To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)


The principles of stoichiometry give a fair idea about the various methods applied for mathematical steps involving chemical reactions where the masses of reactants and products and/or their respective volumes are calculated.

The laws of conservation of matter, conservation of energy and combining weights in chemical reactions constitute the basis of stoichiometric considerations. The applications of these laws to industrial processes is not obvious to the technical worker.

In a chemical change, matter is neither created nor destroyed, but only changed from one form to another. This law requires that "material balance" is maintained in chemical equations.

Law of Definite Proportions

A pure compound is always composed of the same elements combined in a definite proportion by mass.

Law of Multiple Proportions

When two elements combine to form more than one compound, different masses of one element combine with a fixed mass of the other element such that those different masses of the first element are in small whole number ratios to each other.

Related Calculators
Stoichiometry Calculator

Stoichiometry Definition

Back to Top
Anyone, working with chemicals, should know the behavior of substances entering into or resulting from a chemical change along with the relative amounts of theses chemicals which react with one another and the amount of products formed as a result of chemical change.

The calculation of relative amounts of substances in chemical reactions is called Stoichiometry.

Percent Yield of a Reaction

Back to Top
The percentage measurements in chemistry deal with the same underlying concept. The percentage is about checking how many would we have out of hundred. In most of the calculations we make use of the quantity of the provided mass or volume of the molecule or compound and then divide it by the accepted mass or volume of one mole of the molecule or compound.

Percent yield is calculated in an equivalent way by subtracting the actual yield from the theoretical yield and then dividing the theoretical yield. Percent composition is also quite similar, though much simpler. Divide the number of grams of each particular atom in one mole of a substance by its molar mass and we have its percent composition. For all these quantities remember that the answer will be in decimal form so we will need to multiply it by 100 to change it into percentage form.

Example Problem

If 11.2g of Na reacts with 9.00g of H2O and 0.400g of H2 is formed, what is the percent yield of the reaction?


The chemical equation for this problem can be written as

2Na + 2H2O 2NaOH + H2


Key clues that indicate percent yield problems are

a) Amount of both reactants given
b) An amount for the product
c) The words percent yield

In percent yield problems, limiting reactant calculations are frequently performed first to determine the theoretical yields. This is the amount of product that is formed if the reaction goes 100% to completion (this rarely happens in the lab) and what was calculated in exercises 4 and 5.

% of yield = $\frac{Actual\ yield}{Theoritical\ yield}$ $\times 100$

= $\frac{0.400}{0.494}$ $\times 100$ = 81.0 %

Molecular Velocity of Gas

Back to Top
According to Maxwell, the distribution of molecular speeds at a particular temperature is given by the following curve.
Molecular Velocity of Gas
Features of the curve are as follows
  1. Fraction of molecules with too high and too low speeds are very small.
  2. No molecules have zero velocity.
  3. Initially the fractions of molecules increase with increase in velocity, reaches the peak of the curve (corresponding to most probable velocity) and thereafter decreases with increase in velocity.

Percent Composition by Mass

Back to Top
The percent by mass of each element present in a compound is called its elemental analysis. Elemental analysis, also called the mass percent composition, is a catalog of the mass of each element present in 100g of a compound.

To show how the mass percent composition of a compound is related to its chemical formula we begin with a compound whose formula is already known. As mentioned in the previous section, ammonium nitrate (NH4NO3) is an important fertilizer. The molar masses of NH4NO3 and its constituent elements can be used to convert the chemical formula into mass percentages.

The formula tells how many moles of each element are present in 1 mol of the substance. One mole of NH4NO3 contains 2 mol of nitrogen 4 mol of hydrogen and 3 mol of oxygen. The masses of the element contained in 1 mol follows.

2 mol N x 14.007g / mol = 28.014g N
4 mol H x 1.008g / mol = 4.032g H
3 mol O x 15.999g / mol = 47.997g O
Mass of 1 mol NH4NO3 = 80.043g

Each mole of NH4NO3 has a total mass of 80.043g. Of this total, 28.014g is nitrogen. The ratio of these is the mass fraction of nitrogen and the percent nitrogen by mass is the mass fraction multiplied by 100.

(28.014g / 80.043g) = 0.35000 (mass fraction of nitrogen in NH4NO3)
0.35000 x 100 = 35.000% (mass percent of nitrogen in NH4NO3)

Ideal Gas Law Stoichiometry

Back to Top
Many chemical reactions either consume or give off gases. The ideal gas law can be used to relate the volumes of such gases to the amounts of other substances involved in the reaction. The ideal gas equation allows chemists to work stoichiometry problems involving gases. For example, the decomposition of potassium chlorate to potassium chloride and oxygen by heating.

Question: Figure out the volume of oxygen gas produced at 700 torr and 27oC from the decomposition of 25.0g of KClO3


This problem can be solved in three steps


Calculate the number of moles of oxygen gas produced.

$\frac{25.0_{g} KClO_{3}}{1}$ * $\frac{1\ mol\ O_{2}}{122.55_{g} KClO_{3}}$ * $\frac{3\ mol\ O_{2}}{2\ mol\ KClO_{2}}$ = 0.3060 mol O2


Convert the temperature to kelvin and pressure to atmosphere

27 C + 273 = 300K

700 torr / 760 torr/atm = 0.9211 atm


Put everything in ideal gas equation

PV = nRT

v = $\frac{nRT}{p}$

= $\frac{(0.3060\ mol)(0.0821 \frac{L\ atm}{K\ atm})(300\ K)}{0.9211\ atm}$

= 8.18 L

The mean molar mass Mm is a pure operand value without physical meaning, as molecules of gas mixtures do not exist. The mean apparently molar mass of an ideal gas mixture is calculated in compliance with the following equation.

$\rho = \frac{M}{V_{m}}$

Where, $V_{m}$ = Molar volume at 0oC

The mean molar mass of an ideal gas mixture is equal to the sum of the products of volume percentage and molar mass of each individual gas.

With standard temperature and pressure of a gas mixture,

$\rho_{mn} = \frac{M_{m}}{V_{m}}$

Where $\rho_{mn}$ is the mean standard density at 0oC.

Rules Governing the Use of Logarithms

Back to Top
Simple rules govern the use of logarithms for solving most practical problems.

Raising to Powers

To raise a given natural number to an indicated power, the logarithm of the number is multiplied by the exponent indicating the power; the antilogarithm of the result is the required power.

Extracting roots

To extract a required root of a given number divide the logarithm of the number by the index to its root; the antilogarithm of the quotient is the required root.


To multiply natural numbers, add their logarithms and the antilogarithm of their sum is the required product.


To divide natural numbers subtract the logarithm of the divisor from the logarithm of the dividend, the antilogarithm of the difference is the required quotient.

Since this is not a tutorial on logarithms, our discussion has been brief and elementary, being restricted to the need of the present occasion. The interested reader may find in almost any intermediate algebra, trigonometry or business mathematics book, a clear explanation of common logarithm, with adequate tables.

Methods of Using Logarithmic Table

It is obvious that tables of logarithms may be calculated to various degrees of approximation, they may be calculated to 5, 6, 7 or a higher number of decimal places.

Percent Composition

Back to Top
It is often useful to know the compound composition in terms of the masses of its elements. We can obtain this information from the formula of the compound by comparing the mass of each element present in 1 mole of the compound to the total mass of 1 mole of the compound. The mass fraction for each element is calculated as follows.

Mass fraction for a given element = $\frac{Mass\ of\ the\ element\ present\ in\ 1\ mole\ of\ compound}{Mass\ of\ 1\ mole\ of\ compound}$

The composition percentage of any compound is the percentage by weight of each element present in the compound.

Steps for calculation of percentage composition of a compound.
  • Step 1- Calculate the weight of the element in one molecule of the compound.
  • Step 2 - Calculate the molecular mass of the compound by adding up the atomic weight of each element.
  • Step 3 - The percentage of an element is calculated by the relation
% of an element = (Number of parts by weight of the element /molar mass of the compound) x 100

Example 1 : The compound Na2B4O7.10H2O is commonly known as borax. Calculate the percentage of boron (B) in borax.

Answer- Mass of Boron in a mole of borax = 4 x 11g = 44g

Molecular mass of borax = 2 x 23 + 4 x 11 + 7 x 16 + 10 x 18 = 382

Percentage of boron = (44g/382g) x 100 = 11.5%

Example 2 : Find the percentage of water of crystallization in the sample of washing soda Na2CO3.10H2O .

Answer - Molar mass of Na2CO3.10H2O = 2 x 23 + 12 + 3 x 16 + 10 x 18 = 286g

Mass of water = 10 x 18 = 180g

Percentage of water of crystallization = (180g/286g) x100

= 62.93%

Stoichiometry Steps

Back to Top
To solve the problems based on chemical equations we need to keep the following points in mind.
  1. Write down balanced equation in molecular form.
  2. Underline those substances whose weights/volume have been mentioned in the numerical or whose weights/volumes have to be determined.
  3. Write the atomic weights/moles/molar volume of the underlined substances depending upon the nature of the problem.
  4. Write down the actual quantities of the substances which had been underlined. For those substances whose weights/volumes have been calculated just write the sign of interrogation (?).
  5. Calculate the result by unitary method.

Stoichiometry Problems

Back to Top

1. Mass – Mass Relationship Problems

In this type of problem, mass of one of the reactants/products is to be calculated if that of the other reactants/products is given.

2. Mass – volume relationship problems

In this type of problem, mass or volume of one of the reactants or products is calculated from the mass or volume of the other substances. While solving such problems, we should make use of the fundamentals that one mole of each gas occupies a volume of 22.4L at STP.

3. Volume-volume relationship

In this type of problem, the volume of one of the reactants or products is to be calculated from the volume of some other reactants or products.

Solved Examples

Question 1: The chief component of glass is silica (SiO­2). It can be dissolved by the hydrofluoric acid, HF, and according to the following reaction that produces silicon tetra fluoride, SiF­4, a gas at room temperature.

SiO2 + 4HF → SiF4 + 2H2O

How many grams and how many moles of SiF4 can be produced from 63.4g of HF?
The balanced chemical equation for the reaction is

  • Step 1 SiO2 + 4HF → SiF4 + 2H2O
  • Step 2 4 moles 1 mole
  • Step 3 4(1+19) 28.1+ 4x19
80g 104.1g
  • Step 4 63.4g ?

Molecular mass of HF= 20; 20g HF = 1 mole ; 63.4g HF = 3.17 mole

According to the balanced chemical equation,

4moles of HF produce SiF4 = 1mole
3.17 moles HF produce SiF4 = 3.17/4 = 0.793 moles
Molecular mass of SIF4 = 104.1; 1mole of SiF4= 104.1g
0.793 mole of SiF4 = 104.1x0.793 g = 82.5g


Question 2: An impure sample of table salt (NaCl) which weighed 6.5g, gave on treatment with excess silver nitrate solution, 14.35g of silver chloride (AgCl). Calculate the percentage of purity of table salt.
The balanced chemical equation involved is

NaCl + AgNO3 → AgCl + NaNO3
1mole 1 mole
23+35.5 108+35.5
= 58.5g = 143.5g
? 14.53g

It is evident from the above equation:
143.5g of AgCl is obtained from 58.5g NaCl
14.35g of AgCl is obtained from 58.5 x 14.35/143.5 = 5.85g NaCl
Percentage of purity (5.85/6.5) x 100% = 90%

Question 3: How much of KNO3 should be heated to get enough oxygen required to completely burn 56L of hydrogen at STP?
The balanced chemical equation involved in the burning of hydrogen gas is

2H2 + O2 → 2H2O
2moles 1 mole
2x2 =4g 32g
2x22.4 22.4L
= 44.8L
56 L ?

From the above data, it is evident that:
44.8 L of H2 requires 22.4L O2 at STP
56 L of H2 requires 22.4 x 56/ 44.8 = 28 L O2 at STP
The chemical equation involved in the production of O2 from KNO3 is

2KNO3 2KNO2 + O2

2 moles 1 mole
2x 101 g 22.4 L at STP
? 28 L at STP

Now 22.4L of O2 at STP are obtained by heating 202g KNO3
28L of O2 at STP are obtained by heating 202 x 28/ 22.4
= 252.5 g KNO3


Question 4: What volume of carbon dioxide at STP be obtained on treating 1g of marble (CaCO3) with dilute HCl?
The balanced chemical equation involved is:

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

1mole 1 mole
100g 22.4L at STP

From the above data, it is evident that
100g of marble gives 22.4L of CO2 at STP
1g of marble gives 22.4/100 L = 0.224 L of CO2 at STP


Question 5: A certain hydrocarbon required 5 times its volume of oxygen for complete combustion and the resultant gas mixture on being shaken with caustic potash solution showed a contraction equal to thrice the volume of the hydrocarbon taken. Find the formula of hydrocarbon. (All volumes have been measured at same temperature and pressure)
Let the formula of the hydrocarbon is CxHy
The balanced chemical equation for the hydrocarbons combustion is as follows:

CxHy + (x + y/4) O2 → xCO2 + y/2H2O

1 vol. (x+y/4) vol. x vol.
According to the data given, x+y/4 = 5; 4x + y = 20
Contraction in volume on shaking with caustic potash solution (KOH) = volume of evolved CO2.
Therefore, x=3.
Substituting the value of x in equation 4x + y = 20 we will get y=8.

The formula of hydrocarbon is C3H8


More topics in Stoichiometry
The Structure of an Atom Chemical Compounds
Types of Chemical Formula Predicting Products of Chemical Reactions
Conservation of Matter and Energy Solid Liquid and Gases
Volume of Gas at STP Reaction Stoichiometry
Gas Stoichiometry Solution Stoichiometry
Stoichiometry Formulas
NCERT Solutions
NCERT Solutions NCERT Solutions CLASS 6 NCERT Solutions CLASS 7 NCERT Solutions CLASS 8 NCERT Solutions CLASS 9 NCERT Solutions CLASS 10 NCERT Solutions CLASS 11 NCERT Solutions CLASS 12
Related Topics
Chemistry Help Chemistry Tutor
*AP and SAT are registered trademarks of the College Board.