Determine empirical and molecular formulas

Ontario Grade 11 Chemistry

Quantities in Chemical Reactions

Ontario Grade 11 Chemistry — D2.4

A pure compound has a consistent makeup. A small sample and a large sample of the same compound have the same relative amounts of each element. At the particle level, this means the elements occur in a consistent ratio. A chemical formula represents that ratio with element symbols and subscripts. In this lesson, you will use mass information to find the simplest whole-number ratio. If the compound’s molar mass is also known, you can use it to find the molecular formula.

What you will learn

Explain the difference between an empirical formula and a molecular formula.
Determine an empirical formula from element masses or percentage composition.
Use a compound’s molar mass and empirical formula mass to determine its molecular formula.

1. From sample makeup to atom ratio

First, review two ideas. An element symbol identifies an element. A subscript tells how many atoms of that element are represented in a formula. For example, each water molecule, represented by

\mathrm{H_2O}

, contains two hydrogen atoms for every one oxygen atom.

A mole is an amount of substance. The periodic table gives the molar mass of each element: the mass of one mole, usually written in grams per mole. Dividing an element’s mass by its molar mass converts the mass to an amount in moles. Moles let you compare the amounts of different elements on a common basis.

An empirical formula gives the smallest whole-number ratio of elements in a compound. It does not necessarily show the actual number of atoms in one molecule. For example,

\mathrm{CH_2O}

represents a carbon-to-hydrogen-to-oxygen ratio of 1:2:1.

Percentage composition tells what percentage of a compound’s mass comes from each element. If percentages are given without a sample mass, assume a

100.0\,\mathrm{g}

sample for the calculation. Each percentage then has the same numerical value in grams. This is a calculation choice, not a claim that a sample was measured.

n=\frac{m}{M}

An empirical formula shows the simplest whole-number element ratio.
For percentage data, an assumed $100.0\,\mathrm{g}$ sample makes the percentages easy to use as masses.
Convert each element’s mass to moles before comparing amounts.

2. Determine an empirical formula

At the particle level, an empirical formula compares the relative numbers of atoms of each element. Masses cannot be compared directly as atom counts because different elements have different molar masses. Converting the masses to moles gives amounts that can be compared.

Write down the mass of each element. If the problem gives percentages, use the assumed

100.0\,\mathrm{g}

sample to convert them to masses. Divide each mass in grams by the element’s molar mass in grams per mole. The result is an amount in moles.

Divide every mole amount by the smallest mole amount. This gives a ratio whose smallest part is

1

. The values should be close to whole numbers or familiar simple fractions. For example, if one value is close to

1.50

, multiply every part of the ratio by

2

. If a value is close to

1.33

, multiplying every part by

3

may give whole numbers.

Multiply the entire ratio by the same number. This preserves the relative amounts. Write the resulting whole numbers as formula subscripts. Do not write a subscript of

1

. The ratio 1:2:1 is written as

\mathrm{CH_2O}

, not

\mathrm{C_1H_2O_1}

.

If the values are not close to whole numbers or a simple fraction, do not guess. Check the arithmetic, units, and data. Keep several digits during the calculation so early rounding does not distort the ratio.

r_i=\frac{n_i}{n_{\min}}

Use periodic-table molar masses to convert each element mass to moles.
Divide every mole amount by the smallest amount.
If needed, multiply every part of the ratio by the same small whole number.

3. Determine a molecular formula

A molecular formula gives the actual number of each kind of atom in one molecule. It may match the empirical formula, or it may be a whole-number multiple of it. The empirical formula gives the simplest ratio; the compound’s molar mass helps determine the multiplier.

First, calculate the empirical formula mass. Add the molar masses for all atoms shown in the empirical formula, including the number of atoms shown by each subscript. The result is in grams per mole.

Then divide the compound’s given molar mass by the empirical formula mass. The result should be close to a whole number when the data are suitable. That whole number is the multiplier. Multiply every empirical-formula subscript by it to get the molecular formula.

If the multiplier is

6

, for example, multiply each empirical-formula subscript by

6

. If the molar mass is not given, percentage composition can determine an empirical formula, but it cannot by itself determine the molecular formula.

k=\frac{M_{\mathrm{compound}}}{M_{\mathrm{empirical}}}

Add the atomic molar masses represented by the empirical formula to find its formula mass.
Divide the compound’s molar mass by the empirical formula mass to find the multiplier.
Apply the same multiplier to every subscript.

4. Units, rounding, and checks

Keep units beside masses and molar masses. Grams divided by grams per mole gives moles. A chemical formula has no mass unit.

Keep enough digits during intermediate steps. Round only when choosing a whole-number ratio that the data support. Percentages may not total exactly 100% if they have been rounded, so a small difference can be expected.

Check an empirical formula by confirming that its subscripts show the smallest whole-number ratio. Check a molecular formula by confirming that every subscript is the same whole-number multiple of the empirical subscripts. Its formula mass should also agree with the given molar mass within the precision of the data.

\mathrm{MF}=(\mathrm{EF})_k

Carry units through mass-to-mole calculations.
Do not round intermediate mole amounts too early.
Check both the simplest ratio and the molecular-formula multiplier.

Worked example

From percentage composition to both formulas

A compound is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its molar mass is

180\,\mathrm{g/mol}

. Determine its empirical formula and molecular formula.

Assume a sample mass
Treat the percentages as grams in an assumed $100.0\,\mathrm{g}$ sample. The element masses are $40.0\,\mathrm{g}$ carbon, $6.7\,\mathrm{g}$ hydrogen, and $53.3\,\mathrm{g}$ oxygen.
$m_{\mathrm C}=40.0\,\mathrm{g},\quad m_{\mathrm H}=6.7\,\mathrm{g},\quad m_{\mathrm O}=53.3\,\mathrm{g}$
Convert masses to moles
Divide each element mass by its molar mass. The results are about $3.33\,\mathrm{mol}$ carbon, $6.65\,\mathrm{mol}$ hydrogen, and $3.33\,\mathrm{mol}$ oxygen.
$\frac{40.0\,\mathrm{g}}{12.01\,\mathrm{g/mol}}=3.33\,\mathrm{mol\ C};\quad\frac{6.7\,\mathrm{g}}{1.008\,\mathrm{g/mol}}=6.65\,\mathrm{mol\ H};\quad\frac{53.3\,\mathrm{g}}{16.00\,\mathrm{g/mol}}=3.33\,\mathrm{mol\ O}$
Find the simplest ratio
Divide each amount by the smallest amount, about $3.33\,\mathrm{mol}$ . The resulting ratio is approximately 1:2:1, so the empirical formula contains one carbon, two hydrogen, and one oxygen.
$\frac{3.33}{3.33}:\frac{6.65}{3.33}:\frac{3.33}{3.33}\approx1:2:1$
Calculate empirical formula mass
Add the molar masses represented by $\mathrm{CH_2O}$ . The subscript $2$ means that two hydrogen molar masses are included.
$12.01+2(1.008)+16.00=30.03\,\mathrm{g/mol}$
Find the molecular formula
Divide the given molar mass by the empirical formula mass. The multiplier is about $6.0$ . Multiply each empirical subscript by $6$ to obtain the molecular formula.
$\frac{180\,\mathrm{g/mol}}{30.03\,\mathrm{g/mol}}\approx6.0$

Answer: The empirical formula is

\mathrm{CH_2O}

. The molecular formula is

\mathrm{C_6H_{12}O_6}

.

Check: The molecular formula mass is about

180.2\,\mathrm{g/mol}

, which agrees with the given

180\,\mathrm{g/mol}

at its precision. Dividing the molecular subscripts by

6

gives the empirical ratio 1:2:1.

Common mistakes and how to avoid them

Comparing element masses directly to get formula subscripts.

Correction: Convert each mass to moles first. Different elements have different molar masses, so equal masses do not represent equal numbers of atoms.

Rounding a ratio such as 1.50:1 immediately to 2:1.

Correction: Multiply every part of the ratio by

2

to obtain 3:2. Multiplying the whole ratio preserves the relationship.

Multiplying only one subscript when finding a molecular formula.

Correction: Multiply every empirical-formula subscript by the same whole-number multiplier.

Using percentage composition alone to claim a molecular formula.

Correction: Percentages can give an empirical formula. A molecular formula also requires the compound’s molar mass.

Lesson summary

Convert each element mass to moles, then divide by the smallest amount to find the simplest ratio.
Write that whole-number ratio as the empirical formula.
Divide the given molar mass by the empirical formula mass, then multiply every empirical subscript by the resulting whole-number factor.

Check your understanding

Question 1

A compound has an empirical formula of

\mathrm{NO_2}

and a molar mass about twice its empirical formula mass. What is its molecular formula?

$\mathrm{NO_2}$
$\mathrm{N_2O_4}$
$\mathrm{N_2O_2}$
$\mathrm{NO_4}$

Show answer and explanation

\mathrm{N_2O_4}

A multiplier of

2

applies to both subscripts, giving

\mathrm{N_2O_4}

.

Question 2

A sample contains

2.0\,\mathrm{mol}

of element A and

3.0\,\mathrm{mol}

of element B. What is the empirical atom ratio, in the order A:B?

2:3
1:3
3:2
1:2

Show answer and explanation

2:3

Dividing both amounts by the smaller amount,

2.0\,\mathrm{mol}

, gives 1:1.5. Multiplying both parts by

2

gives the simplest whole-number ratio, 2:3.

Key terms

Empirical formula: A formula showing the smallest whole-number ratio of elements in a compound.
Molecular formula: A formula showing the actual number of each kind of atom in one molecule.
Percentage composition: The percentage of a compound’s total mass contributed by each element.
Molar mass: The mass of one mole of a substance, commonly written in grams per mole.
Empirical formula mass: The total molar mass represented by the subscripts in an empirical formula.

Continue through SCH3U

View the complete SCH3U Ontario Grade 11 Chemistry curriculum and lessons

Tutor support for D2.4 · Determine empirical and molecular formulas

Working on D2.4 · Determine empirical and molecular formulas in SCH3U? Bring the question and your own attempt. In a live session, you can write the chemical relationship, check conservation and units, then trace each calculation. Ask about the first step that does not make sense, then try a similar question yourself.

Explore chemistry tutoring · Request an eligible free 30-minute introduction. Your request is matched by subject and language; a tutor must accept before a time is confirmed.

برای مبحث D2.4 · تعیین فرمول تجربی و مولکولی در درس SCH3U، سؤال و تلاش خود را همراه بیاورید. در جلسه می‌توانید رابطهٔ شیمیایی را بنویسید، پایستگی و یکاها را بررسی کنید و هر محاسبه را دنبال کنید. راهنمای معلم شیمی را ببینید یا جلسهٔ معارفهٔ رایگان ۳۰ دقیقه‌ای درخواست کنید. زمان جلسه پس از پذیرش معلم قطعی می‌شود.

About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Chemistry (SCH3U), expectation D2.4. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question