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Stoichiometry Calculations

Stoichiometry is the field of chemistry that studies the relative amounts of reactants and products in chemical reactions. For any balanced chemical equation, whole numbers (stoichiometric coefficients) are used to show the amounts (in moles) of both the reactants and products. Knowing the molecular weight of the compounds involved in the reaction, it is easy to find the mass of these compounds in grams.

So, the first step in stoichiometry calculations is balancing chemical equations. This means looking for stoichiometric coefficients for the reactants and products. It’s important because in a chemical reaction, the quantity of each element does not change (the law of conservation of mass). Thus, each side of the equation must represent the same quantity of atoms of any chemical element. In case of ionic reactions, the same electric charge must be present on both sides of the equation.

There are a number of methods for balancing chemical equations. But in case of complex reactions involving many compounds, it is preferable to balance equations using algebraic methods, based on solving a set of linear equations.

Our stoichiometry calculator uses the Gauss-Jordan elimination algorithm for solving a set of linear equations. The method is modified for finding integer coefficients.

Syntax Guide

• Reactants and products in a chemical reaction are separated by an equal sign (=). The substances are separated by a plus sign (+).

• The formula of a substance should be entered using the upper case for the first character in the element’s name and the lower case for the second character (compare: Co – cobalt and CO – carbon monoxide).

• Indices should be entered as normal numbers after the appropriate elements or groups, e.g. H2O for a water molecule or (NH4)2SO4 for ammonium sulfate.

• Parentheses ( ), square brackets [ ] and braces (curly brackets) { } can be used in the formulas. Nested brackets are also allowed, e.g. [Co(NH3)6]Cl3. The degree of nesting is unlimited, but all the brackets should be balanced.

• In a hydrated compound the middle dot (·) or the asterisk (*) must go before the water molecule formula (e.g. CuSO 4 ·5H 2 O).

• To denote an ion specify its charge in curly brackets after the compound: {+2} or {2+}. Example: H{+}+CO3{2-}=H2O+CO2.

• To include an electron into a chemical equation, use {-}, e.g. Fe{+3}+{-}=Fe.

• Do not enter the state of compounds such as solids (s), liquids (l) or gases (g).

Using Stoichiometry Calculator

When entering a chemical equation manually or pasting the copied equation, it converts automatically to the ‘normal’ form according to the above rules. All the spaces are ignored and symbol → is converted to =. But symbols ↑ and ↓ remain in place.

Note, that both indices and charges can be denoted in the source document using and html tags, e.g. SO 4 -2 , or denoted using the ‘tiny’ symbols, e.g. SO₄⁻². They are also converted automatically to the ‘normal’ form.

A single electron can be denoted as e – (e-) or e⁻ (with ‘tiny’ minus).

Output format

Using the appropriate drop-down menu, one can choose an output format for the balanced chemical equation:

• Html – The balanced equation is represented using html tags for indices and charges. A single electron is denoted as e – . Clicking the ‘Copy to clipboard’ button ( ) you can copy the result ‘as is’, including all the tags, and then you can paste it into any html-page. However, clicking Ctrl-A and Ctrl-C you can copy the result without the tags and paste it into a DOC document keeping duly formatted indices and charges.

• Small indices – The balanced equation is represented using ‘tiny’ symbols for indices and charges. For example, CO₃²⁻ where Unicode characters are used: ₃ = (\u2083), ² = (\u00B2), ⁻ = (\u207B). A single electron is denoted as e⁻.

• Normal – The balanced equation is represented according to the above syntax guide.

Error notifications

There are a number of obvious notifications in case of error detected in the course of initial inspection of the entered equation, like Unexpected character or Brackets not balanced . The following notifications deserve special attention:

Improper equation – The entered equation has chemical elements on the left-hand side that are missing on its right-hand side, or vice versa.

Impossible reaction – The entered equation represents an impossible reaction. For example, the equation (NH 4 ) 2 SO 4 =NH 4 OH+SO 2 has only trivial solution when all the coefficients set to zero.

Multiple independent solutions – The entered equation can be balanced in an infinite number of ways. Usually it is a combination of a few different independent reactions. For example, the equation H+O=H 2 O+HO has no unique solution because, for instance, two solutions are 3H+2O=H 2 O+HO and 4H+3O=H 2 O+2HO which are not multiples of each other. The equation can be separated as H + O = H 2 O and H + O = HO, each of which does have a unique solution.

Element “…” doesn’t exist – An element’s name, entered in accordance with the above syntax guide, does not indicate a real chemical element. Sometimes an immutable group in chemical compounds is replaced with a fictitious element to balance the equation, which otherwise can not be uniquely balanced. In this case, our calculator will balance the chemical equation but will not compute the reaction stoichiometry.

Example of Stoichiometry Calculations

Consider the combustion of methyl mercaptan (CH 4 S or CH 3 SH) that produces sulfur dioxide:

and answer the following questions: a) How many grams of sulfur dioxide (SO 2 ) are formed from complete combustion of 4 g of CH 4 S ? b) How many grams of SO 2 are formed when 4 g of CH 4 S react with 3 g of O 2 ?

With the help of our stoichiometry calculator, we can easily get the result in no time.

First, copy the above equation as it is and paste it into the ‘Equation to Balance’ field of the calculator. It immediately converts to the ‘normal’ form. Then click the ‘Calculate’ button. This field is now highlighted in light pink, indicating that the original equation was unbalanced. In the ‘Balanced Equation’ field you will have the result:

Below is the table of the reactants and products of the reaction with amounts for each compound. The number of moles for the compounds are indicated the same as the stoichiometric coefficients in the equation. The amount in grams for each compound is calculated to be a product of the compound molecular weight and the respective number of moles.

These fields are interactive, and you can change any amount value and all the other fields are automatically recalculated. So let’s get down to solving the above problems.

a) Enter 4 in the grams field for CH 4 S and press the keyboard ‘Enter’ key. You will immediately get 5.326725 g for SO 2 . b) Note that complete combustion of 4 g of CH 4 S requires 7.981831 g of O 2 . So if only 3 g of O 2 react with CH 4 S then oxygen is the limiting reactant . This means that we must enter 3 g for O 2 and get that only 1.503415 g of CH 4 S will react to yield 2.002069 g of SO 2 .

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Stoichiometry Calculator

Enter the chemical equation and the calculator will readily calculate the number of reactants and products involved in it.

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This stoichiometry calculator lets you calculate the relative amounts of reactants and products involved in a chemical reaction.

Our tools helps you to know the exact number of moles or grams of the entities involved in a chemical equation.

What Is Stoichiometry?

In terms of chemistry:

“The technique that helps to calculate the relative amounts of reactants and products in a balanced chemical reaction is known as stoichiometry”

Determining the stoichiometry of chemical reactions aids you to understand the chemistry of any reaction by comparing the amount of all entities present in it.

Stoichiometric Coefficients:

“In balanced chemical reactions, the numbers used to express the quantity of entities are called stoichiometry coefficients.”

Types of Stoichiometry:

Depending upon the concentration of parameters involved in a chemical reaction, following are the types of stoichiometry:

Stoichiometry Example Problems:

Let’s resolve an example to clarify the concept of stoichiometry:

Example # 01:

Suppose you are experiencing a phenomenon like burning of oxygen gas with hydrogen for the formation of water. Now determine the exact mass of oxygen gas that may be required to burn one gram of hydrogen. Also, mention the water mass obtained at the end.

Carrying out stoichiometry conversion as below:

The balanced chemical stoichiometry equations for the water formation reaction is as follows:

$$ 2H_{2} + O_{2} → 2H_{2}O $$

Now you need to write the atomic and molecular masses of each and every atom involved in the reaction:

Atomic masses:

Hydrogen = 1

Oxygen = 16

For calculations and steps, tap the atomic mass calculator .

Here with the help of another molecular formula calculator , you can get to know the steps involved in calculating the molecular mass of any compound.

The above information can also be displayed in the following stoichiometry table:

Change the moles of the elements of the reaction equation.

Now go for determining the actual mass of the compounds:

When 4 grams of the oxygen reacts with the 32 grams of the oxygen, it produces 36 grams of the water molecule. It means that we actually need to burn approximately 1 gram of the hydrogen gas by using the stoichiometry formula.

So at the end, we have:

1 gram of hydrogen reacts with 32/4 = 8 grams of the oxygen

Mass of the water produced = 36/4 = 9 grams

How Solution Stoichiometry Calculator Works?

This section of the content is packed with a complete usage guide of this free stoichiometry converter. Let’s go through it together!

The free stoichiometric equation solver determines the following results:

Determines the amount of reactant as well as the products in terms of moles and grams

References:

From the source of Wikipedia: Stoichiometry, Etymology, Converting grams to moles, Molar proportion , Determining amount of product, Stoichiometric ratio, Limiting reagent and percent yield, Stoichiometry matrix

From the source of Khan Academy: Balancing chemical equations, Calculating amounts of reactants and products, Ideal stoichiometry

From the source of Lumen Learning: Reaction Stoichiometry, Moles to Moles , Grams to Moles

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Reaction Stoichiometry Calculator

Stoichiometry calculator for android phones and tablets

⚐ Stoichoimetry Plus 3.0 is available ⚐

Input a Reaction Equation

Calculation Type: ------Select One------- Reactant Amount Given Product Amount Given

How to use the calculator Show Me

1) Input a reaction equation to the box. No balance necessary. Example: Cu + O2 + CO2 + H2O = Cu2(OH)2CO3

2) Select a Calculation Type. An input table will be created.

If you have information about one or more reactants, select Reactant Amount Given ; Otherwise, select Product Amount Given .

3) Input amount available. Check 'sufficient' box if amount of a reactant is unknown.

4) Click the 'Calculate' button. Calculation results will appear as below.

LiCoO2 = Li0.25CoO2 + Li{+} + e{-} CuSO4*3.5H2O = CuSO4 + H2O K4[Fe(CN)6] + CuSO4 = Cu2[Fe(CN)6] + K2SO4 Al2Si2O5(OH)4 + SiO2 + Mg{+2} = H2O +H{+} + Al7/3Si11/3O10(OH)2Mg1/6

Tip: use formula (O2)0.21*(N2)0.79 for air

Stoichiometry Calculator + Online Solver With Free Steps

What is a stoichiometry calculator.

Stoichiometry is an important branch of chemistry that deals with the relationship between quantities. It deals with obtaining equilibrium in a chemical reaction to stabilize the reaction. The reactants chemically react together to give products . The chemical reaction also contains some by-products.

Stoichiometry defines the relationship between the reactants and products by balancing their components.

Stoichiometry Calculator is an online tool that balances a chemical reaction by equalizing the components of reactants and products resulting in a balanced equation.

It also provides the chemical structures of reactants and products.

The calculator also outputs the equilibrium constant Kc obtained from the balanced equation. It also gives the rate of reaction and the chemical names for the input reactants and products.

In the end, the calculator also provides the user with a wide range of chemical properties of the input reactants and products.

How To Use Stoichiometry Calculator

You can use the Stoichiometry Calculator by following the steps mentioned here.

At first, enter the first reactant of the chemical equation in the Reactant 1 window. The user can enter either the name or the chemical formula for the reactant in this tab.

The first reactant set by default by the calculator is $H_{3} P O_{4}$ which is the chemical formula for phosphoric acid.

Enter the second reactant in the Reactant 2 window of the calculator. The calculator can only input chemical equations with two reactants for the stoichiometric calculations.

The name of the reactant or the chemical formula for it can be entered in this window. The calculator sets the second reactant by default as Na OH which is sodium hydroxide.

The chemical reaction takes place between the reactants and outputs the products. The products are completely different from the reactants as their chemical properties have now been changed.

Enter the first product obtained from the chemical reaction in the Product 1 window. It could be the name of the product or the chemical formula for it.

Both phosphoric acid $H_{3} P O_{4}$ and sodium hydroxide Na OH react to give water $H_{2} O$ which is set by default in the product 1 window by the calculator. Water $H_{2} O$ is the by-product of this reaction.

The by-products are products of a chemical reaction that are not usually desired at the end of the chemical reaction.

Enter the second product of the chemical equation in the window below the title, Product $2$ . This calculator takes in the chemical equations with two reactants and two products.

The chemical reaction between $ H_{3} P O_{4} $ and Na OH gives the product $Na_{3} P O_{4} $. This is the chemical formula for trisodium phosphate which is set by default in the Product 2 window.

Press the Submit button of the Stoichiometry Calculator for it to process the entered chemical equation having the two reactants and two products.

The calculator processes the input reactants and products and displays the output in multiple windows as follows:

Input Interpretation

The calculator processes the input reactants and products and displays the resulting chemical equation in this window. The reactants and products set by default by the calculator give the following chemical equation:

\[ H_{3} P O_{4} \ + \ Na OH \ \longrightarrow \ H_{2} O \ + \ Na_{3} P O_{4} \]

The user will find the input interpretation according to his/her entered reactants and products.

If any reactant or product is not entered , the calculator prompts Not a valid input, please try again .

Balanced Equation

The balanced equation is obtained by a series of steps performed on the entered chemical equation. A balanced equation is defined as an equation with an equal number of atoms in substances on both sides of the equation .

The balanced equation obtained from the chemical equation set by default by the calculator is:

\[ H_{3} P O_{4} \ + \ 3 Na OH \ \longrightarrow \ 3 H_{2} O \ + \ Na_{3} P O_{4} \]

Note that to balance the equation, the user needs three moles of Na OH to react with one mole of $H_{3} P O_{4}$. So, it produces three moles of $H_{2} O$ and one mole of $ Na_{3} P O_{4} $.

This window also shows the option of Show equation details . By pressing on it, the user can view all the steps to balance the input chemical equation.

The calculator displays the chemical structures of all the reactants and products in this window. The structures show the chemical bonds between the atoms of all the compounds in the chemical equation.

Word Equation

This output window shows the word equation for the entered reactants and products. It displays the names of the reactants and products in the form of an equation.

The reactants and products, set by default by the calculator, show the following word equation:

Phosphoric acid + Sodium hydroxide $\longrightarrow$ Water + Trisodium Phosphate

Equilibrium Constant

The equilibrium constant is obtained from the balanced chemical equation . The formula for the equilibrium constant Kc is:

\[ K_c = \frac{ {[Product 1]}^{M_{P_{1}}} \ {[Product 2]}^{M_{P_{2}}} }{ {[Reactant 1]}^{M_{R_{1}}} \ {[Reactant 2]}^{M_{R_{2}}} } \]

$M_{P_{1}}$ is the number of moles of the first product P1 produced in the balanced chemical reaction.

$M_{P_{2}}$ is the number of moles of the second product P2 in the balanced equation.

$M_{R_{1}}$ is the number of moles of the first reactant R1 balanced in the input equation.

$M_{R_{2}}$ is the number of moles of the second reactant R2 balanced in the equation.

The reactants and products, set by default by the calculator, gives the balanced equation as:

\[ H_{3} P O_{4} + 3 Na OH \longrightarrow 3 H_{2} O + Na_{3} P O_{4} \]

The equilibrium constant Kc is obtained from the balanced equation as:

\[ K_c = \frac{ {[ H_{2} O ]}^{3} \ [ Na_{3} P O_{4} ] }{ [ H_{3} P O_{4} ] \ { [Na OH ] }^{3} } \]

Rate of Reaction

The reaction rate is the rate at which the reaction takes place. It is defined as how slow or fast the reactants react to turn into products . The rate of reaction is also obtained from the balanced equation.

The rate of change of reactants and products gives the rate of the reaction. All the reactants and products are divided by $\Delta t$.

The number of moles in the balanced equation is also divided in the rate equation. It is because the greater the number of moles of the reactants or products, the more time it will take for the reaction to take place.

Hence, decreasing the reaction rate. So, the number of moles of the reactants or products is inversely proportional to the rate of reaction.

The rate of reaction for the default set of reactants and products by the calculator is:

\[ Rate = – \frac{ \Delta [ H_{3} P O_{4} ] }{ \Delta t} = – \frac{1}{3} \frac{ \Delta [Na OH] }{ \Delta t} = \frac{1}{3} \frac{ \Delta [H_{2} O] }{ \Delta t} = \frac{ \Delta [Na_{3} P O_{4}] }{ \Delta t} \]

The calculator obtains this equation by assuming the volume constant.

Chemical Names and Formulas

The calculator displays the formula, Hill’s formula and the names of the entered reactants and products in this window.

For the default example, it displays the Hill’s formula for phosphoric acid $H_{3} P O_{4}$ as $H_{3} O_{4} P$.

For sodium hydroxide Na OH, it displays Hill’s formula to be H Na O. For water, $H_{2} O$, Hill’s formula is the same, and for trisodium phosphate $Na_{3} P O_{4}$, Hill’s formula is $Na_{3} O_{4} P$.

For various reactants and products entered by the user, the calculator gives the results accordingly.

Substance Properties

The calculator also outputs some of the chemical properties of the substances entered in the input window. These properties are as follows:

To understand molar mass, one needs to understand the concept of the mole . One mole of a substance contains 6.022 × $10^{23}$ particles.

The molar mass of a substance is the mass of one mole of that substance.

The calculator displays the molar mass of each reactant and product in this window. The molar mass for the default reactants phosphoric acid and sodium hydroxide is 97.994 g/mol and 39.997 g/mol respectively.

The molar mass for the default products water and trisodium phosphate is 18.015 g/mol and 163.94 g/mol respectively.

The calculator also displays the phase or state of the reactants and products at STP. STP stands for standard temperature and pressure .

At STP, both the phase of phosphoric acid and water is liquid. The phase of sodium hydroxide and trisodium phosphate is solid at STP.

Melting Point

The melting point of a substance is defined as the temperature at which a solid turns into liquid . The calculator displays the melting points of the reactants and products.

The melting point for the default reactants, $H_{3} P O_{4}$ and Na OH are 42.4 °C and 323 °C respectively. Similarly, for $H_{2} O$ and $Na_{3} P O_{4}$, the melting points are 0 °C and 75 °C respectively.

Boiling Point

The temperature at which a liquid turns into gas is known as the boiling point of the substance. The calculator also displays the boiling points of the input reactants and products.

So, the boiling points for $H_{3} P O_{4}$, Na OH and $H_{2} O$ are 158 °C, 1390 °C and 99.61 °C respectively.

The density of a substance is defined as the mass per unit volume of a substance. The formula for density is:

\[ Density = \frac{m}{V} \]

Where $m$ is the mass and V is the volume of the substance. The calculator also displays the density of every material.

The densities of $H_{3} P O_{4}$, Na OH, $H_{2} O$ and $Na_{3} P O_{4}$ are $1.685 \ g/cm^{3} $, $2.13 \ g/cm^{3} $, $0.997048 \ g/cm^{3} $ and $2.536 \ g/cm^{3} $ respectively.

Solubility in Water

Solubility in water is defined as how much a substance dissolves in water .

In the default example, the calculator shows Na OH and $Na_{3} P O_{4}$ to be soluble and $H_{3} P O_{4}$ to be very soluble in water.

Surface Tension

The surface tension is defined as the force of material on the surface of a liquid . The calculator also displays the surface tension of the reactants and products.

The surface tension of sodium hydroxide and water is 0.07435 N/m and 0.07435 N/m respectively.

Dynamic Viscosity

The calculator also displays the dynamic viscosity of a fluid. The dynamic viscosity measures the force required by the fluid to overcome friction .

The dynamic viscosity of sodium hydroxide is 0.004 Pa.s at 350 °C and that of water is $8.9 × 10^{-4} \ Pa.s$ at 25 °C.

The odor of a substance is the smell coming from the substance .

In the default chemical equation by the calculator, phosphoric acid, water, and trisodium phosphate are all odorless substances.

Entropy and Enthalpy

The calculator also displays the calculations for entropy and enthalpy for some molecules in the chemical reaction. These are the thermodynamic properties of the particular molecules.

Solved Examples

Following are some solved examples through the Stoichiometry Calculator.

Aluminum reacts with hydrochloric acid to give aluminum chloride and hydrogen gas. For how many moles of $Al$ and HCl, the reaction produces the above products $Al Cl_{3}$ and $H_{2}$ in a balanced equation.

The user enters the equation in the calculator’s input window as follows:

\[ Al \ + \ HCl \ \longrightarrow \ Al Cl_{3} \ + \ H_{2} \]

The calculator shows the above equation in the input interpretation.

In the next window, it shows the above equation in balanced form as follows:

\[ 2 Al \ + \ 6 HCl \ \longrightarrow \ 2 Al Cl_{3} \ + \ 3 H_{2} \]

The calculator also shows the structures of the substances in the chemical equation.

Structure of Al is given as:

\[\mathit{ Al} \]

Structure of HCl is given as:

\[ \mathit{Cl-H} \]

The structure of $AlCl_3$ is given in figure 1 as follows:

Structure of $H_2$ is given as:

\[ \mathit{H – H } \]

The calculator gives the word equation for the input interpretation equation as follows:

Aluminum + Hydrogen Chloride $\longrightarrow$ Aluminum Chloride + Hydrogen

The calculator also displays the reaction thermodynamics for this equation.

The enthalpy of the above chemical reaction is given as follows:

\[ \Delta {H_{rxn}}^{0} \ = \ -1408 \ kJ/mol \ – \ ( – \ 553.8 \ kJ/mol ) \ = \ – \ 854.6 \ kJ/mol \]

The negative sign of enthalpy indicates an exothermic reaction.

The entropy of the chemical reaction is calculated as follows:

\[ \Delta {S_{rxn}}^{0} \ = \ 567 \ J/(mol.K) \ – \ ( 1179 \ J/(mol.K) ) \ = \ – \ 611.6 \ J/(mol.K) \]

The negative sign of entropy of the chemical reaction indicates an exo-entropic reaction.

The equilibrium constant for the balanced equation is given as follows:

\[ K_c = \frac{ {[ Al Cl_{3} ]}^{2} \ {[ H_{2}]}^{3} }{ {[ Al ]}^{2} \ { [HCl] }^{6} } \]

The rate of reaction given by the calculator is,

\[ Rate = – \frac{1}{2} \frac{ \Delta [ Al ] }{ \Delta t} = – \frac{1}{6} \frac{ \Delta [HCl] }{ \Delta t} = \frac{1}{2} \frac{ \Delta [Al Cl_{3}] }{ \Delta t} = \frac{1}{3} \frac{ \Delta [H_{2}] }{ \Delta t} \]

The calculator also provides the chemical names and formulas for the reactants and products.

The IUPAC name for $Al Cl_{3}$ is trichloroalumane. The calculator also shows Hill’s formula for the reactants and products.

The calculator also provides the chemical properties of the reactants and products as shown in table 1.

Ammonia reacts with oxygen gas to produce water and nitric oxide. How many moles of ammonia $NH_{3}$ and oxygen $O_{2}$ are required to produce a balanced equation with water $H_{2} O$ and nitric oxide NO?

The calculator takes the input reactants and products and outputs the input interpretation of the chemical equation as follows:

\[ N H_{3} \ + \ O_{2} \ \longrightarrow \ H_{2} O \ + \ NO \]

The calculator balances the equation and shows the balanced equation as follows:

\[ 4 N H_{3} \ + \ 5 O_{2} \ \longrightarrow \ 6 H_{2} O \ + \ 4 NO \]

In the structures window, the calculator displays the structures of the reactants and products showing bonds between the atoms.

The structure of $ N H_{3} $ is shown in figure 2 as follows:

The structure of $O_{2}$ is given as follows:

\[ \mathit{O=O} \]

The structure for $H_{2} O$ is given in figure 3 as follows:

The structure of NO is given as follows:

\[ \mathit{N=O} \]

The calculator also provides the word equation for the chemical equation as follows:

Ammonia + Oxygen $\longrightarrow$ Water + Nitric Oxide

The Reaction thermodynamics for this equation is also displayed by the calculator.

The enthalpy of the chemical reaction is given as follows:

\[ \Delta {H_{rxn}}^{0} \ = \ -1350 \ kJ/mol \ – \ ( – \ 183.6 \ kJ/mol ) \ = \ – \ 1166 \ kJ/mol \]

The reaction is exothermic as enthalpy is negative.

The Gibbs free energy is also calculated by the calculator as follows:

\[ \Delta {G_{rxn}}^{0} \ = \ – \ 1072 \ kJ/mol \ – \ ( – \ 65.6 \ kJ/mol ) \ = \ – \ 1007 \ kJ/mol \]

The value of Gibbs free energy indicates an exergonic reaction.

\[ \Delta {S_{rxn}}^{0} \ = \ 1263 \ J/(mol.K) \ – \ ( 1797 \ J/(mol.K) ) \ = \ – \ 533.5 \ J/(mol.K) \]

The negative sign of entropy indicates an exo-entropic chemical reaction.

The calculator gives the equilibrium constant for the balanced equation as follows:

\[ K_c = \frac{ {[ H_{2} O]}^{6} \ {[ NO ]}^{4} }{ {[ N H_{3} ]}^{4} \ { [ O_{2} ] }^{5} } \]

The rate of reaction for this equation is given by the calculator as follows:

\[ Rate = – \frac{1}{4} \frac{ \Delta [ N H_{3} ] }{ \Delta t} = – \frac{1}{5} \frac{ \Delta [ O_{2} ] }{ \Delta t} = \frac{1}{6} \frac{ \Delta [ H_{2} O ] }{ \Delta t} = \frac{1}{4} \frac{ \Delta [ NO ] }{ \Delta t} \]

The calculator also provides the chemical names and formulas for the reactants and products. Hill’s formula for ammonia is $H_{3} N$.

The calculator also displays the substance properties of the reactants and products as shown in table $2$.

Therefore, the Stoichiometry Calculator is a powerful tool for determining the chemical properties of a substance.

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Syntax Examples

What is stoichiometry?

Stoichiometry definition: In chemistry, stoichiometry is the calculation of the quantitative relationships between reactants and products in the course of a chemical reaction.

Thanks to the stoichiometric calculation, it is possible to know the amount of reactants that must be used in a reaction to obtain a certain amount of products.

In order to use stoichiometry, it is necessary to first balance the chemical equation for the reaction. This involves determining the correct ratios of the reactants and products in the reaction, so that the number of atoms of each element is the same on both sides of the equation. Once the equation is balanced, it is possible to use the coefficients (the numbers in front of the chemical formulas) to calculate the amounts of reactants and products.

How to do stoichiometry

2 H 2 O → 2 H 2 + O 2

the stoichiometric coefficients would indicate that for every molecule of water (H 2 O) two molecules of hydrogen (H 2 ) and one molecule of oxygen (O 2 ) will be formed.

Calculate the molecular mass of each compound present in the reaction. Continuing with the previous reaction, the molecular mass of each compound involved is as follows: M water = 18 u ; M hydrogen = 2 u ; M oxygen = 32 u. If we consider the stoichiometric coefficients as moles, we will have

The importance of stoichiometry

The importance of stoichiometry in chemistry cannot be overstated. This branch of chemistry is concerned with the relationships between the quantities of substances involved in chemical reactions, and allows chemists to predict the amounts of reactants and products based on the balanced chemical equation for the reaction.

One of the primary applications of stoichiometry is in the field of chemical engineering, where it is used to design and optimize chemical processes. For example, stoichiometry can be used to determine the amount of reactants needed to produce a certain quantity of a product, or to calculate the yield of a reaction. This is critical in the design of industrial processes, where efficiency and cost are key considerations.

Stoichiometry is also important in the study of environmental chemistry, as it allows us to understand the impact of chemical reactions on the environment. By predicting the amounts of reactants and products involved in a reaction, we can assess the potential risks and benefits of a given chemical process and make informed decisions about its use.

In addition to its practical applications, stoichiometry is also a fundamental concept in chemistry that helps students to understand the behavior of chemical systems. By learning about stoichiometry, students can develop their problem-solving skills and gain a deeper understanding of the underlying principles of chemistry.

Overall, the importance of stoichiometry cannot be overstated. Whether in the design of chemical processes, the assessment of environmental impacts, or the study of fundamental chemical principles, stoichiometry is a critical tool that helps us to understand and predict the behavior of chemical systems.

Reaction Stoichiometry Calculator

This calculator will perform reaction stoichiometry calculations.

Instructions

To perform a stoichiometric calculation, enter an equation of a chemical reaction and press the Start button. The reactants and products, along with their coefficients will appear above.

Enter any known value. The remaining values will automatically be calculated.

How To Perform Stochiometric Calculations

The word 'stoichiometry' is derived from the Greek words 'stoikhe' (meaning proportion or balance) and 'eikion' (meaning amount). When we bake a cake, we use basic stoichiometry to determine how much of each ingredient contributes to the final product. By adjusting the proportions of the ingredients, we can create different types of cakes. In a similar way, chemists study the proportion of substances in reactions to find out how they interact with each other.

Balance the Equation

Before you can perform stoichiometric calculations, you must first balance the chemical equation. You can learn how by reading our article on balancing equations or by using our calculator to do it for you.

For example: MgCl2 + NaOH → Mg(OH)2 + NaCl is balanced to get MgCl2 + 2NaOH → Mg(OH)2 + 2NaCl . This means that for every molecule of MgCl2, you need two NaOH to form a single Mg(OH)2 molecule and two NaCl molecules. This is called the mole ratio.

Determine the Molar Mass of Each Reactant and Product

Once you have a balanced equation, determine the molar mass of each compound. This can be done using our molar mass calculator or manually by following our tutorial .

In our prior example:

You now have all the information needed to start doing stoichiometric calculations.

Moles and Grams of the Known or Needed Substance

Every calculation starts with some knowns or inputs. In stoichiometric calculations, this is usually the known amount (in grams or moles) of at least one reactant or product. To convert between moles and grams, multiply moles by the molar mass to get grams, or divide grams by the molar mass to get moles.

For example, lets say we have 100g of MgCl2 and want to convert it to the number of moles: 100/95.211 = 1.05 moles.

Calculate Amount of Needed Substances

Finally, to calculate the amount of the other substances:

You now know the number of moles of each substance needed in a theoretical perfect reaction. If you want to know the number of grams needed of each substance, you can multiply by the molar mass of each substance:

Use an Online Stoichiometry Calculator

To make sure you get the most accurate quickly and easily, you can use our reaction stoichiometric calculator to perform all your calculations. If you're interested in finding limiting reagents when you know the amounts of all reagents, you can use our limiting reactant calculator .

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Stoichiometry Calculator

What is stoichiometry.

Steps to Use Stoichiometry Calculator

Read The procedure to use the online Stoichiometry Calculator is as follows below:

☛ Step 1: Enter the Value in the respective input field

☛ Step 1: Click the “ Submit ” Button to get the optimal solution

☛ Step 1: Finally, Output will be displayed in the new window

Online Free Conversion Tool

Stoichiometry Calculator

Stay tuned, while we are in the process of adding the Stoichiometry Calculator.

Stoichiometry Calculator is a free online tool that displays a balanced equation for the given chemical equation. BYJU’S online stoichiometry calculator tool makes the calculations faster, and it displays the balanced equation in a fraction of seconds.

How to Use Stoichiometry Calculator?

The procedure to use the Stoichiometry calculator is as follows: Step 1: Enter the chemical equation in the input field Step 2: Now click the button “Submit” to get the output Step 3: Finally, the balanced chemical equation will be displayed in the new window

Why do We use Stoichiometry?

Stoichiometry is used to express the quantitative relationship between reactants and products in the chemical reaction . In a balanced equation, the stoichiometric coefficients represent the molar ratios in the reaction. It allows predicting certain values such as product or molar mass of a gas, per cent yield etc.

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Again, consider the balanced equation:

What weight of O 2 will react with 342 g of C 8 H 18 ? (We can study this reaction two chemicals at a time. Because the atoms combine in simple whole number ratios, we can be assured that everything will react in the correct ratio if we weigh out the appropriate amount of chemicals. Four tires are installed on each car (the appropriate ratio) and we can assume if one car and four tires arrive at the factory that a proper car will leave.)

First we assemble all data and their factors. The question cites C 8 H 18 and O 2 , therefore, we need to calculate their molecular weights. Use the periodic table and the molecular formula to determine the molecular weights (O 2 = 32 g/mol and C 8 H 18 = 114 g/mol). The balanced equation indicates that C 8 H 18 and O 2 react in the ratio of 25 moles of O 2 for every 2 moles of C 8 H 18 . The 25 and the 2 are the stoichiometric coefficients of O 2 and C 8 H 18 , respectively.

Because 342 g of C 8 H 18 is the only data actually stated, it becomes the starting point. Apply the method of dimensional analysis.

The only unit left uncancelled is grams O 2 , the unit of oxygen. The answer is 1200 g O 2 .

Try a different variation. Use the same balanced equation. What weight of H 2 O is produced by the reaction of 342 g of C 8 H 18 ? Use the same logic as the first example. Calculate the molecular weight of C 8 H 18 (114 g/mol) and H 2 O (18 g/mol). The ratio in which C 8 H 18 and H 2 O react is 18 moles H 2 O for every 2 moles C 8 H 18 .

Another Example problem

How much iron(III) oxide will be produced by the complete oxidation of 101 g of iron according to the following equation? 4 Fe + 3 O 2 ==> 2 Fe 2 O 3 ?

From the question we deduce that Fe is the given substance and that Fe 2 O 3 is the desired substance.

All of these steps can be combined into a single equation as follows.

I wonder why I have never thought about using buffer capacity when selecting my buffers. Could be it is too hard to calculate by hand, but what the heck, we have computers and 21st century!

Goran Jerković

EBAS - equation balancer & stoichiometry calculator

Single user license: €24.95

Operating systems: XP, Vista, 7, 8, 10, 11

Calculate mass of phosphine PH 3 evolving when 12g of calcium phosphide Ca 3 P 2 is added to 20g of water.

Enter reagents.

Enter masses.

Ready! Read results.

This is a classic stoichiometry problem - limiting reagent question. As in the case of most questions in chemistry, you have to start with balancing chemical equation.

Create new reaction, clicking on the New button on the tool bar. Click once on the add reactant button and once on the add product button to prepare room for all reagents.

In the upper, input frame, enter formulas of both reactants (Ca 3 P 2 , H 2 O - you may select water from the program database if you want) and products (PH 3 and Ca(OH) 2 - note that you don't need any tricks to correctly display the formulas, just enter letter, digits, and parentheses).

Program does its part of the work and displays balanced equation and calculated molar masses.

Now it's time for stoichiometry - enter known amounts of reagents - 12 into mass edit field below edit field with calcium phosphide formula and 20 into edit field below water formula.

Ready! Entered amount of water is displayed in red, to signal it is in excess. Calcium phosphide is a limiting reagent. Stoichiometric amount of phosphine evolved is displayed in the lower, output frame below phosphine formula - and it reads 4.479 g. In fact we have only two significant digits in data, so it should read 4.5g, but program is somewhat generous when it comes to displaying digits.

At this moment we can take a look at the reaction summary - or even print it, just in case:

However, phosphine is a gas - it will be nice to know what is its volume. You may use ideal gas calculator to check it.

Stoichiometry calculator

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COMMENTS

Why Is Stoichiometry Important?
Stoichiometry is important because it explains the relationships of reactants and products in chemical reactions. Stoichiometry means “element measurement” in Greek and is also known as Mass Relations.
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This is a classic stoichiometry problem - limiting reagent question.