Calculating Index Of Hydrogen Deficiency

Stoichiometry and Balancing Reactions

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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 chemical element and metron means mensurate, so stoichiometry literally translated means the measure of elements. In guild to apply stoichiometry to run calculations about chemical reactions, it is of import to starting time empathise the relationships that exist between products and reactants and why they exist, which require agreement how to rest reactions.

Balancing

In chemical science, chemical reactions are oft written every bit 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 pointer that signifies the direction of the reaction. The significance of single and double arrow is important when discussing solubility constants, only we will non go into detail about information technology in this module. To balance an equation, it is necessary that in that location are the aforementioned 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 information technology 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 reacts or is formed relative to i another, through the stoichiometric coefficient. The post-obit equation demonstrates the typical format of a chemical equation:

\[\ce{2 Na(south) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(thou)} \nonumber\]

In the to a higher place equation, the elements nowadays in the reaction are represented past their chemical symbols. Based on the Police force of Conservation of Mass, which states that affair 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 upwardly with often modify in a reaction. In this reaction, sodium (\(Na\)), hydrogen (\(H\)), and chloride (\(Cl\)) are the elements present in both reactants, so based on the law of conservation of mass, they are besides present on the product side of the equations. Displaying each element is important when using the chemical equation to convert betwixt elements.

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 remainder the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients tin be fractions, whole numbers are frequently used and oftentimes preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products. In the balanced equation:

\[\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber\]

nosotros can determine that 2 moles of \(HCl\) will react with 2 moles of \(Na_{(due south)}\) to form 2 moles of \(NaCl_{(aq)}\) and i mole of \(H_{2(one thousand)}\). If we know how many moles of \(Na\) reacted, nosotros can use the ratio of 2 moles of \(NaCl\) to ii moles of Na to determine how many moles of \(NaCl\) were produced or we can employ the ratio of 1 mole of \(H_2\) to 2 moles of \(Na\) to catechumen to \(NaCl\). This is known as the coefficient factor. The counterbalanced equation makes it possible to catechumen information about the alter in i reactant or product to quantitative data about another reactant or product. Understanding this is essential to solving stoichiometric problems.

Instance one

Pb (IV) hydroxide and sulfuric acid react every bit shown beneath. Balance the reaction.

\[\ce{Lead(OH)4 + H2SO4 \rightarrow Lead(SO4)2 +H2O} \nonumber\]

Solution

Offset past counting the number of atoms of each element.

UNBALANCED

Element	Reactant (# of atoms)	Product (# of atoms)
Atomic number 82	1	1
O	8	9
H	vi	2
S	1	2

The reaction is not balanced; the reaction has xvi reactant atoms and but xiv product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must exist added to make the equation counterbalanced. In this example, there are only one sulfur atom present on the reactant side, then a coefficient of 2 should exist added in forepart of \(H_2SO_4\) to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and just nine on the product side, a 4 coefficient should be added in front of \(H_2O\) 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.

\[\ce{ Atomic number 82(OH)four + ii H2SO4 \rightarrow Lead(SO4)2 + 4H2O} \nonumber\]

Counterbalanced

Element	Reactant (# of atoms)	Production (# of atoms)
Atomic number 82	i	1
O	12	12
H	8	eight
S	2	ii

Balancing reactions involves finding least mutual multiples between numbers of elements nowadays on both sides of the equation. In full general, when applying coefficients, add together coefficients to the molecules or unpaired elements final.

A balanced equation ultimately has to satisfy two weather condition.

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

Stoichiometry and Balanced Equations

In stoichiometry, balanced equations make information technology possible to compare different elements through the stoichiometric gene discussed before. This is the mole ratio betwixt two factors in a chemical reaction constitute through the ratio of stoichiometric coefficients. Here is a real world example to show how stoichiometric factors are useful.

Example ii

In that location are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How many political party invitations tin can exist sent?

Solution

The equation for this can exist written every bit

\[\ce{I + 2S \rightarrow IS2}\nonumber\]

where

\(I\) represents invitations,
\(S\) represents stamps, and
\(IS_2\) represents the sent party invitations consisting of one invitation and two stamps.

Based on this, nosotros have the ratio of 2 stamps for one sent invite, based on the counterbalanced equation.

Invitations Stamps Party Invitations Sent

In this example are all the reactants (stamps and invitations) used upward? No, and this is unremarkably the case with chemic reactions. There is oft backlog of one of the reactants. The limiting reagent, the one that runs out first, prevents the reaction from standing and determines the maximum amount of production that can be formed.

Case iii

What is the limiting reagent in this example?

Solution

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

12 I ten (1IS ₂/1I) = 12 IS₂ possible

xx Southward ten (1IS ₂/2S) = ten IS₂ possible

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

Types of Reactions

At that place are 6 bones types of reactions.

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

Tooth Mass

Before applying stoichiometric factors to chemical equations, you lot need to empathise molar mass. Tooth mass is a useful chemical ratio betwixt mass and moles. The atomic mass of each private element equally listed in the periodic table established this human relationship for atoms or ions. For compounds or molecules, you lot 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 H_iiO?

Solution

\[\text{Molar mass} = two \times (1.00794\; 1000/mol) + i \times (xv.9994\; yard/mol) = 18.01528\; g/mol\]

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

Case 5: Combustion of Propane

Propane (\(\ce{C_3H_8}\)) burns in this reaction:

\[\ce{C_3H_8 + 5O_2 \rightarrow 4H_2O + 3CO_2} \nonumber\]

If 200 g of propane is burned, how many thou of \(H_2O\) is produced?

Solution

Steps to getting this answer: Since you cannot summate from grams of reactant to grams of products you must convert from grams of \(C_3H_8\) to moles of \(C_3H_8\) then from moles of \(C_3H_8\) to moles of \(H_2O\). Then convert from moles of \(H_2O\) to grams of \(H_2O\).

Step 1: 200 g \(C_3H_8\) is equal to iv.54 mol \(C_3H_8\) .
Step ii: Since there is a ratio of 4:1 \(H_2O\) to \(C_3H_8\), for every 4.54 mol \(C_3H_8\) there are 18.xviii mol \(H_2O\).
Step 3: Convert eighteen.eighteen mol \(H_2O\) to g \(H_2O\). xviii.eighteen mol \(H_2O\) is equal to 327.27 g \(H_2O\).

Variation in Stoichiometric Equations

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

Density

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

Book x (Mass/Book) = Mass

Mass 10 (Book/Mass) = Volume

Per centum Mass

Percents establish a human relationship every bit well. A percent mass states how many grams of a mixture are of a sure 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.

Instance 6

A substance is five% carbon past 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 one thousand sample x (5 g carbon/100 g sample) = 0.5 g carbon

0.5g carbon 10 (1 mol carbon/12.011g carbon) = 0.0416 mol carbon

Molarity

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

Example 7

How much 5 M stock solution is needed to set up 100 mL of 2 M solution?

Solution

100 mL of dilute solution (1 Fifty/1000 mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(m ml stock solution/1L stock solution) = forty 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 nowadays 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 CO_two and 0.599 g of H_twoO. 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 non worry almost coefficients hither.

\[ \ce{C_xH_yO_z(one thousand) + O_2(yard) \rightarrow CO_2(thousand) + H_2O(k)} \nonumber\]

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

0.599g H₂O (1mol H_iiO/ eighteen.01528g H₂O)(2mol H/ 1mol H_iiO) = 0.0665 mol H in unknown

Calculate the final moles of oxygen past taking the sum of the moles of oxygen in CO₂ and H₂O. This will give yous the number of moles from both the unknown organic molecule and the O₂ so you lot must decrease the moles of oxygen transferred from the O_two.

Moles of oxygen in CO₂:

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

Moles of oxygen in H₂O:

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

Using the Police force of Conservation, we know that the mass before a reaction must equal the mass afterward a reaction. With this we tin can apply the difference of the terminal mass of products and initial mass of the unknown organic molecule to decide the mass of the O₂ reactant.

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

1.466g CO₂ + 0.599g H_twoO - one.000g unknown organic = 1.065g O₂

Moles of oxygen in O₂

i.065g O₂(1mol O₂/ 31.9988g O₂)(2mol O/1mol O_two) = 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 separate past the smallest number.

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

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

Determining Molecular Formulas

To determine a molecular formula, start determine the empirical formula for the compound as shown in the section above and then determine the molecular mass experimentally. Next, split 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 respond to get the molecular formula.

Example 9

In the example above, it was determined that the unknown molecule had an empirical formula of CH₂O.

ane. Discover the tooth mass of the empircal formula CH_iiO.

12.011g C + (1.008 g H) * (2 H) + fifteen.999g O = 30.026 g/mol CH₂O

2. Determine the molecular mass experimentally. For our compound, it is 120.056 thousand/mol.

three. Separate the experimentally determined molecular mass past the mass of the empirical formula.

(120.056 g/mol) / (thirty.026 1000/mol) = three.9984

4. Since 3.9984 is very close to iv, it is possible to safely round upwardly and assume that at that place was a slight fault 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 decision 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 * iv = four

H: 2 * iv = viii

O 1 * four = 4

CH₂O * 4 = C₄H_viiiO₄

6. Cheque your issue by computing the tooth mass of the molecular formula and comparing it to the experimentally determined mass.

molar mass of C₄H_eightO₄= 120.104 g/mol

experimentally determined mass = 120.056 g/mol

% mistake = | theoretical - experimental | / theoretical * 100%

% fault = | 120.104 g/mol - 120.056 yard/mol | / 120.104 g/mol * 100%

% error = 0.040 %

Example x: Circuitous Stoichiometry Problem

An apprentice welder melts downward two metals to brand an alloy that is 45% copper by mass and 55% iron(Two) by mass. The alloy's density is 3.15 g/L. I liter of alloy completely fills a mold of book 1000 cm³. He accidentally breaks off a 1.203 cm³ piece of the homogenous mixture and sweeps it outside where it reacts with acid pelting over years. Assuming the acrid reacts with all the iron(Ii) and non with the copper, how many grams of H₂(grand) are released into the temper because of the amateur's abandon? (Note that the state of affairs is fiction.)

Solution

Footstep ane: Write a balanced equation later on determining the products and reactants. In this situation, since we presume copper does not react, the reactants are but H⁺(aq) and Atomic number 26(southward). The given product is H2(thousand) and based on knowledge of redox reactions, the other product must be Iron² ⁺(aq).

\[\ce{Iron(s) + 2H^{+}(aq) \rightarrow H2(m) + Atomic number 26^{ii+}(aq)} \nonumber\]

Step 2: Write down all the given information

Alloy density = (three.15g alloy/ 1L blend)

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

ten grams of alloy = 55% fe(II) = (55g Fe(s)/100g alloy)

1 liter alloy = 1000cm ³ alloy

alloy sample = 1.203cm³ alloy

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

Stride iv: Offset with the compound you know the near 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 Iron(south) reacted.

1.203cm³ alloy(1liter blend/1000cm ³ blend)(iii.15g alloy/1liter alloy)(55g Fe(s)/100g alloy)(1mol Fe(south)/55.8g Fe(south))=3.74 x 10^-v mol Fe(s)

Brand sure all the units cancel out to give you moles of \(\ce{Fe(s)}\). The above conversion involves using multiple stoichiometric relationships from density, percent mass, and tooth mass.

The balanced equation must at present be used to convert moles of Iron(southward) to moles of H₂(g). Remember that the balanced equation's coefficients land the stoichiometric factor or mole ratio of reactants and products.

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

Footstep v: Check units

The question asks for how many grams of H_two(one thousand) were released so the moles of H_two(thousand) must however exist 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 cantlet.

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

3.74 x x^-5 mol H_ii(g) (two.01588g H₂(thousand)/1mol H₂ (g)) = vii.53 x x^-5 g H₂(one thousand) released

Bug

Stoichiometry and balanced equations brand it possible to use ane piece of data to calculate another. In that location are countless ways stoichiometry can be used in chemistry and everyday life. Try and see if you can use what you learned to solve the following bug.

1) Why are the following equations not considered counterbalanced?

\(H_2O_{(l)} \rightarrow H_{2(yard)} + O_{2(g)}\)
\(Zn_{(s)} + Au^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Ag_{(s)}\)

ii) Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions. Write the balanced chemical equation for this reaction.

iii) Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M solution?

4) If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the limiting reagent and how many moles of water are produced? The unbalanced equation is provided below.

\[\ce{CH4(g) + O2(g) \rightarrow CO2(thousand) + H2O(fifty)} \nonumber\]

5) A 0.777g sample of an organic compound is burned completely. It produces 1.42g CO_twoand 0.388g H_iiO. Knowing that all the carbon and hydrogen atoms in CO₂ and H₂O came from the 0.777g sample, what is the empirical formula of the organic chemical compound?

References

T. E. Brown, H.Due east LeMay, B. Bursten, C. Murphy. Chemistry: The Central Science. Prentice Hall, January eight, 2008.
J. C. Kotz P.Grand. Treichel, J. Townsend. Chemistry and Chemical Reactivity. Brooks Cole, Feb 7, 2008.
Petrucci, harwood, Herring, Madura. General Chemistry Principles & Modern Applications. Prentice Hall. New Jersey, 2007.

Contributors and Attributions

Joseph Nijmeh (UCD), Mark Tye (DVC)