$$\require{mhchem}$$

HOME

General Chemistry Study Guide

Chapter 3. Stoichiometry

---

Yu Wang

1. Molar Mass

1.1. Atomic Mass

OpenStax 2.3 Atomic Structure and Symbolism; 3.1 Formula Mass and the Mole Concept. Brown 2.4 Atomic Weights; 3.4 Avogadro’s Number and the Mole.

Atomic mass is the mass of an atom in atomic mass units (amu)

One atomic mass unit equals to $1/12$ of the mass of one $\ce{^{12}C}$ atom.

On this scale the atomic mass of $\ce{^{1}H}$ is 1.008 amu, NOT exactly 1 amu because the mass of one proton varies in different atoms.

Average atomic mass is the weighted average of all of the naturally occurring isotopes of the element. (The number below every element symbol in the periodic table)

Example: The atomic masses of its two stable isotopes, $\ce{^{63}_{29}Cu}$ (69.09%) and $\ce{^{65}_{29}Cu}$ (30.91%), are 62.93 amu and 64.9278 amu, respectively. What is the average atomic mass?
Answer:
$$62.93\,\text{amu}\times69.09\%+64.9278\,\text{amu}\times30.91\%=63.55\,\text{amu}$$

The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of $\ce{^{12}C}$.

$$1\,\text{mol} = N_A = 6.0221415\times10^{23}$$

$N_A$ is Avogadro's number.

Molar mass is the mass of 1 mole of something in grams. For any element, the atomic mass (amu) equals to the molar mass (grams).

Example: How many grams of Zn are in 0.356 mole of Zn?
Answer:
$$0.356\,\text{mol}\,\ce{Zn}\times\frac{65.39\,\text{g}\,\ce{Zn}}{1\,\text{mol}\,\ce{Zn}} = 23.3\,\text{g}\,\ce{Zn}$$

Requirements

1. Understand the concepts: atomic mass, average atomic mass and molar mass. Understand the defination of mole.
2. Learn how to calculate the mass of element ($m$), number of moles of element ($n$) and number of atoms of element ($N$). Know one of the values, calculate the other two.

1.2. Molecular Mass

OpenStax 3.1 Formula Mass and the Mole Concept. Brown 3.3 Formula Weights.

Molecular mass (also known as molecular weight) is the sum of the atomic masses (in amu) in a molecule.

The molar mass (grams) of a molecule equals to the molecular mass (amu).

Example: The molecular mass of $\ce{H2O}$. $$(1.008\,\text{amu})\times2+16.00\,\text{amu} = 18.02\,\text{amu}$$ How many moles of $\ce{CH4}$ are present in 6.07 g of $\ce{CH4}$?
Answer:
$$\text{molar mass of }\,\ce{CH4}=12.01\,\text{g}+4\times(1.008\,\text{g})=16.04\,\text{g}$$ $$6.07\,\text{g}\,\ce{CH4}\times\frac{1\,\text{mol}\,\ce{CH4}}{16.04\,\text{g}\,\ce{CH4}}=0.378\,\text{mol}\,\ce{CH4}$$

1.3. Formular Mass

OpenStax 3.1 Formula Mass and the Mole Concept. Brown 3.3 Formula Weights.

Formula mass is the sum of the atomic masses (in amu) in the chemical formula of the substance.

Formula mass and molecular mass are the same for molecular compounds. But for ionic compounds, there is no molecular mass, but only formula mass.

The molar mass (grams) of any ionic compound equals to the formula mass (amu).

Example: Calculate the formula mass of $\ce{NaCl}$.
Answer:
The atomic mass of Na is 22.99 amu, Cl is 35.45 amu. The formula mass of NaCl is: $$1\times22.99\,\text{amu} + 1\times35.45\,\text{amu} = 58.44\,\text{amu}$$

from IPython.display import YouTubeVideo
YouTubeVideo("74-X94OP2XI", width=700, height=550)

# Watch the following video for a better understanding 
# on Avogadro's Number, The Mole, Grams, Atoms, Molar Mass.

1.4. Percent Composition

OpenStax 3.2 Determining Empirical and Molecular Formulas; 4.5 Quantitative Chemical Analysis. Brown3.2 Simple Patterns of Chemical Reactivity; 3.3 Formula Weights; 3.5 Empirical Formulas from Analyses.

Percent composition is the percent by mass of each element present in a compound.

Water, $\ce{H2O}$, is the first example. One mole of water is $18.0152\,\text{grams}$. In that compound, there are two moles of $\ce{H}$ atoms and $2\times1.008 = 2.016\,\text{grams}$. That's how many grams of hydrogen are present in one mole of water.

There is also one mole of oxygen atoms weighing $16.00\,\text{grams}$ in the mole of water.

To get the percentage of hydrogen, divide the $2.016$ by $18.015$ and multiply by $100$, giving $11.19\%$.

For oxygen it is $16.00 \div 18.015 = 88.81\%$.

Notice that you can also minus hydrogen's percentage from $100\%$. So, for oxygen it is $100\% - 11.19\%= 88.81\%$.

Experimental Determination of Empirical Formulas

Example: Combustion experiment of ethanol. Combust $11.5\,\text{g}$ ethanol, get $22.0\,\text{g}\,\ce{CO2}$ and $13.5\,\text{g}\,\ce{H2O}$. What is the empirical formula of ethanol?
Answer:
$$\ce{C}\text{: }22.0\,\text{g}\,\ce{CO2}\times\frac{1\,\text{mol}\,\ce{CO2}}{44.0\,\text{g}\,\ce{CO2}}\times\frac{1\,\text{mol}\,\ce{C}}{1\,\text{mol}\,\ce{CO2}}=0.50\,\text{mol}\,\ce{C}$$ $$\ce{H}\text{: }13.5\,\text{g}\,\ce{H2O}\times\frac{1\,\text{mol}\,\ce{H2O}}{18.0\,\text{g}\,\ce{H2O}}\times\frac{2\,\text{mol}\,\ce{H}}{1\,\text{mol}\,\ce{H2O}}=1.50\,\text{mol}\,\ce{H}$$ $$\ce{O}\text{: }\text{mass of}\,\ce{O} = 11.5\,\text{g}\,\text{ethanol} - 0.50\,\text{mol}\,\ce{C}\times\frac{12\,\text{g}\,\ce{C}}{1\,\text{mol}\,\ce{C}}-1.50\,\text{mol}\,\ce{H}\times\frac{1.00\,\text{g}\,\ce{H}}{1\,\text{mol}\,\ce{H}}=4.0\,\text{g}\,\ce{O}$$ $$\text{moles of}\,\ce{O} = 4.0\,\text{g}\,\ce{O}\times\frac{1\,\text{mol}\,\ce{O}}{16.0\,\text{g}\,\ce{O}}=0.25\,\text{mol}\,\ce{O}$$ Divide by $0.25\,\text{mol}$, get the ratio of each element. The empirical formula of ethanol is $\ce{C2H6O}$.

from IPython.display import YouTubeVideo
YouTubeVideo("JeSSucG-CVw", width=700, height=550)

# Watch the following video for a better understanding 
# on how to get empirical forumlas from percent composition.

1.5. Mass Spectrometer

(Just to have a brief idea about this analytical method)

Mass spectrometry (MS) is an analytical technique that ionizes chemical species and sorts the ions based on their mass-to-charge ratio. In simpler terms, a mass spectrum measures the masses within a sample.

Requirements

1. Learn how to calculate molar mass, molecular mass and formula mass of molecular and/or ionic compounds.
2. Learn how to calculate the percent composition of a compound.
3. Learn how to determine empirical formula from combustion experiment.

2. Balance Chemical Reactions

OpenStax 4.1 Writing and Balancing Chemical Equations; 4.3 Reaction Stoichiometry; 4.4 Reaction Yields. Brown 3.1 Chemical Equations; 3.6 Quantitative Information from Balanced Equations; 3.7 Limiting Reactants.

2.1. Balanceing Chemical Equations

A process in which one or more substances is changed into one or more new substances is a chemical reaction.

A chemical equation uses chemical symbols to show what happens during a chemical reaction:

$$\text{reactants}\ce{->}\text{products}$$

Example: $$\ce{2Mg + O2 -> 2 MgO}$$ means 2 $\ce{Mg}$ atoms + 1 $\ce{O2}$ molecule result in 2 formula units of $\ce{MgO}$, or 2 moles of $\ce{Mg}$ + 1 mole of $\ce{O2}$ result in 2 moles of $\ce{MgO}$, NOT 2 grams of $\ce{Mg}$ + 1 gram of $\ce{O2}$ result in 2 grams of $\ce{MgO}$. Actually, 48.6 grams of $\ce{Mg}$ + 32.0 grams of $\ce{O2}$ result in 80.6 grams of $\ce{MgO}$.

Steps to balance chemical equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
   1. Start by balancing those elements that appear in only one reactant and one product.
   2. Balance those elements that appear in two or more reactants or products.
3. Check to make sure that you have the same number of each type of atom on both sides of the equation.

Example: $$\ce{C2H6 + O2 -> CO2 + H2O}$$ Answer:
1. Balance the number of $\ce{C}$ and $\ce{H}$. The coefficients of $\ce{CO2}$ and $\ce{H2O}$ become 2 and 3, respectively. 2. Balance the number of $\ce{O}$, giving the coefficient of $\ce{O2}$ as 3.5. 3. Multiply both sides by 2, giving $$\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}$$

from IPython.display import YouTubeVideo
YouTubeVideo("Z1kucSx-Cdo", width=700, height=550)

# Watch the following video for a better understanding 
# on how to balance an equation.

Amounts of Reactants and Products

1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity
4. Convert moles of sought quantity into desired units

Example: In the reaction $$\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$$ If 856 g of $\ce{C6H12O6}$ is consumed, how much $\ce{CO2}$ is produced?
Answer:
$$\text{mass of }\ce{CO2} = 856\,\text{g}\,\ce{C6H12O6}\times\frac{1\,\text{mol}\,\ce{C6H12O6}}{180.2\,\text{g}\,\ce{C6H12O6}}\times\frac{6\,\text{mol}\,\ce{CO2}}{1\,\text{mol}\,\ce{C6H12O6}}\times\frac{44.01\,\text{g}\,\ce{CO2}}{1\,\text{mol}\,\ce{CO2}}=1.25\times10^3\,\text{g}\,\ce{CO2}$$

from IPython.display import YouTubeVideo
YouTubeVideo("-1xfnq8yGk8", width=700, height=550)

# Watch the following video for a better understanding 
# on how to do stoichiometric calculation of a reaction.

2.2. Limiting Reagent

Limiting reagent is the reactant used up first in the reaction.

Method 1 (More calculations but easier to understand):

1. Calculate the amount of product could be produced by given amount of reactant A; Calculate the amount of product could be produced by given amount of reactant B; ...
2. Whichever reactant gives the least amount of product is the limiting reagent.

Method 2 (A little more tricky but less calculation):

1. Calculate the amount of all reactants in moles.
2. Devide the amount of all reactants with their coefficients.
3. The smallest number you get from step two corresponds to the limiting agent.

from IPython.display import YouTubeVideo
YouTubeVideo("OqlMFEuDews", width=700, height=550)

# Watch the following video for a better understanding 
# on imiting reagent, theoreticall yield, and percent yield.

2.3. Reaction Yield

Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtained from a reaction.

$$\%\,\text{Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$$

Example:
$$\ce{2NH3 + CO2 -> (NH2)2CO + H2O}$$ In one process, 637.2 g of $\ce{NH3}$ are treated with 1142 g of $\ce{CO2}$. (a) Which of the two reactants is the limiting reagent?
Answer:
$$\text{moles of }\ce{NH3}=637.2\,\text{g}\,\ce{NH3}\times\frac{1\,\text{mol}\,\ce{NH3}}{17.03\,\text{g}\,\ce{NH3}}=37.42\,\text{mol}$$ $$\text{moles of }\ce{CO2}=1142\,\text{g}\,\ce{CO2}\times\frac{1\,\text{mol}\,\ce{CO2}}{44.01\,\text{g}\,\ce{CO2}}=25.95\,\text{mol}$$ $$\frac{37.42}{2}=18.71<25.95$$ Thus, $\ce{NH3}$ is the limiting reagent.
(b) Calculate the mass of $\ce{(NH2)2CO}$ formed, i.e. the theoretical yield.
Answer:
$$\text{mass of }\ce{(NH2)2CO} = 637.2\,\text{g}\,\ce{NH3}\times\frac{1\,\text{mol}\,\ce{NH3}}{17.03\,\text{g}\,\ce{NH3}}\times\frac{1\,\text{mol}\,\ce{(NH2)2CO}}{2\,\text{mol}\,\ce{NH3}}\times\frac{60.06\,\text{g}\,\ce{(NH2)2CO}}{1\,\text{mol}\,\ce{(NH2)2CO}}=1124\,\text{g}\,\ce{(NH2)2CO}$$ (c) If 1006 g $\ce{(NH2)2CO}$ was formed, what is the $\%\,\text{Yield}$?
Answer:
$$\%\,\text{Yield}=\frac{\text{1006}}{\text{1124}}\times100\% = 89.5\%$$ (d) How much excess reagent (in grams) is left at the end of the reaction?
Answer:
$$\text{moles of }\ce{CO2}\text{ left}= 25.95\,\text{mol} - \frac{37.42\,\text{mol}}{2}=7.24\,\text{mol}$$ $$\text{mass of }\ce{CO2}\text{ left} = 7.24\,\text{mol}\times\frac{44.01\,\text{g}\ce{CO2}}{1\,\text{mol}\ce{CO2}}=314\,\text{g}\ce{CO2}$$

Requirements

1. Learn how to balance a chemical reaction.
2. Learn to determine limiting reagent, to calculate theoretical yield and $\%\,\text{Yield}$.

---

Please ignore the cell below. It just loads our style for the notebook.

from IPython.core.display import HTML
def css_styling():
    styles = open("custom.css", "r").read()
    return HTML(styles)
css_styling()