formula})}

Glucose has the empirical formula $\ce{CH2O}$ (30.03 g/mol) and a molar mass of 180.16 g/mol, so $n = 6$ and its molecular formula is $\ce{C6H12O6}$ . Different substances can share an empirical formula: acetylene, $\ce{C2H2}$ , and benzene, $\ce{C6H6}$ , are both CH. So mass percents alone give only the empirical formula; you need the molar mass to pin down the molecular formula.

Elemental analysis answers “which elements, in what proportions”: the empirical formula. It can’t tell you density, solubility, or how fast a substance reacts; those take other experiments.

Try the presets first, and notice why the two iron-oxide data sets need a multiplier. Then switch to mass percent and add a molar mass to get a molecular formula. The last preset gives moles directly, so the first step drops out:

Try it: find an empirical formula from masses or mass percents

Molar massg/mol

Example 1: Empirical formula from masses

A 5.00 g sample of an iron oxide contains 3.50 g of Fe; the rest is oxygen. Find its empirical formula.

Step-by-step solution

Find the mass of oxygen. The sample contains only Fe and O, so oxygen is found by difference:
$m(\text{O}) = 5.00 - 3.50 = 1.50\ \text{g}$
Mass → moles. An empirical formula is a ratio of atoms, so compare moles, not masses (in mol):
$\begin{aligned} n(\text{Fe}) &= \frac{3.50}{55.85} = 0.0627 \\ n(\text{O}) &= \frac{1.50}{16.00} = 0.0938 \end{aligned}$
Divide by the smallest.
$\begin{aligned} \text{Fe} : \text{O} &= \frac{0.0627}{0.0627} : \frac{0.0938}{0.0627} \\ &= 1 : 1.496 \end{aligned}$
Multiply until whole. 1.496 ≈ 1.5 is too far from 2 to round, so multiply both by 2:
$\text{Fe} : \text{O} = 2 : 3$
The empirical formula is $\boxed{\ce{Fe2O3}}$ .
Check. Work out the mass percent of Fe in $\ce{Fe2O3}$ and compare it with the data:
$\begin{aligned} \%\text{Fe} &= \frac{2 \times 55.85}{159.70} = 69.94\% \\ \text{data} &: \frac{3.50}{5.00} = 70.0\% \end{aligned}$
They match.

Example 2: Molecular formula from percents and molar mass

A hydrocarbon is 85.6% C and 14.4% H by mass, and its molar mass is 56.1 g/mol. Find its empirical and molecular formulas.

Step-by-step solution

Assume a 100 g sample. Then the percents become grams: 85.6 g of C and 14.4 g of H.
Mass → moles. In mol:
$\begin{aligned} n(\text{C}) &= \frac{85.6}{12.01} = 7.13 \\ n(\text{H}) &= \frac{14.4}{1.008} = 14.3 \end{aligned}$
Find the lowest whole-number ratio. Divide both by the smaller, 7.13:
$\text{C} : \text{H} = 1 : 2.00$
The empirical formula is $\boxed{\ce{CH2}}$ .
Empirical formula mass. In g/mol:
$\begin{aligned} M(\ce{CH2}) &= 12.01 + 2 \times 1.008 \\ &= 14.03 \end{aligned}$
Find the multiple. The molecular formula is a whole-number multiple of the empirical formula:
$\begin{aligned} n &= \frac{56.1}{14.03} = 4.00 \\ \text{formula} &: \boxed{\ce{C4H8}} \end{aligned}$
Check. The molar mass of $\ce{C4H8}$ is 4 × 12.01 + 8 × 1.008 = 56.10 g/mol, matching the given 56.1.

Combustion analysis

A compound of C and H (and possibly O) can be burned completely in excess oxygen: all the C ends up in $\ce{CO2}$ and all the H in $\ce{H2O}$ . Weigh the $\ce{CO2}$ and $\ce{H2O}$ , and you have the moles of C and H in the sample:

n(\text{C}) = n(\ce{CO2})

n(\text{H}) = 2\,n(\ce{H2O})

Oxygen can’t be found this way, because extra oxygen was added for the burning. Get it by difference: sample mass − mass of C − mass of H. For compounds containing N, the nitrogen is measured separately, and the rest of the method is the same.

A shortcut that shows up often: if a hydrocarbon burns to equal moles of $\ce{CO2}$ and $\ce{H2O}$ , then $n(\text{H}) = 2n(\text{C})$ , the empirical formula is $\ce{CH2}$ , and the molecular formula has the form $\ce{C_{n}H_{2n}}$ .

Example 3: Combustion analysis

Complete combustion of 0.255 g of a compound containing only C, H, and O gives 0.560 g of $\ce{CO2}$ and 0.306 g of $\ce{H2O}$ . Find its empirical formula.

Step-by-step solution

Track where each element goes. On combustion, all the sample’s C ends up in $\ce{CO2}$ and all its H in $\ce{H2O}$ . Oxygen also comes from the $\ce{O2}$ in air, so the sample’s O can’t be found from the products, only by difference.
Carbon. One $\ce{CO2}$ contains one C (moles in mol, mass in g):
$\begin{aligned} n(\text{C}) &= \frac{0.560}{44.01} = 0.01272 \\ m(\text{C}) &= 0.01272 \times 12.01 \\ &= 0.1528 \end{aligned}$
Hydrogen. One $\ce{H2O}$ contains two H, so multiply by 2:
$\begin{aligned} n(\text{H}) &= 2 \times \frac{0.306}{18.02} = 0.03396 \\ m(\text{H}) &= 0.03396 \times 1.008 \\ &= 0.0342 \end{aligned}$
Oxygen, by difference.
$\begin{aligned} m(\text{O}) &= 0.255 - 0.1528 - 0.0342 \\ &= 0.0680 \\ n(\text{O}) &= \frac{0.0680}{16.00} = 0.00425 \end{aligned}$
Divide by the smallest, n(O).
$\begin{aligned} \text{C} &: \frac{0.01272}{0.00425} = 2.99 \\ \text{H} &: \frac{0.03396}{0.00425} = 7.99 \\ \text{O} &: 1 \end{aligned}$
The empirical formula is $\boxed{\ce{C3H8O}}$ .
Check. $\ce{C3H8O}$ is 36.03 ÷ 60.09 = 59.96% C by mass, and the data give 0.1528 ÷ 0.255 = 59.9%. They match.

Hydrates

A hydrate such as $\ce{CuSO4.5H2O}$ loses its water when heated. Weigh it before and after: the mass lost is water, and what’s left is the anhydrous salt. Their mole ratio is the x in the formula. Suppose 2.50 g of copper(II) sulfate hydrate is heated until its mass stops changing (constant mass), leaving 1.60 g of anhydrous $\ce{CuSO4}$ :

moles of water: (2.50 − 1.60) ÷ 18.02 = 0.0499 mol;
moles of $\ce{CuSO4}$ : 1.60 ÷ 159.61 = 0.01002 mol;
x = 0.0499 ÷ 0.01002 = 4.98 ≈ 5, so the formula is $\ce{CuSO4.5H2O}$ .

Practice: is this formula already an empirical formula?

Glucose, $\ce{C6H12O6}$
Water, $\ce{H2O}$
Hydrogen peroxide, $\ce{H2O2}$
Ethylene, $\ce{C2H4}$
Dinitrogen tetroxide, $\ce{N2O4}$
Sodium chloride, NaCl
Carbon dioxide, $\ce{CO2}$
Ethanol, $\ce{C2H6O}$
Tetraphosphorus decoxide, $\ce{P4O10}$
Octane, $\ce{C8H18}$

Common mistakes

Mistake 1: “The empirical formula is the molecular formula.” The empirical formula gives only the lowest ratio. $\ce{CH2O}$ could be formaldehyde ( $\ce{CH2O}$ ), acetic acid ( $\ce{C2H4O2}$ ), or glucose ( $\ce{C6H12O6}$ ), all with the same percent composition. You need the molar mass to tell them apart.

Mistake 2: Treating a mass ratio as an atom ratio. In Example 1, Fe and O are 70 : 30 by mass, but 2 : 3 by atoms. Atoms have different masses, so divide by atomic mass to get moles before comparing numbers of atoms.

Mistake 3: Rounding 1.5 up to 2. Rounding Fe : O = 1 : 1.5 to 1 : 2 gives the wrong formula, $\ce{FeO2}$ . Experimental error usually moves a ratio less than 0.1 from a whole number; anything 0.2 or more away needs a multiplier. Also, don’t round intermediate results too early, or 1.33 may turn into 1.3 and hide the factor of 3.

Quick check

Which compound has the greatest mass percent of nitrogen?

A 5.00 g sample of a compound contains 2.00 g of Ca, 0.60 g of C, and 2.40 g of O. What is its empirical formula?

Which compound has the same empirical formula as ethylene, $\ce{C2H4}$ ?

A hydrocarbon burns completely to give equal numbers of moles of $\ce{CO2}$ and $\ce{H2O}$ . What is its empirical formula?

A 2.46 g sample of magnesium sulfate hydrate, $\ce{MgSO4.xH2O}$ , is heated to constant mass, leaving 1.20 g of anhydrous $\ce{MgSO4}$ (120.37 g/mol). The value of x is closest to

Elemental analysis gives the mass percent of each element in a pure substance. Which of the following can be determined directly from these data?

AP-style free response

Two samples of compounds containing only sulfur and oxygen were analyzed. Sample A has a mass of 2.00 g and contains 1.00 g of S. Sample B has a mass of 3.00 g and contains 1.20 g of S.

(a) Could the two samples be the same pure substance? Justify your answer using the law of definite proportions.

(b) Determine the empirical formula of sample A.

(c) Show that the data for sample B are consistent with pure $\ce{SO3}$ .

Model answer and scoring

(a) No. Sulfur is 1.00 ÷ 2.00 = 50.0% of sample A but 1.20 ÷ 3.00 = 40.0% of sample B. By the law of definite proportions, a given compound always has the same mass percent of each element, so the samples are not the same pure substance. (Comparing the total masses or the masses of S alone proves nothing, since samples can be any size.)

(b) Sample A contains 1.00 g of O:

n(\text{S}) = \frac{1.00\ \text{g}}{32.06\ \text{g/mol}} = 0.0312\ \text{mol}

n(\text{O}) = \frac{1.00\ \text{g}}{16.00\ \text{g/mol}} = 0.0625\ \text{mol}

S : O = 1 : 2.00, so the empirical formula is $\ce{SO2}$ .

(c) Sample B contains 1.80 g of O: n(S) = 1.20 ÷ 32.06 = 0.03743 mol and n(O) = 1.80 ÷ 16.00 = 0.1125 mol, so S : O = 1 : 3.01 ≈ 1 : 3. Or work forward: S is 32.06 ÷ 80.06 = 40.0% of $\ce{SO3}$ , matching the data.

Scoring: (a) calculates both mass percents (or mass ratios) and cites constant composition as the reason; (b) converts to moles before finding the ratio; (c) shows agreement through a mole ratio or a mass percent.

Summary

Law of definite proportions: a given compound always has the same mass ratio of its elements; a different ratio means a different pure substance.
Percent composition: mass percent of an element = number of atoms × atomic mass ÷ molar mass × 100%.
Empirical formula: percent to mass (assume 100 g) → mass to moles → divide by the smallest → multiply until whole (.5 → ×2, .33 → ×3, .25 → ×4).
The molecular formula is n times the empirical formula, where n = molar mass ÷ empirical formula mass; you need the molar mass to find it.
Combustion analysis: n(C) = n(CO₂), n(H) = 2n(H₂O), and O by difference. Hydrates: the mole ratio of water lost to anhydrous salt left is x.

AP tip: Multiple-choice questions often ask you to compare mass percents, identify an empirical formula, or work out a formula from combustion products or a hydrate’s mass before and after heating. The numbers are usually friendly, so learn to estimate from atom ratios and atomic masses. On free response, show every step from mass to moles to the whole-number ratio, and say why you multiplied. U1.4 uses the same kind of elemental analysis to decide whether a sample is pure.

Elemental Composition of Pure Substances纯净物的元素组成

Learning goals

Before you start

River water, glacier ice, and lab-made water

Two kinds of particles in pure substances

The law of definite proportions

Percent composition

Empirical formulas

Molecular formulas

Example 1: Empirical formula from masses

Example 2: Molecular formula from percents and molar mass

Combustion analysis

Example 3: Combustion analysis

Hydrates

Common mistakes

Quick check

AP-style free response

Summary

Feedback / Report a problem