The Law of Conservation of Mass

The Law of Conservation of Mass
Lesson 1 1

Antoine-Laurent de Lavoisier
Antoine-Laurant de Lavoisier was an 18th century French scientist and nobleman who is considered the father of modern chemistry. He recognized and named oxygen and hydrogen, was one of the scientists who developed the metric system and through his experiments confirmed that although matter may change its form, the total mass remains constant. Unfortunately, because he was a nobleman he was beheaded during the French Revolution. Antoine-Laurent de Lavoisier The Father of Modern Chemistry

Reactants → Product H2 + O2 H2O The Law of Conservation of Mass.
The idea that the total mass of matter remains constant is know as The Law of Conservation of Mass. It can also be stated as: Mass is never created or destroyed. In an equation, the reactants will equal the products Reactants → Product H2 + O H2O

Very Important Information:
In the lab, mass is measured in ______ or __________. grams kilograms Mass is measured using either a __________ or a _____________. Electronic balance Triple beam balance

Reactant 1 + Reactant 2 Product
According to the Law of Conservation of Mass, the mass of the reactants should equal the mass of the products. Reactant 1 + Reactant Product Which scientist’s results below best support the Law of Conservation of Mass? Scientist Mass of Reactant 1 Mass of Reactant 2 Mass of Product Scientist A 19 g 22 g 42 g Scientist B 19.4 g 22.4 g 41.7 g Scientist C 19.36 g 22.37 g 41.74 g Scientist D g g g

Reactants → Products The Law of Conservation of Mass.
The idea that the total mass of matter remains constant is know as The Law of Conservation of Mass. It can also be stated as: Mass is never created or destroyed. In an equation, the reactants will equal the products Reactants → Products

The Trouble With Gases Many chemical reactions give off a gas. Because of this, the mass of the remaining solid and liquid products is less than the original mass of the reactants. Before Lavoisier, many scientists thought this was proof that mass was destroyed.

The Trouble With Gases Lavoisier understood what was happening, and correctly hypothesized that the difference in mass was equal to the mass of the gas given. So how did he prove it? Lavoisier developed a device to trap the gases that were given off, then he performed many experiments where he showed that the mass of products, including the gas, was equal to the original mass of the reactants.

Understanding Chemical Equations
What is a substance? A substance is an element or a compound The substances that originally react, that is to say that they gain, lose or share electrons, are known as the reactants. The new substances that are produced are known as the products. Reactants Products

Solving for Mass When all but one of the masses are known, solving for the mass of the unknown takes the most elementary of algebra. 3.4g + ?g = 4.2g 2.1g + 8.5g = ?g g + ?g = 12.0g ? = _____ ? = _____ ? = _____ ?g = 4.2g – 3.4 ?g = 12.0g – 3.4g 0.8 g 10.6 g 8.6 g 2.2g + ?g = 5.6g + 2.7g g + 6.7g = ?g + 3.3g ? = _____ ? = _____ ?g = 5.6g + 2.7g – 2.2g ?g = 1.5g + 6.7g – 3.3g 6.1 g 4.9 g

Solve for the unknown masses in your notes.
11

Hydrogen + Oxygen Water
Solving for Mass When this concept is shown with a chemical equation, nothing changes about how it should be solved. Hydrogen Oxygen Water 16.0 g g ? g 16.0 grams of hydrogen reacts with 128 grams of oxygen. Assuming all of the reactants are used up, how much water will be produced? 16.0 g g = 144 g

Make sure you place the numbers with the appropriate substance.
Solving for Mass When you are solving for masses, the most important thing to pay a attention to is the placement of the given masses. Sloppiness and laziness will make an easy question turn into a wrong answer. Na Cl NaCl If 11.5 grams of sodium are mixed with chlorine gas to make 29.3 grams of sodium chloride, how much chlorine gas was used. 11.5 g ? g 29.9 g Make sure you place the numbers with the appropriate substance. 29.9 g – 11.5 g = 18.4 grams

Solve for the unknown masses in each of the chemical equations on your notes.
14

Lesson 2 15

Understanding Coefficients
Coefficients multiply through the entire formula of a compound. 2CO2 = 2C and 4 O 3H2O = 6 H and 3 O 4CaBr2 = 4Ca and 8 Br 5Na2O = 10 Na and 5 O Understanding this simple concept is critical to your ability to balance equations properly.

Determine how many of each element are in the problems on your notes.
17

Understanding Coefficients Na + Cl2  NaCl (unbalanced)
Equations without coefficients are known as skeleton equations Na Cl2  NaCl (unbalanced) reactants product These equations may or may not represent the correct ratio of moles that interact during the reaction shown. Since mass is directly related to the number of particles, if mass does not change, then the number of particles does not change.

Understanding Coefficients 2Na + Cl2  2NaCl (balanced)
To illustrate the Law of Conservation of Mass, we place coefficients in equations to show that mass is not created or destroyed. 2Na Cl2  2NaCl (balanced) reactants product Cl Na Na Cl Cl Na Na These equations are called “balanced equations” because they show the same number of elements on each side.

Yes, this equation is balanced.
Understanding Coefficients You can determine whether or not equations are balanced using a simple accounting method and the skill you just learned. Is the following equation balanced? 4 Fe + 3 O2  2 Fe2O3 Reactants side Products side ___ Fe ___ 4 4 ___ O ___ 6 6 Yes, this equation is balanced.

No, this equation is not balanced.
Understanding Coefficients Is the following equation balanced? Al + O2  Al2O3 Reactants side Products side ___ Al ___ 1 2 ___ O ___ 2 3 No, this equation is not balanced.

Understanding Coefficients You can determine whether or not equations are balanced using a simple accounting method and the skill you just learned. Is the following equation balanced? 2 K + 2 H2O  2 KOH + H2 Reactants side Products side ___ K___ 2 2 ___ H ___ 4 2+2 4 ___ O ___ 2 2 Yes, this equation is balanced.

Al4C3 (s) + H2O (l) → CH4 (g) + Al(OH)3 (s)
Understanding Coefficients Is the following equation balanced? Al4C3 (s) + H2O (l) → CH4 (g) + Al(OH)3 (s) Reactants side Products side ___ Al ___ 4 1 ___ C ___ 3 1 ___ H ___ 2 7 ___ O ___ 1 3 No, this equation is not balanced.

Understanding Coefficients and Balancing Determining if an equation is balanced or not is a very simple task, but once again carelessness will cause an incorrect answer. Do NOT pretend to understand better than you really do. Make a table and get it RIGHT! Is the following equation balanced? N O2  N2O5 Reactants side Products side ___ N ___ 2 2 ___ O ___ 6 5 No, this equation is not balanced.

Understanding Coefficients Is the following equation balanced? For some of you, it will be easier to just ignore the polyatomic ions and focus on the other elements. 2LiI Pb(NO3)2  2LiNO PbI2 Reactants side Products side ___ Li ___ 2 2 ___ I ___ 2 2 ___ Pb ___ 1 1 Yes, this equation is balanced.

Determine if the equations on your notes are balanced or not using the accounting method.
26

_____ Li _____ _____ N _____ 6 6 3 3 6 6 6 3 2 4 2 4 2 2 2 1
When you are given sets of coefficients, all you need to do is plug them in to see if they are balanced or not. Once again, this is a test of your patience more than anything else. ___Li ___N2 ___Li3N A 6, 2, 2 B 3, 1, 2 C 3, 2, 1 D 6, 1, 2 _____ Li _____ _____ N _____ 6 6 3 3 6 6 6 3 2 4 2 4 2 2 2 1

Determine which set of coefficients on your notes will balance the equations.
28

Always check every answer choice!
When you are given a molecule to place into the equation, the same principle applies. C6H12O6 H2O2 2H2O 2CO CH O CO ____ _____ C _____ _____ H _____ _____ O _____ A 2CO B H2O2 C 2H2O D C6H12O6 1 7 1 1 1 1 3 1 4 12 4 4 4 4 2 4 8 4 4 4 4 4 4 Always check every answer choice!

_____ C _____ _____ H _____ _____ O _____ 1 1 4 4 2
When you are given a molecule to place into the equation, the same principle applies. However, in this case there is actually a short cut! CH O CO ____ _____ C _____ _____ H _____ _____ O _____ A 2CO B H2O2 C 2H2O D C6H12O6 2C + 2O 1 1 2H + 2O 4 4 H’s 4H + 2O 4 2 2 O’s 6C + 12H + 6O Third, determine which answer meets this need. Second, figure out what more you need. First, add up everything that is already there.

Determine which substance will balance the equation on your notes.
31

Lesson 3 32

Antoine-Laurent de Lavoisier
The Law of Conservation of Mass states that the sum of the reactant’s masses is equal to the sum of the products’ masses This is how we can solve for the mass of an unknown and why we balance chemical equations. Antoine-Laurent de Lavoisier The Father of Modern Chemistry

Make sure you place the numbers with the appropriate substance.
Solving for Mass When you are solving for masses, the most important thing to pay a attention to is the placement of the given masses. Sloppiness and laziness will make an easy question turn into a wrong answer. 2H O H2 O If 4.0 grams of hydrogen are mixed with oxygen gas to make 36.0 grams of water, how much oxygen gas was used. 4.0 g ? g 36.0 g Make sure you place the numbers with the appropriate substance. 36.0 g – 4.0 g = 32.0 grams

Solve for the mass of the substances on your notes.
64.0 g ? g 88.0 g 14.0 g 6.0 g ? g 35

Understanding Coefficients and Balancing Determining if an equation is balanced or not is a very simple task, but once again carelessness will cause an incorrect answer. Do NOT pretend to understand better than you really do. Make a table and get it RIGHT! Is the following equation balanced? N O2  N2O5 Reactants side Products side ___ N ___ 2 2 ___ O ___ 6 5 No, this equation is not balanced.

N2 + O2  N2O5 ___ N ___ 2 2 ___ O ___ 2 5
Balancing from Scratch Now that we fully understand what it means to be balanced, we will learn how to balance an equation from the beginning. Do NOT panic. You will see that while it can take some time, the steps are not hard. Balance the following equation. N O2  N2O5 Reactants side Products side ___ N ___ 2 2 ___ O ___ 2 5 Step 1: Determine how many of each element are present.

N2 + O2  N2O5 ___ N ___ 2 2 ___ O ___ 2 5
Balancing from Scratch Now that we fully understand what it means to be balanced, we will learn how to balance an equation from the beginning. Do NOT panic. You will see that while it can take some time, the steps are not hard. Balance the following equation. N O2  N2O5 Reactants side Products side ___ N ___ 2 2 ___ O ___ 2 5 Step 2: Identify which element (if any) is keeping the equation from being balanced.

N2 + O2  N2O5 5 2 ___ N ___ 2 2 ___ O ___ 2 5
Balancing from Scratch Now that we fully understand what it means to be balanced, we will learn how to balance an equation from the beginning. Do NOT panic. You will see that while it can take some time, the steps are not hard. Balance the following equation. N O2  N2O5 5 2 Reactants side Products side ___ N ___ 2 2 ___ O ___ 2 5 Step 3: Add coefficients to make this element balanced.

N2 + O2  N2O5 5 2 ___ N ___ 2 4 ___ O ___ 10 10
Balancing from Scratch Now that we fully understand what it means to be balanced, we will learn how to balance an equation from the beginning. Do NOT panic. You will see that while it can take some time, the steps are not hard. Balance the following equation. N O2  N2O5 5 2 Reactants side Products side ___ N ___ 2 4 ___ O ___ 10 10 Step 4: Recalculate how many of EVERY element you now have in the equation.

N2 + O2  N2O5 2 5 2 ___ N ___ 4 2 4 4 ___ O ___ 10 10 10 10
Balancing from Scratch Now that we fully understand what it means to be balanced, we will learn how to balance an equation from the beginning. Do NOT panic. You will see that while it can take some time, the steps are not hard. Balance the following equation. N O2  N2O5 2 5 2 Reactants side Products side ___ N ___ 4 2 4 4 ___ O ___ 10 10 10 10 Step 5: Add coefficients to make any other element that has been unbalanced correct and recalculate.

N2 + O2  N2O5 2 5 2 ___ N ___ 4 4 ___ O ___ 10 10
Balancing from Scratch Now that we fully understand what it means to be balanced, we will learn how to balance an equation from the beginning. Do NOT panic. You will see that while it can take some time, the steps are not hard. Balance the following equation. N O2  N2O5 2 5 2 Reactants side Products side ___ N ___ 4 4 ___ O ___ 10 10 Step 6: STOP ADDING COEFFICENTS once it is balanced. Many students find this to be the hardest step.

Pb(NO3)2 + NaI  NaNO3 + PbI2 ___ Pb ___ 1 1 ___ I ___ 1 2 ___ Na ___
Balancing from Scratch Now lets go through another one together. Balance the following equation. Pb(NO3) NaI  NaNO PbI2 Reactants side Products side ___ Pb ___ 1 1 ___ I ___ 1 2 ___ Na ___ 1 1 ___ NO3 ___ 2 1 Step 1: Determine how many of each element are present. (Remember that it is often easier to think of polyatomic ions as 1 substance.)

Pb(NO3)2 + NaI  NaNO3 + PbI2 ___ Pb ___ 1 1 ___ I ___ 1 2 ___ Na ___
Balancing from Scratch Now lets go through another one together. Balance the following equation. Pb(NO3) NaI  NaNO PbI2 Reactants side Products side ___ Pb ___ 1 1 ___ I ___ 1 2 ___ Na ___ 1 1 ___ NO3 ___ 2 1 Step 2: Identify an element that is keeping the equation from being balanced. Just do one at a time.

Pb(NO3)2 + NaI  NaNO3 + PbI2 2 ___ Pb ___ 1 1 ___ I ___ 1 2
Balancing from Scratch Now lets go through another one together. Balance the following equation. Pb(NO3) NaI  NaNO PbI2 2 Reactants side Products side ___ Pb ___ 1 1 ___ I ___ 1 2 ___ Na ___ 1 1 ___ NO3 ___ 2 1 Step 3: Add coefficient(s) to make this element balanced.

Pb(NO3)2 + NaI  NaNO3 + PbI2 2 ___ Pb ___ 1 1 ___ I ___ 2 2
Balancing from Scratch Now lets go through another one together. Balance the following equation. Pb(NO3) NaI  NaNO PbI2 2 Reactants side Products side ___ Pb ___ 1 1 ___ I ___ 2 2 ___ Na ___ 2 1 ___ NO3 ___ 2 1 Step 4: Recalculate how many of EVERY element you now have in the equation.

Pb(NO3)2 + NaI  NaNO3 + PbI2 2 2 ___ Pb ___ 1 1 1 1 ___ I ___ 2 2 2 2
Balancing from Scratch Now lets go through another one together. Balance the following equation. Pb(NO3) NaI  NaNO PbI2 2 2 Reactants side Products side ___ Pb ___ 1 1 1 1 ___ I ___ 2 2 2 2 ___ Na ___ 2 2 2 1 ___ NO3 ___ 2 2 2 1 Step 5: Add coefficients to make any other element that has been unbalanced correct and recalculate.

Pb(NO3)2 + NaI  NaNO3 + PbI2 2 2 ___ Pb ___ 1 1 ___ I ___ 2 2
Balancing from Scratch Now lets go through another one together. Balance the following equation. Pb(NO3) NaI  NaNO PbI2 2 2 Reactants side Products side ___ Pb ___ 1 1 ___ I ___ 2 2 ___ Na ___ 2 2 ___ NO3 ___ 2 2 Step 6: STOP ADDING COEFFICENTS once it is balanced. It is very important that you know when to stop.

Balancing from Scratch
When you are asked this on a multiple choice test, you will most likely be asked in the following way: Al HCl  AlCl H2 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 It is important that you understand that you can never know just one coefficient. You must make sure that the entire equation is balanced before you can know the answer.

When you are asked this on a multiple choice test, you will most likely be asked in the following way: Al HCl  AlCl H2 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 ___ Al ___ 1 1 ___ H ___ 1 2 ___ Cl ___ 1 3 Step 1: Determine how many of each element are present. (Remember that it is often easier to think of polyatomic ions as 1 substance.)

Balancing from Scratch Hydrogen and chlorine are unbalanced!
When you are asked this on a multiple choice test, you will most likely be asked in the following way: Al HCl  AlCl H2 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 ___ Al ___ 1 1 Hydrogen and chlorine are unbalanced! ___ H ___ 1 2 ___ Cl ___ 1 3 Step 2: Identify an element that is keeping the equation from being balanced. Just do one at a time.

2 ___ Al ___ 1 1 1 1 ___ H ___ 2 1 2 2 ___ Cl ___ 2 1 3 3
Balancing from Scratch When you are asked this on a multiple choice test, you will most likely be asked in the following way: Al HCl  AlCl H2 2 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 ___ Al ___ 1 1 1 1 ___ H ___ 2 1 2 2 ___ Cl ___ 2 1 3 3 Step 3: Add coefficient(s) to make this element balanced.

When you are asked this on a multiple choice test, you will most likely be asked in the following way: Al HCl  AlCl H2 2 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 Notice that we have a problem. There is not a coefficient that will multiply times 3 to equal 2. Chlorine cannot be balanced this way which means the first coefficient we added was incorrect! ___ Al ___ 1 1 ___ H ___ 2 2 ___ Cl ___ 2 3

When you are asked this on a multiple choice test, you will most likely be asked in the following way: Al HCl  AlCl H2 2 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 Balancing Tip: When you are experiencing troubles like this, it often makes things easier if you place a 2 in front of the compound containing the odd number of atoms. ___ Al ___ 1 1 2 1 ___ H ___ 1 1 2 2 ___ Cl ___ 1 1 3 6 Repeat Step 3: It is okay to make a mistake and start over. This will happen many times as you first begin balancing.

When you are asked this on a multiple choice test, you will most likely be asked in the following way: Al HCl  AlCl H2 6 2 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 ___ Al ___ 1 1 2 2 ___ H ___ 6 1 2 2 ___ Cl ___ 6 1 6 6 Steps 4 & 5: Recalculate how many of EVERY element you now have in the equation and continue to add coefficients until it is balanced.

When you are asked this on a multiple choice test, you will most likely be asked in the following way: Al HCl  AlCl H2 6 2 3 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 ___ Al ___ 1 1 2 2 ___ H ___ 6 6 6 2 ___ Cl ___ 6 6 6 6 Steps 4 & 5: Recalculate how many of EVERY element you now have in the equation and continue to add coefficients until it is balanced.

When you are asked this on a multiple choice test, you will most likely be asked in the following way: 2 Al HCl  AlCl H2 6 2 3 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 ___ Al ___ 1 2 2 2 ___ H ___ 6 6 6 6 ___ Cl ___ 6 6 6 6 Steps 4 & 5: Recalculate how many of EVERY element you now have in the equation and continue to add coefficients until it is balanced.

Balancing from Scratch Don’t forget to answer the question!
When you are asked this on a multiple choice test, you will most likely be asked in the following way: 2 Al HCl  AlCl H2 6 2 3 When the above equation is balanced, the coefficient for aluminum chloride is – A 1 B 2 C 3 D 4 ___ Al ___ 2 2 Don’t forget to answer the question! ___ H ___ 6 6 ___ Cl ___ 6 6 Step 6: STOP ADDING AND CHANGING COEFFICENTS once it is balanced, and check your answers when you are done!

Sometimes you might get coefficients that balance, but are not the correct set. 4 2 LiI Pb(NO3)2  LiNO PbI2 2 4 2 2 ___ Pb ___ 2 1 2 1 These coefficients are NOT correct because even though they are balanced, they are not the smallest numbers that can be used. Notice that they can all be reduced by a factor of 2. ___ I ___ 4 2 4 2 ___ Li ___ 4 2 2 4 ___ NO3 ___ 4 2 4 2 Now the chemical equation is balanced with the correct set of coefficients. Always make sure that your coefficients are in their most reduced form.

Balance the chemical equations on your notes.
2 1 2 Balance the chemical equations on your notes. 2 1 2 1 1 2 1 2 2 1 1 1 1 1 1 2 2 1 1 12 11 4 3 2 4 6 1 2 1’s will be placed in the blanks on the screen, but it is not necessary that you write 1’s. 2 1 2 1 2 7 4 6 1 3 2 2 2 1 2 1 2 60

The Law of Conservation of Mass

Similar presentations

Presentation on theme: "The Law of Conservation of Mass"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

The Law of Conservation of Mass

Similar presentations

Presentation on theme: "The Law of Conservation of Mass"— Presentation transcript:

Similar presentations

About project

Feedback