Stoichiometry is the branch of chemistry dealing with relationships of combining elements, especially quantitatively. It’s the calculation of the quantities of chemical elements or compounds involved in chemical reactions.It’s the calculation of the quantities of chemical elements or compounds involved in chemical reactions.

 

Essentially, all stoichiometry problems can be broken down into three steps:

1. Take the given quantity (i.e. mass. volume) and convert to moles

2. Use a mole to mole ratio to find the number of moles of the desired compound

3. Answer the question - convert the moles of the desired compound to the appropriate quantity (i.e. mass, volume)

                     

 

*MASS-MASS STOICHIOMETRY

This problem is commonly known as a mass-mass since you are given a mass compound and you are required to find the mass of another compound.

 

                

 

 

 

 

First, we have to determine the molar mass of Fe2O3 since it is needed to convert to moles (step one). Dividing the given mass by the molar mass always converts it to moles (step one). In the second conversion factor (step two), the coefficients in the reaction determines the ratio.The coefficients of the reaction are only used when solving a stoichiometry problem. To convert moles of Fe to grams you need to use the molar mass (step three). It's worth nothing that in a mass-mass problem, the first and third steps are opposites of each other. That is, in step one you convert grams to moles (divide by molar mass) and in the third step you convert moles to grams (multiply by molar mass).

                      

 

 

 

 

 

*Volume-Volume stoichiometry

 

 

(1) Convert volme to moles using PV = nRT and the initial set of T and P 

(2) Use a ratio and proportion to determine moles of other substance involved in problem 

(3) Use PV = nRT with new T and P as well as moles of substance from step 2. You will calculate a new volume.

 

 

Volume-volume stoichiometry is also known as Gas stoichiometry. The relationship between 1 mole of gas at STP and the molar volume of 22.414 L is important. Consider the reaction below:

 

 

 

 

 

 

 

Note that the first and last step in a volume-volume problem will cancel each other. This is because the first step, converting to liters of oxygen to moles, requires a division by 22.414. In the third step, conversion of moles of ammonia to liters, requires multiplication by 22.414. These two steps cancel each other and render step two (mole to mole ratio) the only important step. It needs to be stressed that this only happens in a volume-volume problem.

 

 

 

 

 

 

 

 

 

First, divide a mole of O2 by the molar volume of 22.414 multiply it by the coefficient of NH3 divided by the coefficient of O2 multiply it by 22.414 divide it by a mole of NH3. Multiply 22.5L O2 to all the numerator and divide it by the product of the denominator.

 

Note: 1 mole of gas at STP is 22.414

 

 

 

*Mass-volume stoichiometry

 

In Mass-Volume stoichiometry, either the volume of  the gas or the mass of the compound will be given, if the volume of a gas is given the mass of a compound will be asked, but if the mass of a compound will be given the volume of the gas will be asked.

 

 

 

 

 

 

 

 

 

*Problem:

*Determine the volume of carbon dioxide gas that will be produced from 112.5 grams of iron at STP.

The concept of STP is important to understand. STP stands for Standard temperature and pressure. Standard temperature and pressure is a set of conditions (273.15 K and 1 atm) at which 1 mole of any ideal gas will occupy 22.414 L. As a conversion factor:

1 mole gas = 22.414 L

 

First, find the equivalent of 1 mole of Fe in grams and multiply it by the coefficient of CO2 over the coefficient of Fe and multiply it by 22.414 (STP) over a mole of CO2. Multiply 112.5g of Fe by the product of all the numerators and divide it by the products of all the denominators.

 

 

 

 

 

 

 

 

 

 

Sources: http://www.chemistry-reference.com/stoichiometry/index.asp

 

 

written by: Glen Edgardo Malaca