# 4.1: Stoichiometry

- Page ID
- 51452

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Learning Objective

- Master Stoichiometric Relationships
- Determine relationships of quantities of reactants consumed and products produced
- express these in both number of chemical entities (moles) or masses of chemical entities

## Introduction

A balanced chemical equation gives the identity of the reactants and the products as well as the accurate number of molecules or moles of each that are consumed or produced in a chemical reaction. ** Stoichiometry **is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. The stoichiometric coeficient (section 3.1) represents the lowest whole number of each reactant or product consumed or produced as related to each other and the ratio of the coefficients represents the relative rates at which the species are consumed or produced. This, for each specific reaction, the coefficients represent an equivalence statement. Consider the following balanced equation describing the formation of Analine (C

_{6}H

_{5}NH

_{2}) from from nitrobenzene (C

_{6}H

_{5}NO

_{2}).

\[4C_6H_5NO_2 + 9Fe +4H_2O \rightarrow 4C_6H_5NH_2 + 3Fe_3O^*_4\]

From the last Chapter we know this relates the number of particles to each other, but we can not count atoms or molecules. As stated in section 3.1 the word sotichiometry comes from the Greek words **stoicheion **(meaning “element”) and

**(meaning “measure”) and in this Chapter we will use the balanced chemical equation to relate measurable properties like the mass or volume of a sample to predict the quantities of other reactants or products consumed or produced. We will also be able to predict the maximum theoretical yield for a reaction and the quantities of any excess reagents.**

*metron**Note, Fe_{3}O_{4} is the mineral magnetite, and would be called Iron(II,III) oxide as it has both Iron +2 and Iron +3 in its crystal structure (you will not be responsible for naming it in this class).

## Counting by Measurement

**So how do we "count" chemical entities involved in a chemical reaction?**

We count chemical entities in a reaction by measuring macroscopic observables like mass, volume, temperature and pressure. What we measure depends on the state of matter, but we need to relate a measurable property to the number of particles. In this Chapter we will learn how to do this for solid substances and for solutes in a solution. In the gas phase Chapter we will use the ideal gas law (PV=nRT) to do this for gas phase reactants.

We will start with the simplest types of stoichiometric equations, those involving masses.

## Calculations Involving Mass Measurements

Mass measurements are the easiest as you use the molar mass to convert between grams and moles. In Figure \(\PageIndex{1}\) you see three arrows, two yellow and one red. Each of these involves a conversion factor. The yellow ones are conversions involving the molar masses, which relate the mass (grams) of a substance to its moles, and the red one involves the balanced chemical equation, which relates the stoichiometric coefficients of the various species to each other.

**Technique:** This is a three step process which should be done in one equation which uses three conversion factors.

- Conversion Factor #1: Use molar mass to convert
**mass**of known material**to moles**. - Conversion Factor #2: Use coefficients of balanced reaction equation to convert
**moles**of known material**to moles**of desired material. - Conversion Factor # 3: Use molar mass to convert
**moles**of desired material**to mass**of desired material.

**Set up as Dimensional Analysis Problem and solve with one equation (not three)!**

### How many grams of oxygen are produced when 4.0 g butane (C_{4}H_{10}) combusts?

Figure \(\PageIndex{2}\) solves this problem with one equation, where each step is a different conversion factor (equivalence statement). Each of these steps is detailied in the bullet list below

- Conversion Factor #1:
**Mass**\(\rightarrow\)**Mole**. Divide by molar mass of butane. Note, this has significant digits, and the units are included in the numerator and denominator as they are different, but the identity is the same and so not repeated in the denominator (it is implicitly understood that it is the same chemical entity, which in this case is butane) - Conversion Factor #2:
**Mole**\(\rightarrow\)**Mole**. Use coefficients of balanced reaction equation to convert moles of known material to moles of desired material. These are exact numbers that come from the balanced equation and so do not have significant Figures, but since they are relating two different entities, the identity of both the numerator and denominator are explicitly stated (oxygen in the numerator, butane in the denominator). - Conversion Factor # 3:
**Mole**\(\rightarrow\)**Mass**. Multiply by molar mass to convert moles oxygen to grams oxygen. Note, this has significant digits, and the units are included in the numerator and denominator as they are different, but the identity is the same and so not repeated in the denominator (it is implicitly understood that it is the same chemical entity, which in this case is oxygen butane)

Cancel units in the above equation to be sure your answer has the desired units and only round off your final answer to the correct number of significant digits, do not round off during intermediate steps.

**How many significant digits should your molar masses be?** This can vary from instructor to instructor, but the rule of thumb is to use one more significant digit than the measured values of your problem statement. Since this problem had 4.0 g of butane, which has 2 significant digits, we used molar masses of 3 significant digits, but rounded the final answer to two. If at all possible, never use a molar mass with fewer significant digits than the measured values you are working with, as stated in the problem.

**Do you Always start with the Mass-to-Mole step?** * NO! *If the question was what mass of oxygen is produced if you combust 4 moles of butane, you do not need to do the first step. Note, both propane and carbon dioxide are gases at ambient conditions and so you can not measure their mass directly. Instead, you need to measure the pressure, temperature and volume of the gas to determine the number of moles. We will learn how to do this when we study gasses, but for now we will use masses in our calculations.

## Worked Problems

Consider the formation of Analine (C_{6}H_{5}NH_{2}) from from nitrobenzene (C_{6}H_{5}NO_{2}), iron and water.

\[4C_6H_5NO_2 + 9Fe +4H_2O \rightarrow 4C_6H_5NH_2 + 3Fe_3O_4\]

Lets look at some of the questions you could answer. (AND BE CAREFUL TO READ THE QUESTIONS!)

- How many moles of iron would you need to produce 100 mole of analine?
- How many grams of iron would you need to produce 100 mole of analine?
- How many grams of iron would you need to produce 100 grams of analine?
- How many grams of analine could 100 g nitrobenzene produce?
- How many moles of Fe
_{3}O_{4}could 100 g of nitrobenzene produce? - How many moles of Fe
_{3}O_{4}would be produced if 100 g of analine was produced? - ...and many more permutations....SO READ THE QUESTION CAREFULLY!

Table \(\PageIndex{1}\); molar masses of chemicals in Equation 4.1.2

C_{6}H_{5}NO_{2} |
Fe | H_{2}O |
C_{6}H_{5}NH_{2} |
Fe_{3}O_{4} |

123.105g/mol | 55.845g/mol | 18.016g/mol | 93.121g/mol | 231.535 g/mol |

### Mol-to-Mol

**Question 1: How many moles of iron is needed to produce 100.0 mol of Analine?**

Implicit in this question is that nitrobenzene and water are in excess. Note: a good technique is to write down the equation and under each species identify what is given, and what is being asked

\[4C_6H_5NO_2 +\underbrace{9Fe}_{moles=?} +4H_2O \rightarrow \underbrace{4C_6H_5NH_2}_{\text{100 moles}} + 3Fe_3O_4\]

\[100 \cancel{mol \; C_6H_5NH_2}\left ( \frac{9 \; mole \; Fe}{4 \cancel{\; mole \; C_6H_5NH_2}} \right )=\]

### Mass to Mass

Question 2: How many grams of analine could 100 g nitrobenzene produce?

We are now solving a problem where we can actually measure the quantity of reactants and products involved. We start by writing an equation, and under each species write the given values, while identifying what we are solving for.

\[\underset {mass=100.0g \\ fw=\frac{123.105g}{mol}} {4C_6H_5NO_2} +9Fe +4H_2O \rightarrow \underset{mass=?g \\ fw=\frac{93.121g}{mol}}{4C_6H_5NH_2} + 3Fe_3O_4\]

We were given 100 g of nitrobenzene and wanted the mass of analine that could be produced, so the first thing I did was write that down under each species. I also needed the molar mass of nitrobenzene and analine, and so wrote that under those. Now on the equation I have everything I need to solve (Video \(\PageIndex{2}\)

\[100 \; \cancel{gC_6H_5NO_2}\frac{\color{red}{\cancel{mole \; 4C_6H_5NO_2}}}{123.105\cancel{g}}\left ( \frac{4\color{blue}{ \cancel{\;mol \;C_6H_5NH_2}}}{4 \; \color{red}{\cancel{mol \; C_6H_5NO_2}}} \right )\frac{93.121g \;C_6H_5NH_2}{\color{blue}{\cancel{mol}}}=\]

### Mass to Mole

Question 3: This time we want to know how many moles \(Fe_3O_4\) would be formed from 100 g nitrobenzene. We start by writing down what we are given and what we are solving for (100g nitrobenzene given and moles of \(Fe_3O_4\) sought for) and below that we right any additional information we need to solve this problem. That is, we need the molar mass of nitrobenzene in order to calculate the moles. We do not need the molar mass of \(Fe_3O_4\) because the question is looking for how many moles, not how many grams.

\[\underbrace{4C_6H_5NO_2}_{\frac{mass=100g}{fw=\frac{123.105g}{mol}}} +9Fe +4H_2O \rightarrow 4C_6H_5NH_2 + \underbrace{3Fe_3O_4}_{moles=?}\]

\[100 \cancel{g \; C_6H_5NO_2}\frac{\color{red}{\cancel{mole \; C_6H_5NO_2}}}{123.105\cancel{g}}\left ( \frac{3 \;mol \;Fe_3O_4}{4 \; \color{red}{\cancel{mol \;C_6H_5NO_2}}} \right )=?\]

Exercise \(\PageIndex{1}\)

What is true regarding the following statements made for the following balanced chemical reaction?

4 Al(s) + 3 O_{2}(g) --> 2 Al_{2}O_{3}(s)

A) for every 8 moles of aluminum and 6 moles of oxygen consumed, 4 moles of aluminum oxide are produced.

B) for every 4 molecules of aluminum and 3 molecules of oxygen consumed, 2 molecules of aluminum oxide are produced.

C) for every 4 grams of aluminum and 3 grams of oxygen consumed, 2 grams of aluminum oxide are produced.

D) A and B only

E) A, B, and C

**Answer**-
D) A and B only. Stoichiometric ratios in a balanced chemical equations represent a particles/particle ratio that can be scaled up to a mole/mole ratio. So, instead of looking at one molecule at a time, which is not very feasible, we can count by the mole (think by the dozen) of particles at time. Coefficients never represents a mass/mass to ratio because each species in a chemical reaction has a different molar mass. Note: A is still true. We are just doubling the amount of moles, but the ratio is still the same.

## Contributors and Attributions

Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

- Anonymous
- Ronia Kattoum (exercise & learning goals)