# Stoichiometry

## Balanced Equation & Molar Mass

First, you would need to balance the equation you have, to make sure your input and output equal each other. Then, you would have to calculate the molar masses of each of the reactants. To find the molar masses you would find the molar masses of one element it would also be considered the molar mass. For example: C3H8 = 3(12.011) + 8(1.008) = 44.097, O2 = 31.998, CO2 = 44.009, and 18.015.

## IUPAC Name

To find the IUPAC name, first figure out if the is ionic or covalent. For ionic, you would have to add -ride to the end of the element unless it a polyatomic. For covalent, you would use prefix in the front of the element.

Ex. C3H8 - Propane, O2 - Oxygen, CO2 - Carbon Dioxide, H2O - Water

## Type of Reaction

Propane and Oxygen is a Combustion reaction.

## Mole to Mole Conversions

To find a mole to mole conversion, you would need to set both ratios equal to each other which is called proportion then you would cross multiply them together.

Ex 1. You have 11.30 moles of C3H8. How many moles of CO2 can you make?

Ans. (11.3 x 3) / 1 = 33.9 mole of CO2

Ex. 2 You want to make 12.1 moles of H2O. How many moles of O2 do you need?

Ans. (12.1 x 5) / 4 = 15.1 mole of O2

## Mass to Mass Conversions

1. Find the given. Which is the grams of mole A.
2. Then find the molar mass of the mole A and divided it by mole A.
3. Then find the ratio of mole B to mole A and multiply it to step 2.
4. Then multiply the molar mass of mole B to step 3.

Ex. How many grams of CO2 can you make when you use 8.12 grams of C3H8?

Ans. 8.12 grams of C3H8 / 44.097 mole of C3H8 = .184 grams per mole of C3H8 X (3 mole CO2 / 1 mole C3H8) = .552 X 44.009 mole of CO2 = 24.31 grams of CO2

## Limiting & Excess Reactant & Theoretical Yield

1. Find the two reactant that has mass with it.
2. Set up the same equation with mass of each and put it in different equation.
3. The lower number of the same amount of product is the limiting reactant.
4. The higher one is excess reactant.
5. To find the Theoretical Yield is the limiting reactant.

Ex.

How much carbon dioxide can you make if you react 5.25 g of C3H8 with 12.18 g of O2?

Ans.

C3H8: 15.72 g of CO2 Excess Reactant

O2: 10.05 g of CO2 Limiting Reactant

Theoretical Yield = O2

## Percent Yield

To find the percent yield, you put the given value and divide that by the limiting reactant and then multiply by 100 to get the percentage .

Ex. 8.4 g of CO2 were recovered from the experiment. What is your percent yield?

Ans. (8.4 / 10.1) X 100 = 83.17%