Today, we learned how to do stoichiometry calculations involving particles, moles, and mass. These calculations are somewhat similar to mole conversions except, instead of converting between particles, moles, or mass of just one element in the problem, you can convert between particles, moles, or mass of more than one element. *But remember, they are much harder than mole conversioins because you have to use your knowledge of all the mole conversions we've learned in the past, plus what we learned last day about the mole ratio to do these stoichiometry calculations.
Stoichiometry
a branch of chemistry that deals with the quantitative relationships that exist between the reactants and products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of whole numbers. For example, in a reaction that forms ammonia (NH3), exactly one molecule of nitrogen (N2) reacts with three molecules of hydrogen (H2) to produce two molecules of NH3:
N2 + 3H2 → 2NH3
A chemical equation is an expression of a chemical process. For example:
AgNO3(aq) + NaCl(aq) ---> AgCl(s) + NaNO3(aq)
In this equation, AgNO3 is mixed with NaCl. The equation shows that the reactants (AgNO3 and NaCl) react through some process (--->) to form the products (AgCl and NaNO3). Since they undergo a chemical process, they are changed fundamentally.
Often chemical equations are written showing the state that each substance is in. The (s) sign means that the compound is a solid. The (l) sign means the substance is a liquid. The (aq) sign stands for aqueous in water and means the compound is dissolved in water. Finally, the (g) sign means that the compound is a gas.
Coefficients are used in all chemical equations to show the relative amounts of each substance present. This amount can represent either the relative number of molecules, or the relative number of moles (described below). If no coefficient is shown, a one (1) is assumed.
To solve questions as conversions, you need to have conversion factors. Conversion factors relate units to one another. For example, there are 100 cents in 1 dollar, or 100 cents = $1. To use this as a conversion factor, write it as a fraction:
Since 100 cents = 1 dollar, this fraction is equal to 1. That means you can multiply it by anything else, and still have the same quantity that you started with. For example, if you want to convert $5.27 to cents, you can multiply it by the fraction. The dollars cancel out, leaving you with 525 cents:
We can also flip the fraction to get a new conversion factor:
As long as the top and bottom are equal to one another, and we're careful to cancel out units to see what we have left, we can string conversion factors like this together to convert from a starting number to a final answer.
That brings us to stoichiometry. Stoichiometry is the study of relationships in chemical reactions. in other words, it's the study of what amounts of things are equal to amounts of other things in chemical reactions.
We all know that an exothermic reaction gives off energy in the form of heat, light, or sound, and that an endothermic reaction absorbs energy. If the reaction takes more energy to break the bonds that hold molecules together, it is endothermic. If it takes less energy, the reaction is exothermic.
ENTHALPY, or "H", is the sum of all kinetic and potential energies in a system. You will see this term used many times.
Change in enthalpy is denoted as 'delta' H.
Exothermic reactions result in a negative enthalpy, or negative "H". Also, endothermic reactions are characterized by a positive heat flow, or positive "H".
An example of an endothermic reaction is photosynthesis. Plants convert CO2 and water into oxygen and glucose from the energy from the sun. The general formula of an endothermic reaction is: A + B + Energy --> AB
The burning of any substance is an example of an exothermic reaction. The general form of an exothermic reaction is: A + B --> AB + Energy
Energy diagrams tell us about the relative rate of a chemical reaction. They are also called potential energy diagrams or chemical reaction energy reactions. Here is an energy diagram of a exothermic reaction:
This is an energy diagram of an endothermic reaction:
where: (a) is the activation energy, which is the energy required to initiate a chemical reaction, (b) is the change between the reactants and products, and (c) is the change in energy, or enthalpy
The purpose of the lab is to observe a variety of chemical reactions and to interpret and explain obsevations with balanced chemical equations and classify each reactions asa one of the four main types.
There was a reaction where we added solid copper(II) sulfate pentahydrate to a test tube and heated it over a Bunsen burner. Over time, the copper(II) sulfate pentahydrate changed colors (blue to white)
In this reaction, we took the results of the reaction above and added drops of water to it. The copper (II) sulfate pentahydrate changed from white to blue color.
Basically, there are four types of reactions in this lab: synthesis, decomposition, single replacement and double replacement.
There are 6 types of reactions. They are Synthesis, Decomposition, Single Replacement, Double Replacement, Combustion, and Neutralization. Synthesis:
Where 2 or more reactants combine together to form 1 product.
It takes this general form:
A+B--->C.
Decomposition:
This is a reaction where one reactant is decomposed into 2 or more products.
The general form is
A---> B+C
Single Replacement:
This one is more tricky! It takes the general form of
A+ BC---> AC +B (if A is a metal) or A+ BC----> B +AC (if A is a non-metal).
The metals switch with each other and non-metals swith with each other. But that's not all we have to worry about..
Double displacement: This is when the anions and cations of two different molecules switch places, forming two entirely different compounds. These reactions are in the general form: AB + CD ---> AD + CB
Combustion: A combustion reaction is when oxygen combines with another compound to form water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An example of this kind of reaction is the burning of napthalene: C10H8 + 12 O2 ---> 10 CO2 + 4 H2O
Neutralization: This is a special kind of double displacement reaction that takes place when an acid and base react with each other. The H+ ion in the acid reacts with the OH- ion in the base, causing the formation of water. Generally, the product of this reaction is some ionic salt and water: HA + BOH ---> H2O + BA
***Note***
Remember that the single element on the reactants side must be above the element it switches with on the ACTIVITY SERIES chart. This is because some elements are more reactive than others.
Refer to the following reaction:
1Cu + 2AgNO3 ---> 2Ag + 1Cu(NO3)2
Cu switches with the other metal, Ag. This is a reaction because copper is above silver on the ACTIVITY SERIES chart.
Here is the activity series chart :
This link will lead you to a bunch of online worksheets. It also provides a lot of information about all the topics we've learned or will be learning in chemistry. Check it out! http://www.chalkbored.com/lessons/chemistry-11.htm
Net Ionic Equations:
A Net Ionic Equation is a chemical equation for a reaction which lists only those species participating in the reaction. To write a Net Ionic Reaction, follow these 3 steps:1) Start by simply writing the overall balanced chemical reaction. This is also called the Molecular Equation.
2) Then, you break apart the soluble molecules into the two ions that are formed (one positive and one negative). You will have to use the solubility rules to do this, they can be found online. If something is insoluble, it should not be broken apart. Write the reaction out with all of the separated ions. This is called the Total Ionic Equation.
3) Then, you simplify by canceling things out if they appear on both sides of the reaction, resulting in the Net Ionic Equation.
For example, let look at the reaction of calcium carbonate with hydrochloric acid to form calcium chloride, water, and carbon dioxide:
Molecular Equation:
CaCO3(s) + 2HCl(aq) ---> CaCl2(aq) + H2O(l) + CO2(g)
Total Ionic Equation:
CaCO3(s) + 2H+(aq) + 2Cl-(aq) ---> Ca2+(aq) + 2Cl-(aq) + H2O(l) + CO2(g)
Net Ionic Equation:
CaCO3(s) + 2H+(aq) ---> Ca2+(aq) + H2O(l) + CO2(g)
Notice that the Cl- was canceled out from the Net Ionic Equation, because it really isn't playing an important part of this reaction. It is just there to balance out the charge because you can't have an ion just by itself -- you must always pair an ion with another one of opposite charge so that the overall charge is zero.
Physical States
- (s) – solid
- (l) – liquid
- (g) – gas
- (aq) – aqueous : dissolved in water
Example:
H2(g) is read hydrogen gas
Writing unbalanced equations
- Write correct formulas for reactants and products.
- Include a yield arrow between them.
- Include states when they are known or given.
- Be careful of diatomics (Dr. HOBrFINCl)
Examples
- Solid mercury (II) oxide decomposes to produce liquid mercury and gaseous oxygen.
- Write formulas and check charges
Hg2+O2- Hg + O
- Add states
HgO(s) Hg(l) + O(g)
- Check HOBrFINCl
HgO(s) Hg(l) + O2(g)
Example 2
- Solid carbon reacts with gaseous oxygen to form gaseous carbon dioxide.
C(s) + O2(g) CO2(g)
You try it!
• Solid magnesium metal reacts with liquid water to form solid magnesium hydroxide and hydrogen gas.
- Solid aluminum dichromate decomposes to form solid chromium (II) oxide, gaseous nitrogen, and gaseous water.
- Gaseous ammonia reacts with gaseous oxygen to produce nitrogen monoxide and gaseous water
Balancing Chemical Equations
- Balanced equations have equal numbers of atoms on both sides of the equation.
- Once formulas have been written, the subscripts cannot be changed. This would change the substances formed.
- Balancing is done by adding coefficients, large numbers in front of the reactants or products. 2 H2O = 4 Hydrogen and 2 oxygen atoms.
- Coefficients in balanced equations must always be whole numbers in the lowest ratio possible.
Writing Balanced Equations
Fe(s) + O2(g) Fe2O3(s)
-Start with the most complex molecule and add coefficients.
4Fe(s) + 3O2(g) 2Fe2O3(s)
-Check that the coefficients are in the lowest ratio.
Balancing equations
-Write the unbalanced equation, leave a space in front of each molecule to add a coefficient.
___Fe(s) + ___O2(g) ___Fe2O3(s)
1 Fe 2
2 O 3
-Underneath the arrow, write the elements and the count for the reactant and product sides.
-As coefficients are added, change the count on each side until the numbers on each side are equal.
Balancing chemical equations isn't difficult, once you know the way to do it.
Start by finding out how many atoms of each type are on each side of the equation. Some teachers recommend making a little table listing the numbers of each atom for the left hand side and for the right hand side.
Next, look for an element which is in only one chemical on the left and in only one on the right of the equation.
To balance that element, multiply the chemical species on the side which doesn't have enough atoms of that type by the number required to bring it up to the same as the other side. The number is called the coefficient.
BUT
If you have to multiply by, say, 2 1/2, do so, THEN multiply EVERYTHING on each side of the equation by two to get rid of the half.
Now look for the next element or species that is not balanced and do the same thing.
Repeat until you are forced to balance the hydrogen and oxygens.
If there is a complex ion, sometimes called a polyatomic ion, on each side of the equation that has remained intact, then that can often be balanced first, as it is acts as a single species. The ions NO3- and CO32- are examples of a complex ion.
A VERY useful rule is to leave balancing oxygen and hydrogen to the last steps as these elements are often in more than one chemical on each side , and it is not always easy to know where to start. Some people also say you should leave any atom or species with a valancy of one one until the end, and also generally leave anything present as an element to the end.
Example 1
Unbalanced equation:- H2SO4 + Fe ---> Fe2(SO4)3 + H2
Balance the SO4 first (as it is a complex ion and it is in one chemcial species on each side)
3H2SO4 + Fe ---> Fe2(SO4)3 + H2
Now balance the Fe (which is also in one chemical on each side)
3H2SO4 + 2Fe ---> Fe2(SO4)3 + H2
Finally, balance the hydrogen (although it is in one chemical species on each side, it is usually a good idea to leave it until last)
Balanced Equation:- 3H2SO4 + 2Fe ---> Fe2(SO4)3 + 3H2
Example 2
Unbalanced Equation:- Al + O2 ---> Al2O3
We can start with either the Al or the O, but we will start with Al, as we normally leave O to the end. Even though it wouldn't matter where we started in this case.
Put a two in front of the Al on the left
2Al + O2 ---> Al2O3
Put a 1 1/2 in front of the O2 on the left
2Al + 1 1/2O2 ---> Al2O3
We don't like halves, so multiply everything on BOTH sides by two
4Al + 3O2 ---> 2Al2O3
Gases expand and contract (or change volume) in temperature and pressure.
STP= 1 atmosphere (atm) of pressure, and temperature at 0 degrees Celsius of 273.15K
At STP, there are 22.4 L per 1 mole Therefore we have this mole conversion: 22.4L of gas1 mol of gas 1 mol of gas OR 22.4L of gas
Dilution: reducing the concentration of a chemical
When a solution is diluted, more solvent is added to it.
Since M = n ÷ L, and n (the moles of solute) is the same for the original solution and the new diluted solution, it follows that M1L1 = M2L2
where M1=original concentration of solution
L1=original volume of solution
M2=new concentration of solution after dilution
L2=new volume of solution after dilution
To calculate the new concentration (M2) of a solution given its new volume (L2) and its original concentration (M1) and original volume (L1):
M2 = (M1 x L1) ÷ L2
To calculate the new volume (L2) of a solution given its new concentration (M2) and its original concentration (M1) and original volume (L1):
L2 = (M1 x L1) ÷ M2
Examples
1. M2=(M1L1) ÷ L2
Calculate the new concentration (molarity) if enough water is added to 100mL of 0.25M sodium chloride to make up 1.5L.
M2=(M1L1) ÷ L2
M1 = 0.25M
L1 = 100mL = 100 ÷ 1000 = 0.100L (volume must be in litres)
L2 = 1.5L
[NaCl(aq)]new = M2 = (0.25 x 0.100) ÷ 1.5 = 0.017M
(or 0.0.017 mol/L or 0.0.17mol L-1)
2. L2=(M1L1) ÷ M2
Calculate the volume to which 500mL of 0.02M coppper sulfate solution must be diluted to make a new concentration of 0.001M.
L2=(M1L1) ÷ M2
M1 = 0.02M
L1 = 500mL = 500 ÷ 1000 = 500 x 10-3L = 0.500L (since there are 1000mL in 1L)
Introduction The solution is the homogeneous mixture of two or more substances. The solute is the substance being dissolved and is in smaller quantity. The solvent is the substance that dissolves the solute and is in larger quantity.
Molarity (or Molar Concentration) Molarity is the number of moles of solute in one litre of a solution. We use "M" to denote molar concentration and it has the units of "moles/L".
