# Unit 7: Stoichiometry

### A set of review materials for Unit 7: Stoichiometry

## The Mole

We need this because when we are talking about chemical reactions, the coefficients we make in our balanced reaction are ratios between the numbers of molecules of different compounds. But counting molecules is almost impossible. Instead we use moles as a way of counting the **enormous** number of atoms needed to have an amount that we can physically handle.

When doing the math, use Avogadro's Number as a conversion factor.

Check out this video if you need more help:

## Molar Mass

*super*easy, the molar mass of an element is the average atomic mass of that element as found on the periodic table. The unit for molar mass is grams per mole (g/mol).

To calculate the molar mass of a compound, you simply will need to add up the average atomic mass of each atom present in the compound.

## Gram <=> Mole Conversions

- Grams => moles, divide by the molar mass.
- Moles => grams, multiply by the molar mass.

You can also combine this with mole to particle and particle to mole conversions!

## Percent Composition

Originally, we had to do very careful experiments to determine how much of a sample of copper oxide was copper and how much was oxygen. When we did, we noticed that the percentages were always either 89% copper and 11% oxygen **or** 80% copper and 20% oxygen. After a lot of careful math, you can determine that these percentages correspond to the two different compounds formed between copper and oxygen: Copper (I) Oxide and Copper (II) Oxide. These percentages are how we originally developed chemical formulas and determined the subscripts in all compounds.

Calculating the percentage composition from the formulas is not too challenging if you remember that the math for **any** percent is (part/whole) x 100. In this instance, the 'part' is the part of the molar mass that is element A and the 'whole' is the molar mass of the whole thing

## Empirical Formula

Empirical formulas can be calculated from a percent composition by following these steps:

- Assume a 100 g sample (change % to g)
- Convert the grams for each element to moles (divide by molar mass)
- Reduce the ratio (divide by the smallest number)
- optional: scale to whole numbers (1/2 => multiply by 2. 1/3 or 2/3 => multiply by 3. 1/4 or 3/4 => multiply by 4)

## Molecular Formula

**not**in it's simplest ratio. The relationship between a molecular formula and an empirical formula is always some integer multiple.

Empirical x n = Molecular

As such, the relationship between the molar mass of the empirical formula and the molecular formula will always be the exact same integer.

Empirical Mass x n = Molecular Mass

Molecular formulas can be calculated from the empirical formulas if you know the molar mass of the molecular formula.

## The Mole Ratio (Using a Chemical Reaction)

## Putting It All Together (Stoichiometry)

- How many grams of hydrogen peroxide are needed to make 15 grams of water?
- How many grams of silver are made when 16 grams of copper reacts?
- How many grams of oxygen are needed to react with 20 grams of hydrogen?

At the most, Stoichiometry is a three step process:

- Grams of compound A to moles of compound A (divide by molar mass)
- Moles of compound A to moles of compound B (multiply by the mole ratio)
- Moles of compound B to grams of compound B (multiply by the molar mass)

## Percent Yield

As such, we are going to use percent yield to compare the amount that we actually made to the amount that we predicted that we should have made using this formula.

(Actual / Theoretical) x 100 = Percent Yield