Master MCAT molecules and stoichiometry with clear explanations, worked examples, and practice questions with detailed answers to boost your score.
(Note: This guide is part of our MCAT General Chemistry series.)
Table of Contents
Part 1: Introduction to molecules and stoichiometry
Part 2: Moles
a) Definition of a mole
b) Molecular mass
Part 3: Formation of molecules
a) Representations of molecular formulas
b) Types of chemical reactions
Part 4: Stoichiometry
a) Balancing equations
b) Determining theoretical yield
Part 5: High-yield terms
Part 6: Passage-based questions and answers
Part 7: Standalone questions and answers
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Part 1: Introduction to molecules and stoichiometry
Molecules may be difficult to conceptualize because they cannot be easily seen. For instance, you may be able to see individual grains of table salt, but you cannot visualize a molecule of salt: the smallest possible unit of salt that can be identified as such. Even then, the salt molecule can be further broken down into atoms such as sodium and chlorine atoms.
Stoichiometry allows us to quantify and understand such quantities by employing conversion factors centralized around the unit of mole. In chemistry, we often rely on stoichiometry to relate quantities of reactants and products. Thus, stoichiometry allows us to balance chemical equations according to laws of conservation.
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Part 2: Moles
a) Definition of a mole
How do we quantify reactants and products? The amount of a material that is present may be defined in terms of mass, numbers of particles, or some other unit.
It may be difficult to count the numbers of particles within a quantity of reactants or products. This is because the particles are very small and may be atoms or molecules. (For a review on the structure of the atom, be sure to refer to our guide on atomic and nuclear physics.) Instead, a unit called a mole (mol) is used to define these quantities.
Figure 1 Different ways of expressing the same quantity of a substance
A mole can also be defined in the context of gaseous substances. One mole of a gaseous compound occupies 22.4 L under standard temperature and pressure (STP). In chemistry, standard temperature and pressure refer to the following: a temperature of 273.15 K and an absolute pressure of 100 kPa (equal to 1 bar).
b) Molar mass
The atomic or molecular mass of a substance is used to convert the number of moles that are present to the mass in grams. For this reason, it is also referred to as molar mass or molecular weight.
Determining the molar mass of an atomic element is relatively simple. For the case of monoatomic elements, the molar mass of the element is simply the atomic mass. The atomic mass can be found right below the element’s symbol on the periodic table. For instance, Na (sodium) has an atomic mass of 22.99 AMU. Thus, we can use this number to convert the mass of a known quantity of sodium to the number of moles in that sample.
Figure: Converting 15.00 g of sodium (Na) to the corresponding quantity of moles.
Let’s take oxygen as an example. The atomic mass of oxygen is 15.999, approximately 16 AMU. However, oxygen in its natural state is present as a diatomic molecule (O2). Thus, to determine its molar mass, we must multiply this quantity by the number of atoms present in the molecule.
Figure 2 Converting 15.00 g of sodium (Na) to the corresponding quantity of moles
Note that there are two sodium atoms present. The atomic mass of sodium is 22.99 AMU. To account for the mass contributed by sodium: 2x22.99=45.98.
Note that there is one sulfur atom present. The atomic mass of sulfur is 32.07 AMU. To account for the mass contributed by sulfur: 1x32.07=32.07.
Note that there are four oxygen atoms present. The atomic mass of oxygen is 16.00 AMU. To account for the mass contributed by oxygen: 4x16.00=64.00
Thus, the total molecular mass of the compound is given by:
45.98 + 32.07 + 64.00 = 142.05 g/mol
Figure 3 Converting 15.00 g of sodium sulfate (Na2SO4) to the corresponding quantity of moles.
Using both Avogadro’s number and the molar mass of a substance, we can perform calculations to represent the quantity of a substance in grams, moles, or the number of particles.
Figure 4 Using Avogadro’s number to determine the number of molecules in a sample.
This information can also be used to calculate the density (ρ) of a material, as the density of a sample is equal to its mass divided by the volume it occupies. In other words,
ρ = m/V
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Part 3: Formation of molecules
a) Representations of molecular formulas
Molecular formulas can be represented as Lewis structures, structural formulae, and empirical formulas.
Lewis structures are visual depictions of atoms and bonds formed between atoms. (To learn more about this, be sure to refer to our guide on bonds and interactions.) Lewis structures begin with an elemental symbol of an atom. Dots surround the element to depict its valence electrons. Each side around the element gets a maximum of two dots to express an electron pair, but before a pair can form, each side of the element must have a dot to express the atom’s bonding capacity.
Figure 5 Writing a Lewis structure
Structural formulas are depictions of how atoms in compounds are bonded to one another. These are the commonly found “illustrations” of molecules. Straight lines between atoms represent chemical bonds. Using structural formulas, we can predict what the overall shape of the molecule may be.
Structural formulas incorporate single, double, and triple bonds. Each bond is drawn using a single solid line.
Figure 6 Structural formulas for water (H2O) and carbon dioxide (CO2).
To derive the empirical formula for glyceraldehyde, the lowest ratios of carbon, hydrogen, and oxygen would be determined. There are 3 atoms of carbon, 6 atoms of hydrogen, and 3 atoms of oxygen. The greatest common factor between these numbers is 3. Thus, the lowest ratio of 3:6:3 would be 1:2:1. The empirical formula for glyceraldehyde would be CH2O.
Figure 7 Molecular and empirical formula representations of glyceraldehyde.
b) Types of chemical reactions
How might molecules be formed and rearranged? Broadly speaking, there are five different types of chemical reactions: synthesis, single-displacement, double-displacement, decomposition, and combustion reactions.
A single replacement reaction occurs when one element replaces an element of a compound: A + BC → AC + B. For example, potassium reacts with water and yields potassium hydroxide and hydrogen gas: 2K + 2H2O → KOH + H2. Here, the hydrogen atom in H2O was replaced with potassium to form KOH.
A double replacement reaction occurs when the anions or cations of two compounds are exchanged, thus forming new compounds: AB + CD → AD + CB. For instance, when sodium chloride reacts with silver fluoride, sodium fluoride and silver chloride form. The anions, chlorine and fluorine, “swapped” places to form two new products. NaCl + AgF → NaF + AgCl.
A decomposition reaction occurs when a single product dissociates to two or more constituent reactants: AB → A + B. For instance, sodium chloride can decompose into sodium and chlorine. 2NaCl → 2Na + Cl2.
Figure 8 Four key types of chemical reactions
Finally, a combustion reaction occurs between a combustible reactant and oxygen. Combustion results in the production of heat, along with carbon dioxide and water in the case of complete combustion. In incomplete combustion, or combustion that takes place in the presence of insufficient oxygen, carbon monoxide and other carbon compounds may be formed—a substance we recognize as ash.
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Part 4: Stoichiometry
a) Balancing equations
Let’s begin to tie these concepts together. Recall that stoichiometry ensures that a chemical equation is balanced. In general, chemical equations list all reactants on the left side and all products on the right side. An arrow separates the two and indicates that a reaction takes place.
When writing equations, it is important to ensure they are balanced, that is, having the same quantities of each type of atom listed on both the reactant and the product sides. In addition, we can also see that conservation of charge also holds true. We do this by changing the coefficients of the compounds involved.
Figure 9 The unbalanced reaction for synthesis of table salt
First, we need to consider how many atoms each element has. There is one atom of sodium and two atoms of chlorine on the reactants side. On the products side, there is one atom of sodium and one atom of chlorine making up sodium chloride. While there are two atoms of chlorine on the reactants side, there is only one atom of chlorine on the products side. In order to conserve mass, we must balance the chemical equation so that the number of atoms of each element is equal both before and after a reaction.
Chemical equations can be balanced by adjusting stoichiometric coefficients, which describe the amounts of reactants and products relative to one another. Coefficients are located in normal font size, in front of an elemental symbol.
In this example, the stoichiometric coefficients in front of each of the reactants and products must be adjusted to balance the equation. This can be accomplished by placing a coefficient equal to 2 in front of both the Na and NaCl.
Figure 10 The balanced reaction for synthesis of table salt.
What information do these stoichiometric coefficients provide? They provide the molar ratio, or the relationship between the amounts of reactants consumed and products consumed. In other words: if one mole of chlorine gas (Cl2) is fully consumed, we can expect the reaction to yield two moles of salt (NaCl). This is because the molar ratio between this reactant and product is 1:2.
Be careful not to confuse coefficients with subscripts. Subscripts, the lowered numbers after an elemental symbol, indicate how many atoms are in that respective molecule. The total number of atoms in a molecule, therefore, can be determined by multiplying the coefficient by the subscript. For instance, if there are 2 chlorine molecules as indicated by the coefficient and 2 inherent atoms of that molecule as indicated by the subscripts, there are 4 atoms of that molecule in the context of the particular chemical equation.
Figure 11 Using subscripts and coefficients to determine the number of atoms.
b) Determining theoretical yield
In real life, there is a limit on our reactants or products. Once the limiting reactant of a reaction is fully consumed, the reaction will stop. How can we determine when this will occur, and which reactant is the limiting reactant?
Stoichiometry plays a significant role in this process. When provided a balanced chemical reaction and the quantities (in grams or moles) of each reactant involved, we will have enough information to solve for the limiting reactant. To do so,
Start by assuming one of the reactants to be the limiting reactant.
Then, determine the amount of product that would form. Be sure to use a balanced chemical equation, and consider molar ratios.
Repeat steps 1-2 for each of the reactants.
The reactant that yields the lowest amount of hypothetical product is the limiting reactant.
It’s that simple! Let’s work through an example.
Since this is the reaction for which we have already written a balanced equation, we can use the equation written in the previous section.
Figure 13 Math demonstrating example of how to calculate the limiting reactant
If sodium were the limiting reactant, 0.331 moles of NaCl would be produced. If chlorine were the limiting reactant, we would end up with 0.119 moles of NaCl. Chlorine is thus our limiting reactant since less product is produced by assuming chlorine is the limiting reactant. In this reaction, only 0.119 moles of NaCl will be produced.
It’s worth noting that these calculations are only theoretical. Our calculated value of 0.119 moles of NaCl is a theoretical yield. Yet, in actuality, the experimental yield varies from the theoretical yield. Due to nonideal conditions, waste from the experimental process, and less than 100% reactivity, the experimental yield tends to be less than the theoretical yield. Thus, the efficiency, or percent yield, of a reaction is useful to consider.
The higher a percent yield value is, the more product is produced.
A calculated mass percentage represents how much of a product is made up by an element of interest.
Let’s calculate the percent mass of hydrogen and oxygen in water.
Figure 14 Math demonstrating example of how to calculate the mass percentage of a molecule, using water as an example.
Thus, the mass percent of hydrogen in water is 11.12%. For every water molecule, 11.12% will be made up of hydrogen. Further, the mass percent of oxygen in water is 88.88%.
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Part 5: High-yield terms
Avogadro’s number (NA): quantifies the number of particles in one mole, where NA = 6.022 x 1023
Standard temperature and pressure (STP): a temperature of 273.15 K and an absolute pressure of 100 kPa (equal to 1 bar).
Molar/atomic/molecular mass: used to convert the number of moles that are present to the mass in grams
Lewis structures: visual depictions of atoms and bonds formed between atoms
Structural formulas: illustrations that predict the shape of a molecule
Empirical formulas: written formulas that express the lowest whole-number ratios among elements of a compound
Synthesis reaction: occurs when two or more reactants come together to form a single product: A + B → AB
Single replacement reaction: occurs when one element replaces an element of a compound: A + BC → AC + B
Double replacement reaction: occurs when the anions or cations of two compounds are exchanged, thus forming new compounds: AB + CD → AD + CB
Decomposition reaction: occurs when a single product dissociates to two or more constituent reactants AB → A + B
Complete combustion: results in the production of carbon dioxide and water from hydrocarbons and oxygen
Incomplete combustion: results in the production of ash, carbon monoxide, carbon dioxide, and water from hydrocarbons and oxygen
Stoichiometric coefficients: indicate the amounts of reactants and products relative to one another
Molar ratio: numerical relationship between the amounts of reactants consumed and products consumed
Subscripts: lowered numbers after an elemental symbol that indicate how many atoms are in that respective molecule
Limiting reactant: reactant that is fully consumed first in a chemical reaction
Percent yield: (experimental yield ÷ theoretical yield) x 100
Mass percentage: (mass of element ÷ mass of compound) x100
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Part 6: Passage-based questions and answers
Adult participants were assigned to one of two experimental groups: either a placebo group or the caffeine-stimulated group. Both groups underwent sleep deprivation for 50 hours. The caffeine-stimulated group was asked to consume four gum pellets containing caffeine every 2 hours. Each gum pellet contained 200 mg of caffeine. The cognitive task administration timetable is shown in Figure 1.
Figure 1 Caffeine administration and testing session timetable
The molar mass of caffeine is 194.194 grams per mole. The molar mass of dichloromethane is 84.93 grams per mole.
CREATOR AND ATTRIBUTION PARTY: AIDMAN, E., BALIN, M., JOHNSON, K. ET AL. CAFFEINE MAY DISRUPT THE IMPACT OF REAL-TIME DROWSINESS ON COGNITIVE PERFORMANCE: A DOUBLE-BLIND, PLACEBO-CONTROLLED SMALL-SAMPLE STUDY. SCI REP 11, 4027 (2021). THE ARTICLE’S FULL TEXT IS AVAILABLE HERE: HTTPS://WWW.NATURE.COM/ARTICLES/S41598-021-83504-6. THE ARTICLE IS NOT COPYRIGHTED BY SHEMMASSIAN ACADEMIC CONSULTING. DISCLAIMER: SHEMMASSIAN ACADEMIC CONSULTING DOES NOT OWN THE PASSAGE PRESENTED HERE. CREATIVE COMMON LICENSE: HTTP://CREATIVECOMMONS.ORG/LICENSES/BY/4.0/. CHANGES WERE MADE TO ORIGINAL ARTICLE TO CREATE AN MCAT-STYLE PASSAGE.
Question 1: Which of the following correctly provides empirical and molecular formulas for caffeine?
A) Empirical: C8H10N4O2; Molecular: C8H10N4O2
B) Empirical: C4H5N2O; Molecular: C8H10N4O2
C) Empirical: C8H10N4O2; Molecular: C4H5N2O
D) Empirical: CHNO; Molecular: C4H5N2O
A) 0.65 g CH2Cl2
B) 7.2 g CH2Cl2
C) 0.11 g CH2Cl2
D) 3.42 g of CH2Cl2
Question 3: A single gum pellet dissolved in 300 grams of dichloromethane yields 0.0005 moles of the fully dissolved product. What is the percent yield for the dissociation reaction?
A) 25%
B) 0.05%
C) 50%
D) 100%
Question 4: An unknown quantity of caffeine was fully dissolved in dichloromethane. The reaction required 60 grams of dichloromethane. Assuming 100% yield, what was the original quantity of caffeine?
A) 15 g C8H10N4O2
B) 260 g C8H10N4O2
C) 40 g C8H10N4O2
D) 0.4 * 10-5 C8H10N4O2
Answer key for passage-based questions
1. Answer choice B is correct. As provided in the passage, the molecular formula of caffeine is C8H10N4O2. Empirical formulas use the lowest ratios of subscripts based on the molecular formula (choices A and C are incorrect). The lowest possible ratios of 8:10:4:2 would be 4:5:2:1 since the greatest common factor among 8, 10, 4, 2 is 2 (choice B is correct). A molecular formula describes the actual composition of the compound (choices C and D are incorrect).
2. Answer choice A is correct. First, convert the grams of caffeine in the gum to moles: 0.5 g C8H10N4O2* (1 mol C8H10N4O2/ 194.194 g) = 0.0025 moles of caffeine. It is known that the stoichiometric ratio between dichloromethane and caffeine is 3:1. Thus, this quantity of caffeine would react with 0.0025 x 3 = 0.0075 moles of dichloromethane.
This quantity of dichloromethane can be converted to grams by using the molar mass of dichloromethane: 0.0075 mol CH2Cl2* (84.93 g / 1 mol CH2Cl2) = 0.65 grams.
Note that while the use of a calculator is not permitted on the MCAT, the molar masses of caffeine and dichloromethane can be approximated as 200 g/mol and 100 g/mol, respectively. These approximations yield a final answer of 0.75 grams, which is sufficient to identify the correct answer option.
3. Answer choice C is correct. To determine percent yield, we must first determine the limiting reactant for solvation reaction. First, assume that caffeine is the limiting reactant. A single pellet contains 200 mg, or 0.2 g, of caffeine. This is equal to 0.2÷194.14 ≈
0.2÷200 = 0.001 moles of caffeine. This quantity would yield 0.001 moles of the dissolved product. Secondly, assume that dichloromethane is the limiting reactant. 300 grams of dichloromethane is equal to 100÷84.93 ≈
300÷100 = 3 moles of dichloromethane. According to the molar rations presented in the passage, this quantity would yield 1 mole of the dissolved product. Since the theoretical yield calculated by assuming caffeine is the limiting reagent is smaller, we would assume this to be the theoretical yield.
Percent yield is calculated by dividing experimental yield (0.0005 moles) by theoretical yield (0.001 moles). Thus, the percent yield of this reaction is 50%. (choice C is correct).
4. Answer choice C is correct. The provided quantity of dichloromethane can be converted to moles: 60 g x (1 mol/100 g) = 0.6 moles. The molar ratio between dichloromethane and caffeine is 3:1; thus, 0.6÷3=0.2 moles of caffeine were present in the original reaction. This quantity of caffeine can be converted to grams: 0.2 mol x (200 g/mol) = 40 grams (choice C is correct).
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Part 7: Standalone questions and answers
A) 2 x 1023
B) 1.5 x 1015
C) 12 x 1023
D) 4 x 1023
A) Water is not a product of this chemical reaction
B) 12.00 g H2O
C) 0.57 g H2O
D) 4.9034 g H2O
Question 3: Carbon dioxide reacts with water to produce glucose and oxygen. What are the stoichiometric coefficients of this reaction?
A) 1, 1, 1, 1
B) 6, 6, 6, 6
C) 6, 6, 1, 6
D) 1, 2, 6, 1
Question 4: Sodium chloride reacts with silver fluoride to form a solid precipitate. What type of reaction has occurred?
A) Single replacement
B) Double replacement
C) Synthesis
D) Disproportionation
Answer key for standalone questions
1. Answer choice C is correct. The provided sample of 200 grams is equivalent to: 200 x (1 mol/300 g) = ⅔ moles of calcium phosphate. Each mole of calcium phosphate contains three moles of calcium atoms; this information is provided by the subscript number 3 after the chemical symbol Ca. Thus, the sample contains ⅔ x 3 = 2 moles of calcium. This can be converted to the number of atoms by using Avogadro’s number: 2 x 6.022 x 1023=12 x 1023 (choice C is correct).
2. Answer choice D is correct. A complete combustion reaction results in the formation of water and carbon dioxide. The balanced reaction is as follows: 1C3H8 + 5O2 → 3CO2 + 4H2O. Thus, the dimensional analysis of this reaction is as follows: 3.00 g C3H8 * (1 mol C3H8/44.1 g) * (4 mol H2O/1 mol C3H8)(18.02 g H2O/1 mol H2O) = 4.9034 g H2O (choice D is correct). Note that although the use of a calculator is not permitted on the MCAT, the molar mass of propane can be approximated as 50 g/mol, and the molar mass of water can be approximated as 20 g/mol. Use of these approximations would result in an answer of 4.8 grams of water, which is sufficient to select the correct answer choice.
3. Answer choice C is correct. First, write a chemical equation based on the provided reactants and products: CO2 + H2O → C6H12O6 + O2. Then, balance the chemical equation, ensuring that the number of atoms of each element is equal on both sides. This results in the following: 6CO2 + 6H2O → C6H12O6 + 6O2. The numbers in front of each product or reactant provide the stoichiometric coefficients: 6, 6, 1, 6 (choices A and D are incorrect). Stoichiometric coefficients are usually reduced to their least common multiples (choice B is incorrect).
4. Answer choice B is correct. The correct reaction is written as follows: AgF + NaCl → AgCl + NaF. One of the products, AgCl, forms an insoluble compound, which precipitates from the reaction. (For more information on precipitates, be sure to refer to our guide on solutions and gases.) The two cations have switched places in this reaction (choices A and C are incorrect). As a result, this is a double replacement reaction (choice B is correct). A disproportionation reaction is a type of redox reaction—for more information on this, be sure to refer to our guide on oxidation and reduction reactions (choice D is incorrect).
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