STOICHIOMETRY

Mei 10, 2017

What do cookies and chemistry have in common? Many things, it turns out! A balanced chemical equation is the recipe for a reaction: it contains a list of all the reactants (the ingredients) and products (the cookies) as well as their relative proportions. Using a balanced chemical equation to calculate amounts of reactants and products is called stoichiometry. It is a super technical-sounding word that simply means using ratios from the balanced equation. In this article, we will discuss how to use mole ratios to calculate the amount of reactants needed for a reaction. Stoichiometry is the part of chemistry that studies amounts of substances that are involved in reactions. You might be looking at the amounts of substances before the reaction. You might be looking at the amount of material that is produced by the reaction. Stoichiometry is all about the numbers.

The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). Stoichiometry deals with calculations about the masses (sometimes volumes) of reactants and products involved in a chemical reaction. It is a very mathematical part of chemistry, so be prepared for lots of calculator use.

Jeremias Benjaim Richter (1762-1807) was the first to lay down the principles of stoichiometry. In 1792 he wrote:

"Die stöchyometrie (Stöchyometria) ist die Wissenschaft die quantitativen oder Massenverhältnisse zu messen, in welchen die chymischen Elemente gegen einander stehen."
[Stoichiometry is the science of measuring the quantitative proportions or mass ratios in which chemical elements stand to one another.]

All reactions are dependent on how much stuff you have. Stoichiometry helps you figure out how much of a compound you will need, or maybe how much you started with. We want to take the time to explain that reactions depend on the compounds involved and how much of each compound is needed.

What do you measure? It could be anything. When you're doing problems in stoichiometry, you might look at...

Mass of Reactants (chemicals before the reaction)
Mass of Products (chemicals after the reaction)
Chemical Equations
Molecular Weights of Reactants and Products
Formulas of Various Compounds

The most common stoichiometric problem will present you with a certain amount of a reactant and then ask how much of a product can be formed. Here is a generic chemical equation:

2 A + 2B ---> 3C

Here is a typically-worded problem (often called a "mass-mass" problem):

Given 20.0 grams of A and sufficient B, how many grams of C can be produced?

You will need to use molar ratios, molar masses, balancing and interpreting equations, and conversions between grams and moles. If you struggled with those in class, welcome to the club. Go back and review them if you need to, because if you can't do that stuff, you can't do stoichiometry.

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balanced reactions.

Balancing

In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The reactants are displayed on the left side of the equation and the products are shown on the right, with the separation of either a single or double arrow that signifies the direction of the reaction. The significance of single and double arrow is important when discussing solubility constants, but we will not go into detail about it in this module. To balance an equation, it is necessary that there are the same number of atoms on the left side of the equation as the right. One can do this by raising the coefficients.

Reactants to Products

A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as well as their states, and the proportion for how much of each particle is create relative to one another, through the stoichiometric coefficient. The following equation demonstrates the typical format of a chemical equation:

2 N a (s) + 2 H C l (a q) \to 2 N a C l (a q) + H 2 (g)

In the above equation, the elements present in the reaction are represented by their chemical symbols. Based on the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and products, though the elements they are paired up with often change in a reaction. In this reaction, sodium (

N a

), hydrogen (

H

), and chloride (

C l

) are the elements present in both reactants, so based on the law of conservation of mass, they are also present on the product side of the equations. Displaying each element is important when using the chemical equation to convert between elements.

C l

Stoichiometric Coefficients

In a balanced reaction, both sides of the equation have the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products. In the balanced equation:

2 N a (s) + 2 H C l (a q) \to 2 N a C l (a q) + H 2 (g)

we can determine that 2 moles of

H C l

will react with 2 moles of

N a_{(s)}

to form 2 moles of

N a C l_{(a q)}

and 1 mole of

H_{2 (g)}

. If we know how many moles of

N a

we start out with, we can use the ratio of 2 moles of

N a C l

to 2 moles of Na to determine how many moles of

N a C l

were produced or we can use the ration of 1 mole of

H_{2}

to 2 moles of

N a

to convert to

N a C l

. This is known as the coefficient factor. The balanced equation makes it possible to convert information about one reactant or product to quantitative data about another element. Understanding this is essential to solving stoichiometric problems.

Example 1

Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.

P b (O H) 4 + H 2 S O 4 \to P b (S O 4) 2 + H 2 O

Solution:

Start by counting the number of atoms of each element.

UNBALANCED

Element	Reactant (# of atoms)	Product (# of atoms)
Pb	1	1
O	8	9
H	6	2
S	1	2

The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a coefficient of 2 should be added in front of

H_{2} S O_{4}

to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4 coefficient should be added in front of

H_{2} O

where there is a deficiency of oxygen. Count the number of elements now present on either side of the equation. Since the numbers are the same, the equation is now balanced.

P b (O H) 4 + 2 H 2 S O 4 \to P b (S O 4) 2 + 4 H 2 O

BALANCED

Element	Reactant (# of atoms)	Product (# of atoms)
Pb	1	1
O	12	12
H	8	8
S	2	2

Balancing reactions involves finding least common multiples between numbers of elements present on both sides of the equation. In general, when applying coefficients, add coefficients to the molecules or unpaired elements last.

A balanced equation ultimately has to satisfy two conditions.

The numbers of each element on the left and right side of the equation must be equal.
The charge on both sides of the equation must be equal. It is especially important to pay attention to charge when balancing redox reactions.

Stoichiometry and Balanced Equations

In stoichiometry, balanced equations make it possible to compare different elements through the stoichiometric factor discussed earlier. This is the mole ratio between two factors in a chemical reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show how stoichiometric factors are useful.

Example 2

There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How many party invitations can be sent?

Solution:

The equation for this can be written as

I + 2 S \to I S 2

where

$I$
represents invitations,
$S$
represents stamps, and
$I S_{2}$
represents the sent party invitations consisting of one invitation and two stamps.

Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation.

Invitation Stamps Party Invitations Sent

In this example are all the reactants (stamps and invitations) used up? No, and this is normally the case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the one that runs out first, prevents the reaction from continuing and determines the maximum amount of product that can be formed.

Example 3

What is the limiting reagent in this example?

Solution:

Stamps, because there was only enough to send out invitations, whereas there were enough invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can be solved using stoichiometric factors.

12 I x (1IS₂/1I) = 12 IS₂ possible

20 S x (1IS₂/2S) = 10 IS₂ possible

When there is no limiting reagent because the ratio of all the reactants caused them to run out at the same time, it is known as stoichiometric proportions.

Types of Reactions

There are 6 basic types of reactions.

Combustion: Combustion is the formation of CO₂ and H₂O from the reaction of a chemical and O₂
Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a complex product.
Decomposition: Decomposition is when complex reactants are broken down into simpler products.
Single Displacement: Single displacement is when an element from on reactant switches with an element of the other to form two new reactants.
Double Displacement: Double displacement is when two elements from on reactants switched with two elements of the other to form two new reactants.
Acid-Base: Acid- base reactions are when two reactants form salts and water.

Molar Mass

Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table established this relationship for atoms or ions. For compounds or molecules, you have to take the sum of the atomic mass times the number of each atom in order to determine the molar mass.

Example 4

What is the molar mass of H₂O?

Solution:

Molar mass = 2 \times (1.00794 g / m o l) + 1 \times (15.9994 g / m o l) = 18.01528 g / m o l

Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of products or vice versa.

Example 5: Combustion of Propane

Propane (

C_{3} H_{8}

) burns in this reaction:

C 3 H 8 + 5 O 2 \to 4 H 2 O + 3 C O 2

If 200 g of propane is burned, how many g of

H_{2} O

is produced?
Solution:

Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products you must convert from grams of

C_{3} H_{8}

to moles of

C_{3} H_{8}

then from moles of

C_{3} H_{8}

to moles of

H_{2} O

. Then convert from moles of

H_{2} O

to grams of

H_{2} O

Step 1: 200 g $C_{3} H_{8}$ is equal to 4.54 mol $C_{3} H_{8}$
Step 2: Since there is a ratio of 4:1 $H_{2} O$ to $C_{3} H_{8}$ , for every 4.54 mol $C_{3} H_{8}$ there are 18.18 mol $H_{2} O$
Step 3: Convert 18.18 mol $H_{2} O$ to g $H_{2} O$ . 18.18 mol $H_{2} O$ is equal to 327.27 g $H_{2} O$

Variation in Stoichiometric Equations

Almost every quantitative relationship can be converted into a ratio that can be useful in data analysis.

Density

Density ( $ρ$ ) is calculated as mass/volume. This ratio can be useful in determining the volume of a solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse relationship would be used.

Volume x (Mass/Volume) = Mass

Mass x (Volume/Mass) = Volume

Percent Mass

Percents establish a relationship as well. A percent mass states how many grams of a mixture are of a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams are of the stated element or compound. This is useful in determining mass of a desired substance in a molecule.

Example 6

A substance is 5% carbon by mass. If the total mass of the substance is 10 grams, what is the mass of carbon in the sample? How many moles of carbon are there?

Solution:

10 g sample x (5 g carbon/100 g sample) = 0.5 g carbon

0.5g carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon

Molarity

Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical equations and dilutions.

Example 7

How much 5 M stock solution is needed to prepare 100 mL of 2 M solution?
Solution:

100 mL of dilute solution (1 L/1000 mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(1000 ml stock solution/1L stock solution) = 40 mL stock solution.

These ratios of molarity, density, and mass percent are useful in complex examples ahead.

Determining Empirical Formulas

An empirical formula can be determined through chemical stoichiometry by determining which elements are present in the molecule and in what ratio. The ratio of elements is determined by comparing the number of moles of each element present.

Example 8: Combustion of Organic Molecules

1.000 gram of an organic molecule burns completely in the presence of excess oxygen. It yields 0.0333 mol of CO₂ and 0.599 g of H₂O. What is the empirical formula of the organic molecule?
Solution:

This is a combustion reaction. The problem requires that you know that organic molecules consist of some combination of carbon, hydrogen, and oxygen elements. With that in mind, write the chemical equation out, replacing unknown numbers with variables. Do not worry about coefficients here.

C x H y O z (g) + O 2 (g) \to C O 2 (g) + H 2 O (g)

Since all the moles of C and H in CO₂ and H₂O, respectively have to have came from the 1 gram sample of unknown, start by calculating how many moles of each element were present in the unknown sample.

0.0333mol CO₂ (1mol C/ 1mol CO₂) = 0.0333mol C in unknown

0.599g H₂O (1mol H₂O/ 18.01528g H₂O)(2mol H/ 1mol H₂O) = 0.0665 mol H in unknown

Calculate the final moles of oxygen by taking the sum of the moles of oxygen in CO₂ and H₂O. This will give you the number of moles from both the unknown organic molecule and the O₂ so you must subtract the moles of oxygen transferred from the O₂.

Moles of oxygen in CO₂:

0.0333mol CO₂ (2mol O/1mol CO₂) = 0.0666 mol O

Moles of oxygen in H₂O:

0.599g H₂O (1mol H₂O/18.01528 g H₂O)(1mol O/1mol H₂O) = 0.0332 mol O

Using the Law of Conservation, we know that the mass before a reaction must equal the mass after a reaction. With this we can use the difference of the final mass of products and initial mass of the unknown organic molecule to determine the mass of the O₂ reactant.

0.333mol CO₂(44.0098g CO₂/ 1mol CO₂) = 1.466g CO₂

1.466g CO₂ + 0.599g H₂O - 1.000g unknown organic = 1.065g O₂

Moles of oxygen in O₂

1.065g O₂(1mol O₂/ 31.9988g O₂)(2mol O/1mol O₂) = 0.0666mol O

Moles of oxygen in unknown

(0.0666mol O + 0.0332 mol O) - 0.0666mol O = 0.0332 mol O

Construct a mole ratio for C, H, and O in the unknown and divide by the smallest number.

(1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 mol O) => 1mol C: 2 mol H: 1 mol O

From this ratio, the empirical formula is calculated to be CH₂O.

Determining Molecular Formulas

To determine a molecular formula, first determine the empirical formula for the compound as shown in the section above and then determine the molecular mass experimentally. Next, divide the molecular mass by the molar mass of the empirical formula (calculated by finding the sum the total atomic masses of all the elements in the empirical formula). Multiply the subscripts of the molecular formula by this answer to get the molecular formula.

Example 9

In the example above, it was determined that the unknown molecule had an empirical formula of CH₂O.
1. Find the molar mass of the empircal formula CH₂O.

12.011g C + (1.008 g H) * (2 H) + 15.999g O = 30.026 g/mol CH₂O

2. Determine the molecular mass experimentally. For our compound, it is 120.056 g/mol.
3. Divide the experimentally determined molecular mass by the mass of the empirical formula.

(120.056 g/mol) / (30.026 g/mol) = 3.9984

4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a slight error in the experimentally determined molecular mass. If the answer is not close to a whole number, there was either an error in the calculation of the empirical formula or a large error in the determination of the molecular mass.

5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular formula.

CH₂O * 4 = ?

C: 1 * 4 = 4

H: 2 * 4 = 8

O 1 * 4 = 4

CH₂O * 4 = C₄H₈O₄

6. Check your result by calculating the molar mass of the molecular formula and comparing it to the experimentally determined mass.

molar mass of C₄H₈O₄= 120.104 g/mol

experimentally determined mass = 120.056 g/mol

Example 10: Complex Stoichiometry Problem

An amateur welder melts down two metals to make an alloy that is 45% copper by mass and 55% iron(II) by mass. The alloy's density is 3.15 g/L. One liter of alloy completely fills a mold of volume 1000 cm³. He accidently breaks off a 1.203 cm³ piece of the homogenous mixture and sweeps it outside where it reacts with acid rain over years. Assuming the acid reacts with all the iron(II) and not with teh copper, how many grams of H₂(g) are released into the atmosphere because of the amateur's carelessness? (Note that the situation is fiction.)

Solution:

Step 1: Write a balanced equation after determining the products and reactants. In this situation, since we assume copper does not react, the reactants are only H⁺(aq) and Fe(s). The given product is H2(g) and based on knowledge of redox reactions, the other product must be Fe²⁺(aq).

F e (s) + 2 H + (a q) \to H 2 (g) + F e 2 + (a q)

Step 2: Write down all the given information

Alloy density = (3.15g alloy/ 1L alloy)

x grams of alloy = 45% copper = (45g Cu(s)/100g alloy)

x grams of alloy = 55% iron(II) = (55g Fe(s)/100g alloy)

1 liter alloy = 1000cm³ alloy

alloy sample = 1.203cm³ alloy

Step 3: Answer the question of what is being asked. The question asks how much H2(g) was produced. You are expected to solve for the amount of product formed.

Step 4: Start with the compound you know the most about and use given ratios to convert it to the desired compound.

Convert the given amount of alloy reactant to solve for the moles of Fe(s) reacted.

1.203cm³ alloy(1liter alloy/1000cm³ alloy)(3.15g alloy/1liter alloy)(55g Fe(s)/100g alloy)(1mol Fe(s)/55.8g Fe(s))=3.74 x 10^-5 mol Fe(s)

Make sure all the units cancel out to give you moles of Fe(s). The above conversion involves using multiple stoichiometric relationships from density, percent mass, and molar mass.

The balanced equation must now be used to convert moles of Fe(s) to moles of H₂(g). Remember that the balanced equation's coeffiecients state the stoichiometric factor or mole ratio of reactants and products.

3.74 x 10^-5 mol Fe (s) (1mol H₂(g)/1mol Fe(s)) = 3.74 x 10^-5 mol H₂(g)

Step 5: Check units

The question asks for how many grams of H₂(g) were released so the moles of H₂(g) must still be converted to grams using the molar mass of H₂(g). Since there are two H in each H₂, its molar mass is twice that of a single H atom.

molar mass = 2(1.00794g/mol) = 2.01588g/mol

3.74 x 10^-5 mol H₂(g) (2.01588g H₂(g)/1mol H₂ (g)) = 7.53 x 10^-5 g H₂(g) released

Komentar

shabrina10 Mei 2017 pukul 23.06
balanced equations make it possible to compare different elements through the stoichiometric factor. what is the stoichiometrric factor?
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Arifahrachel11 Mei 2017 pukul 05.30
Can you give an example about the application of stoichiometry in everyday life?
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Unknown11 Mei 2017 pukul 06.03
Please explain the laws that govern the stoichiometry!
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febri yulianti13 Mei 2017 pukul 09.55

Please give another example besides the above example
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Unknown20 Mei 2017 pukul 00.09
Is there a stoichiometric relationship with Ph?
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NIDA UL AZMI21 Mei 2017 pukul 01.04

In stoichiometry, a balanced equation makes it possible to compare different elements through stoichiometric factors, please explain the factor!
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Gita Sitepu 22 Mei 2017 pukul 22.12
Please give an example from Acid-Base.
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Unknown22 Mei 2017 pukul 22.56
Please explain about the limiting reaction ??
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