Chemical Equation Balancing Tool

Use this highly interactive tool to practice or teach balancing chemical equations.

The compound is entered in the long box and its numerical coefficient in the short box.

The Reactants column is for the compounds on the left side of the equation; the Products column is for the compounds on the right side of the equation.

Click, or tap, the Get Equation button to get a chemical equation to be balanced.

Note on Balancing Chemical Equations  ∨

The method of balancing any chemical equation is mostly trial-and-error.

So, a certain amount of practice in balancing different chemical equations is necessary in order to get the hang of it.

Keep an eye on the compounds having the most number of atoms and/or numerical subscripts.

Changing their numerical coefficient affects the number of many atoms.

Practice Balancing Your Chemical Equations Here

Reactants

 

 

 

 

Products

 

 

 

 

 

 

 

 

 

Empirical Formula

Learning Objectives

When you finish reading this text, you will be able to:

• define what empirical formula is


• determine a compound's empirical formula


• determine a compound's chemical formula using its empirical formula and its formula mass.

Definition

The empirical formula of a compound indicates the smallest integer ratio of the compound's constituent atoms.

It simply tells us the mole-to-mole ratio of the atoms present in a compound.

Calculation of Empirical Formula

To be able to calculate the empirical formula of a compound, you need to:

Step 1	Determine the individual masses of the compound's constituent atoms.
Step 2	Convert these masses into moles.
Step 3	Divide each of these amounts in moles by the one with the smallest value.
Step 4	If the results are not whole numbers, then multiply these values by an integer that will yield the smallest whole numbers.

Example 1 and Example 2 show the step-by-step calculation of empirical formula from the known masses of atoms, and from the percent composition, of a compound, respectively.

Example 3 shows how to calculate the empirical formula of a hydrate.

Example 1 Determining the Empirical Formula of an Unknown Compound From the Masses of Atoms

Example 1  

A sample of a certain compound contains the following: 0.747 g of Na, 0.195 g of C, and 0.780 g of O.

Using the step by step guide outlined above, we will determine the empirical formula of the given compound.

Step 1 Masses of the compound's constituent atoms are given in the problem.

\begin{align}Na\ \ \ 0.747\ g\\C\ \ \ 0.195\ g\\O\ \ \ 0.780\ g\end{align}

Step 2 Convert these masses into moles by dividing each mass by the molar mass of its corresponding atom.

`mo\l\e Na:`

`(0.747 g)/(22.98977 g//mo\l\e) =  0.0325 mo\l\e`

`mo\l\e C:`

`(0.195 g)/(12.011 g//mo\l\e)=  0.0162 mo\l\e`

`mo\l\e O:`

`(0.780 g)/(15.9994 g//mo\l\e)=  0.0488 mo\l\e`

Step 3 Divide these amounts in mole by the one with the smallest value.

`Na:`

`(0.0325 mo\l\e)/(0.0162 mo\l\e) = 2.0`

`C:`

`(0.0162 mo\l\e)/(0.0162 mo\l\e) = 1.0`

`O:`

`(0.0488 mo\l\e)/(0.0162 mo\l\e) = 3.0`

Step 4 Since the results are whole numbers, our calculation is done.

The empirical formula of the compound is

Na₂CO₃.

Example 2 Determining the Empirical Formula of an Unknown Compound From Percent Composition

When the individual masses of the compound's atoms are not given explicitly, they can still be obtained using percent composition.

Example 2  

An organic halide has a percent composition of 12.82% C, 56.76% Cl and 30.42% F.

Let us find its empirical formula.

Step 1 We can determine the masses of the constituent atoms by multiplying their weight percentages by any amount we desire.

For convenience, we will use 100-g sample. In this way, the weights in grams are numerically equal to the weight percentages:

\begin{align}0.1282\ x\ 100\ g\ =\ 12.82\ g\ \ C\ \\0.5676\ x\ 100\ g\ =\ 56.76\ g\ \ Cl\\0.3042\ x\ 100\ g\ =\ 30.42\ g\ \ F\ \end{align}

Step 2 Convert these masses into moles by dividing each mass by the molar mass of its corresponding atom.

`mo\l\e C:`

`(12.82 g)/(12.011 g//mo\l\e) =  1.0674 mo\l\e`

`mo\l\e Cl:`

`(56.76 g)/(35.453 g//mo\l\e) =  1.6010 mo\l\e`

`mo\l\e F:`

`(30.42 g)/(18.9984 g//mo\l\e) =  1.6012 mo\l\e`

Step 3 Divide these amounts in mole by the one with the smallest value.

`C:`

`(1.0674 mo\l\e)/(1.0674 mo\l\e) = 1.0`

`Cl:`

`(1.6010 mo\l\e)/(1.0674 mo\l\e) = 1.5`

`F:`

`(1.6012 mo\l\e)/(1.0674 mo\l\e) = 1.5`

Step 4 Here, we obtain non-integer results:

C₁Cl_1.5F_1.5

So we multiply them with an integer that will give the smallest whole number. That number is 2.

C_{1 x 2}Cl_{1.5 x 2}F_{1.5 x 2}

Hence, the empirical formula is

C₂Cl₃F₃.

Example 3 Determining the Empirical Formula of a Hydrate

A hydrate is a compound with water molecules adsorbed on its surface. The problem is about finding the number of water molecules bound to the compound.

Example 3  

A 5.00-g sample of the hydrated CoCl₂ was analyzed and found to contain 2.27 g of H₂O.

We are going to determine the compound's empirical formula.

Step 1 Instead of using the masses of the atoms, we use the masses of CoCl₂ and H₂O.

`H_2O\ \ \ \ \ \ 2.27\ g`

`CoCl_2\ \ \ 5.00\ g - 2.27\ g = 2.73\ g`

Step 2 To convert these masses into moles, we divide them by their respective formula masses instead.

`mo\l\e\ \ H_2O:`

`(2.27\ g)/(18.0152\ g//mo\l\e)=0.1260\ mo\l\e`

`mo\l\e\ \ CoCl_2:`

`(2.73\ g)/(129.8392\ g//mo\l\e)=0.0210\ mo\l\e`

Step 3 Divide these amounts in mole by the one with the smaller value.

`H_2O:`

`(0.1260\ mo\l\e )/(0.0210\ mo\l\e) = 6.0`

`CoCl_2:`

`(0.0210\ mo\l\e )/(0.0210\ mo\l\e) = 1.0`

Step 4 Since we obtain whole numbers, the empirical formula is

(CoCl₂)₁(H₂O)₆

which can be rewritten as

CoCl₂·6H₂O

The compound is cobalt(II) chloride hexahydrate.

Use of Empirical Formula

Empirical formula is used in determining the chemical formula of a compound.

A compound's chemical formula is a multiple of its empirical formula.

Depending on the compound, the chemical formula may consist of one or more empirical formula units. (See table below.)

Table of Some Compounds With Their Chemical and Empirical Formulas

Show Table  ∨

Compound	Formula	Empirical Formula
benzene	C6H6	CH
acetylene	C2H2	CH
mercury(I) nitrate	Hg2(NO3)2	HgNO3
ammonium oxalate	(NH4)2C2O4	NH4CO2

From the table, the first two examples are molecular compounds and the last two examples are ionic compounds.

Looking at their formulas, we can see that:

• benzene has 6 empirical formula units, (CH)₆


• acetylene has 2 empirical formula units, (CH)₂


• mercury(I) nitrate has 2 empirical formula units, (HgNO₃)₂


• and ammonium oxalate has 2 empirical formula units, (NH₄CO₂)₂

Facts you should be aware of regarding the empirical formula of compounds:

Compounds having different chemical formulas can have the same empirical formulas as shown by the first two compounds in the table.

For the majority of ionic compounds, their empirical formulas are their chemical formulas also. The last two compounds given in the table are just some of the very few exceptions to this.

For a compound whose chemical formula consists of more than one empirical formula unit, deriving its correct chemical formula requires a knowledge of its empirical formula and its formula mass.

The formula mass is divided by the empirical formula mass to obtain the number of empirical formula units (efu):

`efu=(f\o\r\m\u\la\ mass)/(em\p\irical\ \f\o\r\m\u\la\ mass)`

where:

• formula mass, usually determined by experiment, is the sum of the atomic masses of the compound's constituent atoms


• empirical formula mass is the sum of the atomic masses of the atoms in the empirical formula

---

NOTE: Formula mass is a generic term for the molar mass of molecular and ionic compounds.

---

The subscripts of the empirical formula are multiplied by this calculated number of efu in order to obtain the correct chemical formula of the compound.

Example 4 Determining the Molecular Formula of an Unknown Molecular Compound

Show Example 4  ∨

An organic compound was found by analysis to contain the following composition: 54.53% C, 9.15% H, 36.32% O. Its molecular mass was determined to be 88.106 amu.

We'll determine its molecular formula.

Step 1 Following the method outlined in Step 1 of Example 2, we obtain the following masses:

\begin{align}C\ \ \ 54.53\ g\\H\ \ \ \ \ 9.15\ g\\O\ \ \ 36.32\ g\end{align}

Step 2 To convert these masses into moles, we divide them by their respective molar masses.

`mo\l\e\ \ C:`

`(54.53\ g)/(12.011\ g//mo\l\e)=4.5400\ mo\l\e`

`mo\l\e\ \ H:`

`(9.15\ g)/(1.0079\ g//mo\l\e)=9.0783\ mo\l\e`

`mo\l\e\ \ O:`

`(36.32\ g)/(15.9994\ g//mo\l\e)=2.2701\ mo\l\e`

Step 3 Divide these amounts in mole by the one with the smallest value.

`C:`

`(4.5400\ mo\l\e)/(2.2701\ mo\l\e) = 2.0`

`H:`

`(9.0783\ mo\l\e)/(2.2701\ mo\l\e) = 4.0`

`O:`

`(2.2701\ mo\l\e)/(2.2701\ mo\l\e) = 1.0`

Step 4 Since we obtained whole numbers, the empirical formula is

C₂H₄O.

Step 5 Now, we calculate the number of empirical formula units by dividing the given molecular mass, 88.106 amu, by the empirical formula mass which is 2(12.011) + 4(1.0079) + 15.9994 = 44.053 amu.

`efu=(f\o\r\m\u\la\ mass)/(em\p\irical\ \f\o\r\m\u\la\ mass)`

`\ \ \ \ \ \ =88.106/44.053`

`\ \ \ \ \ \ =2`

Hence, the empirical formula is to be multiplied by 2,

(C₂H₄O)₂.

The molecular formula, therefore, is

C₄H₈O₂.

Percent Composition Calculator

In order to calculate the percent composition of a compound, you need to determine the weights of its constituent elements. For details and examples, see my previous post Percent Composition.

After determining the weight percentages of a compound's constituent elements, you can now calculate the individual weights of the elements contained in any given weight of that compound.

Percent Composition Calculator does just that. It provides you a quick check on your calculations whether you're solving or composing percent composition problems.

Just give the correct formula, weight and percent purity (if any) of the compound:

Formula. For the most commonly encountered binary inorganic compounds, a message is provided below the formula entry line if you enter their incorrect formula.

Weight. You can use any unit for the weight, meaning if your unit of weight is in grams then the computed weights of the compound's components will also be in grams.

% Purity. It is optional. If you don't enter a value, the application assumes that the compound is pure.

You need to round off the results to the required number of significant digits.

Percent Composition Calculator

Enter the following information about the compound:

Formula:

 

Weight:

 

% Purity:

(optional)

 

 

 

 

Percent Composition

Percent composition of a compound refers to the mass percentages (also expressed as % w\w) of the constituent elements of the compound. It indicates the individual weights of the elements in any given weight of the compound.

For a pure compound, percent composition gives its elemental analysis; for an impure compound, percent composition gives the percent purity of the compound.

---

Self-Assessment Test

You may take this test before and/or after reading this text.

Calculate the percent by weight of ?

---

Calculation

If the individual weights of the constituent elements of a compound are known, then calculation of percent composition is straightforward.

Just divide the weight of each of the constituent elements by the weight of the compound and then multiply by 100:

`((weight of e\l\ement)/(weight of compound))  x  100 = % weight`

Example 1  

A certain oxide of copper is found by experiment to contain 2.157 g of Cu and 0.543 g of O. Calculate its percent composition.

Step 1 To find the total weight of the compound, add the weights of the individual elements.

\begin{align}Cu\ \ \ 2.157\ g\\O\ \ \ 0.543\ g\\\_\_\_\_\_\_\_\_\_\\2.700\ g\end{align}

Step 2 In order to calculate the percent weights of the constituent elements, divide each weight of the elements by the weight of the compound and then multiply by 100.

`% Cu:`

`((2.157 g)/(2.700 g))  x  100 =  79.89%`

`% O:`

`((0.543 g)/(2.700 g))  x  100 =  20.11%`

Check that the total percentage is equal to 100.

If the weights are unknown, then percent composition can still be calculated from the known chemical formula of a compound.

It is made possible by considering the amounts of a compound and its components in terms of moles.

Example 2  

Let's calculate the percent composition of ammonium sulfate, (NH₄)₂SO₄:

Step 1 Determine the number of moles of the constituent elements contained in 1 mole of the compound.

1 mole of (NH₄)₂SO₄ contains:

• 2 moles N
• 8 moles H
• 1 mole S
• 4 moles O

Step 2 Convert these number of moles to weight in grams by multiplying them by the molar weights of the elements.

`weight of N:`

`2 mo\l\es  x  (14.0067 g)/(mo\l\e)  =  28.0134 g`

`weight of H:`

`8 mo\l\es  x  ( 1.0079 g)/(mo\l\e)  =    8.0632 g`

`weight of S:`

`1 mo\l\e  x  (32.0666 g)/(mo\l\e)  =  32.0666 g`

`weight of O:`

`4 mo\l\es  x  (15.9994 g)/(mo\l\e)  =  63.9976 g`

Step 3 Add all the weights of the elements. This total is the weight of 1 mole of the compound. It is also known as the formula weight of the compound.

\begin{align}28.0134\ g\\8.0632\ g\\32.0666\ g\\63.9976\ g\\\_\_\_\_\_\_\_\_\_\\132.1408\ g\end{align}

Step 4 Divide each weight of the elements by the formula weight of the compound.

`% N:`

`((28.0134 g)/(132.1408 g))  x  100 =  21.20%`

`% H:`

`((8.0632 g)/(132.1408 g))  x  100 =    6.10%`

`% S:`

`((32.0666 g)/(132.1408 g))  x  100 =  24.27%`

`% O:`

`((63.9976 g)/(132.1408 g))  x  100 =  48.43%`

Their total percentage must be equal to 100.

Uses

Percent composition is used to calculate the weight of a particular element in any given weight of a compound.

Example 3  

Let's calculate the individual weights of the constituent elements present in 5.0 g of (NH₄)₂SO₄.

Step 1 Find the percent composition of the compound.

As calculated in Example 2 , the percent composition of (NH₄)₂SO₄ is:

\begin{align}N\ \ \ \ 21.20\ \%\\H\ \ \ \ \ \ 6.10\ \%\\S\ \ \ \ \ 24.27\ \%\\O\ \ \ \ \ 48.43\ \%\end{align}

Step 2 To determine the individual weights of the elements, multiply the mass percentage of each element by the given weight of the compound.

`weight of N:`

`5.0 g   x   0.2120 =  1.1  g`

`weight of H:`

`5.0 g   x   0.0610 =  0.3  g`

`weight of S:`

`5.0 g   x   0.2427 =  1.2  g`

`weight of O:`

`5.0 g   x   0.4843 =  2.4  g`

Step 3 Check that the total of the weights of the individual elements is equal to the given weight of the compound.

\begin{align}N\ \ \ \ 1.1\ \ g\\H\ \ \ \ 0.3\ \ g\\S\ \ \ \ \ 1.2\ \ g\\O\ \ \ \ \ 2.4\ \ g\\\_\_\_\_\_\_\_\_\_\\5.0\ \ g\end{align}

Percent composition is also used in the determination of the empirical formula of a compound. This is another topic that will be covered in a subsequent post.

Formula Weight Calculation

This text teaches you how to calculate the formula weight of a substance.

Before you begin, please take note that you should have a basic understanding of the following concepts:

• ions and molecules
• mole
• hydrated compounds

---

Self-Assessment Test

You may take this test before and/or after reading this text.

Click START button to begin the test comprising of four questions.

---

Definition of Formula Weight

Formula weight is the sum of the atomic weights of all the atoms indicated in the chemical formula of a substance. It is the weight in grams of 1 mole of that substance, usually expressed as grams per mole. Hence, it is also sometimes called the molar weight.

The quantity of the atoms in a given formula is indicated by a numerical subscript after each atom. If no subscript is present after an atom, it is assumed to have a numerical subscript of one.

Formula Weight or Molecular Weight?

Formula weight is a generic term that may be used for both substances that exist as ions and molecules. Whereas, molecular weight is applicable only to substances that exist as molecules.

Uses of Formula Weight

Formula weight is used in the following:

• stoichiometry problems
• mole calculation
• percent composition calculation
• empirical formula calculation

Note: The following calculations are based on atomic weights with three significant digits.

Example Calculation 1

Formula weight of water, H2O :

H

O

2

1

x

x

1.008

15.999

=

=

2.016

15.999

18.015

Example Calculation 2

Formula weight of ammonium carbonate, (NH4)2CO3 :

N

H

C

O

2 x 1

2 x 4

1

3

x

x

x

x

14.007

1.008

12.011

15.999

=

=

=

=

28.014

8.064

12.011

47.997

96.086

Example Calculation 3

Formula weight of calcium chloride hexahydrate, CaCl2.6H2O :

Ca

Cl

H

O

1

2

6 x 2

6 x 1

x

x

x

x

40.078

35.453

1.008

15.999

=

=

=

=

40.078

70.906

12.096

95.994

219.074

Compounds With Planar Shape

Featured in this post are compounds having planar shape other than trigonal planar.

The following compounds are all planar, with the last two having more than one central atom.

xenon tetrafluoride, XeF4

benzene molecule, C6H6

CH2=NH

Geometry of Xenon Tetrafluoride, XeF₄

The Lewis structure of xenon tetrafluoride, XeF4, below shows that the compound's central atom has four bonding electron pairs and two lone electron pairs.

According to the VSEPR theory, the arrangement that affords the maximum distance between each of the mentioned electron pairs is a square bipyramidal.





The axial positions of the two lone electron pairs offer the least net repulsion than it would have been if one or both of these non-bonding electron pairs were in the equatorial positions.



In the illustration shown above, a square bipyramidal outline is superimposed on the xenon tetrafluoride molecule.

• shape of xenon tetrafluoride: square planar
• F-Xe-F bond angle: 90°

Geometry of Benzene, C₆H₆

The resonance structures of benzene are shown below; the third structure is a suggested representation of the delocalization state of the pi electrons.





Each of the six carbon atoms can be taken as a central atom about which the geometry is trigonal planar.





The result is a hexagonal planar molecule with one carbon atom and one hydrogen atom jutting out at each of the six vertices.

Geometry of CH₂=NH

The CH2=NH molecule has two central atoms, carbon and nitrogen (see the Lewis structure below).





About each center, the geometry is trigonal planar.





However, the shape about the nitrogen atom is angular due to the presence of a lone electron pair.



All of the atoms lie in one plane.

Compounds With Bent Shape

The following chemical species have bent geometry:

water molecule, H2O

nitrogen dioxide molecule, NO2

nitrite ion, NO2-

sulfur dioxide molecule, SO2

The illustrations and discussion below show how the presence of non-bonding electron pair(s) on a molecule's central atom affects the geometry of the molecule and the bond angle of its atoms.

Geometry of Water, H₂O

As shown in its Lewis structure below, the central atom of water has four pairs of electrons: two bonding pairs and two non-bonding or lone pairs.

These pairs of electrons are presumed to take a tetrahedral arrangement in space.





Here, the water molecule is depicted as being superimposed in a tetrahedral outline.





The lone pairs of electrons are relatively closer to the nucleus of the central atom and they tend to crowd the two bonding pairs together so that the H-O-H bond angle is less than the ideal tetrahedral angle of 109.5°.

• shape of water: bent or angular
• H-O-H bond angle: 104.5°

Geometry of Nitrogen Dioxide, NO₂

The central atom of nitrogen dioxide, nitrogen, has two sigma bond electron pairs and an unpaired non-bonding electron as shown in its Lewis electron dot structure below.





In order for this group of electrons to be farthest from each other, it is predicted that they take a trigonal planar arrangement.





Shown in the next illustration is the nitrogen dioxide molecule superimposed on a trigonal planar outline.





The observed O-N-O bond angle of 134.1° in nitrogen dioxide molecule, being closer to trigonal angle of 120°, confirms this prediction.

The opening out of the O-N-O bond angle is due to the less crowding affected by the half filled non-bonding orbital of the central atom.

This half-filled orbital accounts for the nitrogen dioxide (NO₂) being paramagnetic.

• shape of nitrogen dioxide: bent or angular
• O-N-O bond angle: 134.1°

Geometry of Nitrite Ion, NO₂^-

An addition of an electron to the central atom of nitrogen dioxide (NO₂) creates the nitrite anion, NO₂^-.





In the above diagram, the resonance structures of nitrite ion is shown. The oxygen atoms in red partial shades are atoms with negative formal charge.

As is the case with the nitrogen dioxide, the arrangement of the two sigma bond electron pairs and the lone electron pair of nitrite ion's central atom lies on a trigonal plane.

The only difference is the observed O-N-O bond angle of 115° for the nitrite ion. This bond angle now being much closer but less than the trigonal angle of 120° is attributed to the filling up of the non-bonding orbital of the nitrite ion's central atom.

The lone electron pair is relatively closer to the central atom and occupies more space than the adjacent bonding electron pairs do.

• shape of nitrite ion: bent or angular
• O-N-O bond angle: 115°

Geometry of Sulfur Dioxide, SO₂

Referring to the contributing Lewis structures of sulfur dioxide below, there is a -1 formal charge on one of the oxygen atoms and +1 formal charge on the sulfur atom which cancel each other out, leaving the molecule with zero net charge.

Given the two sigma bond electron pairs and one lone electron pair on the central atom of sulfur dioxide, as shown in the figure above, SO₂ is predicted to have a trigonal planar geometry.

A trigonal planar outline is shown below superimposed on the sulfur dioxide molecule.

Given this geometry, it is expected that SO₂ molecule has an angular shape. This is supported by the fact that sulfur dioxide has a dipole moment.

The O-S-O bond angle is expected to be somewhat less than the trigonal angle of 120° due to the presence of a lone electron pair on the central atom.

• shape of sulfur dioxide: bent or angular
• O-S-O bond angle: less than 120°