# General Chemistry/Stoichiometry

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.]

## Molar Calculations

### Your Tool: Dimensional Analysis

Luckily, almost all of stoichiometry can be solved relatively easily using dimensional analysis. Dimensional analysis is just using units, instead of numbers or variables, to do math, usually to see how they cancel out. For instance, it is easy to see that:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle grams \times \dfrac{moles}{grams} \times \dfrac{atoms}{moles} = atoms}**

It is this principle that will guide you through solving most of the stoichiometry problems (chemical reaction problems) you will see in General Chemistry. Before you attempt to solve a problem, ask yourself: what do I have now? where am I going? As long as you know how many (units) per (other units), this will make stoichiometry significantly easier.

### Moles to Mass

*How heavy is 1.5 mol of lead? How many moles in 22.34g of water?* Calculating the mass of a sample from the number of moles it contains is quite simple. We use the **molar mass** (mass of one mole) of the substance to convert between mass and moles. When writing calculations, we denote the molar mass of a substance by an upper case "M" (e.g. M(Ne) means "the molar mass of neon"). As always, "n" stands for the number of moles and "m" indicates the mass of a substance. To find the solutions to the two questions we just asked, let's apply some dimensional analysis:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 1.5 \; mol \; Pb \; \times \dfrac{207.2 \; g \; Pb}{1 \; mol \; Pb} = 310.8 \; g \; Pb}**

Can you see how the units cancel to give you the answer you want? All you needed to know was that you had 1.5 mol Pb (lead), and that 1 mol Pb weighs 207.2 grams. Thus, multiplying 1.5 mol Pb by 207.2 g Pb and dividing by 1 mol Pb gives you 310.8 g Pb, your answer.

### Mass to Moles

But we had one more question: "How many moles in 22.34g of water?" This is just as easy:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 22.34 \; g \; H_2 O \; \times \dfrac{1 \; mol \; H_2 O}{18 \; g \; H_2 O} = 1.24 \; mol \; H_2 O}**

Where did the 18 g H_{2}O come from? We looked at the periodic table and simply added up the atomic masses of two hydrogens and an oxygen to get the molecular weight of water. This turned out to be 18, and since all the masses on the periodic table are given with respect to 1 mole, we knew that 1 mol of water weighed 18 grams. This gave us the relationship above, which is really just (again) watching units cancel out!

### Calculating Molar Masses

Before we can do these types of calculations, we first have to know the molar mass. Fortunately, this is not difficult, as the molar mass is exactly the same as the **atomic weight** of an element. A table of atomic weights can be used to find the molar mass of elements (this information is often included in the periodic table). For example, the atomic weight of oxygen is **16.00** amu, so its molar mass is 16.00 g/mol.

For species with more than one element, we simply add up the atomic weights of each element to obtain the molar mass of the compound. For example, sulfur trioxide gas is made up of sulfur and oxygen, whose atomic weights are 32.06 and 16.00 respectively.

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{matrix} \hbox{M(SO}_3\hbox{)} &=& 32.06 + 3 \times 16.00 \\ \ &=& 80.06 \hbox{g}/\hbox{mol} \\ \end{matrix}}**

The procedure for more complex compounds is essentially the same. Aluminium carbonate, for example, contains aluminium, carbon, and oxygen. To find the molar mass, we have to be careful to find the *total* number of atoms of each element. Three carbonate ions each containing three oxygen atoms gives a total of nine oxygens. The atomic weights of aluminium and carbon are 26.98 and 12.01 respectively.

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{matrix} \hbox{M}(\hbox{Al}_2(\hbox{CO}_3)_3) &=& 2 \times 26.98 + 3 \times 12.01 + 9 \times 16.00 \\ \ &=& 233.99 \hbox{g}/\hbox{mol} \\ \end{matrix}}**

## Empirical Formulae

The empirical formula of a substance is *the simplest ratio of the number of moles of each element in a compound*. The empirical formula is ambiguous, e.g. the formula CH could represent CH, C_{2}H_{2}, C_{3}H_{3} etc. These latter formulae are called **molecular formulae**. It follows that the molecular formula is always a *whole number multiple* of the empirical formula for a compound.

Calculating the empirical formula is easy if the relative amounts of each element in the compound are known. For example, if a sample contains 1.37 mol oxygen and 2.74 mol hydrogen, we can calculate the empirical formula. A good strategy to use is to divide all amounts given by the smallest **non-integer** amount, then multiply by whole numbers until the simplest ratio is found. We can make a table showing the successive ratios.

Hydrogen | Oxygen | |
---|---|---|

2.74 | 1.37 | divide by 1.37 |

2 | 1 | ANSWER |

The empirical formula of the compound is H_{2}O.

Here's another example. A sample of piperonal contains 1.384 mol carbon, 1.033 mol hydrogen and 0.519 mol oxygen.

Carbon | Hydrogen | Oxygen | |
---|---|---|---|

1.384 | 1.033 | 0.519 | divide by 0.519 |

2.666 | 2 | 1 | multiply by 3 |

8 | 6 | 3 | ANSWER |

The empirical formula of piperonal is C_{8}H_{6}O_{3}.

### Converting from Masses

Often, we are given the relative composition by mass of a substance and asked to find the empirical formula. These masses must first be converted to moles using the techniques outlined above. For example, a sample of ethanol contains 52.1% carbon, 13.2% hydrogen, and 34.7% oxygen by mass. Hypothetically, 100g of this substance will contain 52.1 g carbon, 13.2 g hydrogen and 34.7 g oxygen. Dividing these by their respective molar masses gives the amount in moles of each element (as we learned above). These are 4.34 mol, 13.1 mol, and 2.17 mol respectively.

Carbon | Hydrogen | Oxygen | |
---|---|---|---|

4.34 | 13.1 | 2.17 | divide by 2.17 |

2 | 6 | 1 | ANSWER |

The empirical formula of ethanol is C_{2}H_{6}O.

### Molecular Formula

**Beware:**In the case of H

_{2}O, the whole number multiple is 1, so its empirical formula is the same as its molecular formula. This is not always the case!

As mentioned above, the molecular formula for a substance equals the count of atoms of each type in a molecule. This is always a *whole number multiple* of the empirical formula. To calculate the molecular formula from the empirical formula, we need to know the **molar mass** of the substance. For example, the empirical formula for **benzene** is CH, and its molar mass is 78.12 g/mol. Divide the actual molar mass by the mass of the empirical formula, 13.02 g/mol, to determine the multiple of the empirical formula, "n". The molecular formula equals the empirical formula multiplied by "n".

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{matrix} \hbox{M(CH)} &=& 13.02 \hbox{ g/mol} \\ \hbox{M(benzene)} &=& 78.12 \hbox{ g/mol}\\ \hbox{M(benzene)} / \hbox{M(CH)} &= (78.12\ g/mol)/(13.02\ g/mol)&= 6 \\ \end{matrix}}**

This shows that the molecular formula for benzene is 6 times the empirical formula of CH. The molecular formula for benzene is C_{6}H_{6}.

## Solving Mass-Mass Equations

A typical mass-mass equation will give you an amount in grams and ask for another answer in grams.

- To solve a mass-mass equation, follow these rules

- Balance the equation if it is not already.
- Convert the given quantity to moles.
- Multiply by the molar ratio of the demanded substance over the given substance.
- Convert the demanded substance into grams.

For example, given the equation **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle Cu^{2+} + 2 AgNO_3 \to Cu(NO_3)_2 + 2 Ag^+}**
, find out how many grams of silver (Ag) will result from 43.0 grams of copper (Cu) reacting.

- Convert the given quantity to moles.

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 43.0g~Cu \times \frac{1~mol~Cu}{63.55g~Cu}}**

- Multiply by the molar ratio of the demanded substance and the given substance.

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 43.0g~Cu \times \frac{1~mol~Cu}{63.55g~Cu} \times \frac{2~mol~Ag}{1~mol~Cu}}**

- Convert the demanded substance to grams.

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle 43.0g~Cu \times \frac{1~mol~Cu}{63.55g~Cu} \times \frac{2~mol~Ag}{1~mol~Cu} \times \frac{107.86g~Ag}{1~mol~Ag} = 1.46 \times 10^2~g~Ag}**

Notice how dimensional analysis applies to this technique. All units will cancel except for the desired one (grams of silver, in this case). |

## Summary

To solve a stoichiometric problem, you need to know what you already have and what you want to find. Everything in between is basic algebra.

- Key Terms

*Molar mass:*mass (in grams) of one mole of a substance.*Empirical formula:*the simplest ratio of the number of moles of each element in a compound*Molecular formula:*the actual ratio of the number of moles of each element in a compound

In general, all you have to do is keep track of the units and how they cancel, and you will be on your way!