## MASS to MASS Stoichiometry

Stoichiometry is a branch of chemistry that deals with the relative quantities of 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. It can be used to find quantities such as the amount of products (in mass, moles, volume, etc.) that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product). Stoichiometry calculations can predict how elements and components diluted in a standard solution react in experimental conditions. Stoichiometry is founded on the law of conservation of mass: the mass of the reactants equals the mass of the products. Mole method is the stoichiometric coefficients in a chemical equation that can be interpreted as the number of moles of each substance.

A mass to mass problem is one in which the mass of a reactant or product is given. You are then asked to calculate the mass of another reactant required or the mass of another product formed. Here are the following steps to solve the mass to mass stoichiometry:

Find the amount of Copper produced when 2.08 grams of Iron reacts with Copper sulfate.

1. Write the balanced equation for the reaction.

Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

2. Convert the grams of the reactant to number of moles.

2.08 g Fe x

55.85 g Fe

3. Use the mole ratio from the balanced equation to calculate the number of moles of product formed.

0.0372 mol Fe x

1 mol Fe

4. Convert the moles of products to grams of products.

0.0372 mol Cu x

1 mol Cu

Ratio and Proportion:

55.85 g Fe 63.55 g Cu

55.85 g Fe 55.85 g Fe

X gram Cu = 2.37 grams of Copper

Unit Factor:

2.08 g Fe x

55.85 g Fe 1 mol Fe 1 mol Cu

Donna Clarisse L. Calma

IV-Zara

References:

http://en.wikipedia.org/wiki/Stoichiometry

http://www.digitalillusions.ca/applewoodscience/LessonsOnLiine/Chemistry/

**Mass to Mass Stoichiometry**A mass to mass problem is one in which the mass of a reactant or product is given. You are then asked to calculate the mass of another reactant required or the mass of another product formed. Here are the following steps to solve the mass to mass stoichiometry:

Find the amount of Copper produced when 2.08 grams of Iron reacts with Copper sulfate.

1. Write the balanced equation for the reaction.

Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

2. Convert the grams of the reactant to number of moles.

2.08 g Fe x

__1 mol Fe__= 0.0372 mol Fe55.85 g Fe

3. Use the mole ratio from the balanced equation to calculate the number of moles of product formed.

0.0372 mol Fe x

__1 mol Cu__= 0.0372 mol Cu1 mol Fe

4. Convert the moles of products to grams of products.

0.0372 mol Cu x

__63.55 g Cu__= 2.37 grams of Copper1 mol Cu

Ratio and Proportion:

__2.08 g Fe__=__X gram Cu__55.85 g Fe 63.55 g Cu

__(2.08 g Fe)(63.55 g Cu)__= (__X gram Cu)( 55.85 g Fe)__55.85 g Fe 55.85 g Fe

X gram Cu = 2.37 grams of Copper

Unit Factor:

2.08 g Fe x

__1 mol Fe__x__1 mol Cu__x__63.55 g Cu__= 2.37 grams of Copper55.85 g Fe 1 mol Fe 1 mol Cu

Donna Clarisse L. Calma

IV-Zara

References:

http://en.wikipedia.org/wiki/Stoichiometry

http://www.digitalillusions.ca/applewoodscience/LessonsOnLiine/Chemistry/

## MASS to VOLUME Stoichiometry

The word STOICHIOMETRY came from the Greek words:

The Mass to Volume problem is a problem in which the mass of a reactant or product is given and you are asked to calculate the volume of another reactant or a volume of a product. Here is an example to show you the steps involved in this problem:

* Propane, C3H8 reacts completely with oxygen to form carbon dioxide and water vapor. If 15 grams of propane is reacted with an excess of oxygen and the water vapor is collected and measured at STP, what volume of water vapor will be collected?

* First, write the balanced chemical equation.

C3H8 + 5 O2 3 CO2 + 4 H2O

This equation states that if one mole of propane reacts with five moles of oxygen, it produces three moles of carbon dioxide and four moles of dihydrogen monoxide.

* Then, you can either use Unit Factor or Ratio And Proportion.

- Unit Factor:

15g C3H8 x 44.11g C3H8 x 1 mol C3H8 x 1 mol H2O = 30 L H2O

- Ratio and proportion:

(1 mol C3H8)(44.11g C3H8) = (4 mol H20)(22.4 L H2O)

(15 g C3H8)(4 mol H2O)(22.4 L H2O) = (L H2O)(1 mol C3H8)(44.11 g C3H8)

1344 = (L H2O)(44.1)

(44.1) (44.1)

30 = L H2O

VOLUME: 30 L H2O

Leomar T. Santiago

IV-Zara

REFERENCES:

- http://www.chemteam.info/Stoichiometry/mass-volume-stoichiometry.html

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

- http://www.chemteam.info/Stoichiometry/What-is-Stoichiometry.html

*stoicheion*and*metron*. Stoicheion means “element” and metron means “measure”. Stoichiometry is a branch of chemistry that deals with calculations about the volumes, masses and moles of reactants and products involved in a chemical reaction. It is a very mathematical part of chemistry. It is the science of measuring the quantitative proportions and mass ratios of chemical substances in a reaction. Stoichiometry calculations is able predict how substances diluted in a sample solution react in experimental conditions. It can be simplified as a branch of chemistry that deals with the relative quantities of reactants and products involved in a chemical reactionThe Mass to Volume problem is a problem in which the mass of a reactant or product is given and you are asked to calculate the volume of another reactant or a volume of a product. Here is an example to show you the steps involved in this problem:

* Propane, C3H8 reacts completely with oxygen to form carbon dioxide and water vapor. If 15 grams of propane is reacted with an excess of oxygen and the water vapor is collected and measured at STP, what volume of water vapor will be collected?

* First, write the balanced chemical equation.

C3H8 + 5 O2 3 CO2 + 4 H2O

This equation states that if one mole of propane reacts with five moles of oxygen, it produces three moles of carbon dioxide and four moles of dihydrogen monoxide.

* Then, you can either use Unit Factor or Ratio And Proportion.

- Unit Factor:

__1 mol C3H8____4 mol H2O____22.4 L H20__15g C3H8 x 44.11g C3H8 x 1 mol C3H8 x 1 mol H2O = 30 L H2O

- Ratio and proportion:

__15 g C3H8____L H2O__(1 mol C3H8)(44.11g C3H8) = (4 mol H20)(22.4 L H2O)

(15 g C3H8)(4 mol H2O)(22.4 L H2O) = (L H2O)(1 mol C3H8)(44.11 g C3H8)

1344 = (L H2O)(44.1)

__1344__=__(L H2O)(44.1)__(44.1) (44.1)

30 = L H2O

VOLUME: 30 L H2O

Leomar T. Santiago

IV-Zara

REFERENCES:

- http://www.chemteam.info/Stoichiometry/mass-volume-stoichiometry.html

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

- http://www.chemteam.info/Stoichiometry/What-is-Stoichiometry.html

## VOLUME to VOLUME Stoichiometry

Stoichiometry - is a branch of chemistry that deals with the relative quantities of 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

Stoichiometry can be used to find quantities such as the amount of products (in mass, moles, volume, etc.) that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product). Stoichiometry calculations can predict how elements and components diluted in a standard solution react in experimental conditions. Stoichiometry is founded on the law of conservation of mass: the mass of the reactants equals the mass of the products.

Volume-Volume Problems:

N2 + 3H2 ---> 2NH3

What volume of hydrogen is necessary to react w/ five liters of nitrogen to produce ammonia?

So you are trying to find the volume of hydrogen. Before you start, know that 1 mole of a gas is equal to 22.4 L, that's the STP (standard temperature and pressure). So start out with the given (5L N2). Then go straight to the STP which gives you one mole of N2. For every one mole of N2, there are 3 moles of H2. 1 mole H3 is equal to 22.4 L. Now multiply it out!

__1 mole N2__

__3 moles H2__

__22.4 L H2__

5L N2 x 22.4 L N2 x 1 mole N2 x 1 mole H2 = 15 L H2

You can just simplify it and only use mole ratio as a multiple:

__3 moles H2__

5L N2 x 1 mole N2 = 15 L H2

Matthew W. Garcia

IV-Zara

**Limiting Reactants in our Daily Lives **

Chemists expect to know, which reactant will run out first? In any chemical reaction, we can pick one reagent as a prospect for the limiting reagent; calculate how many moles of that reagent we have, and calculate how many grams of the other reagent we need to react both to completion. We’ll discover one of the two things. Either we have an excess of the first or second reagent. The one we have in excess is the “excess reagent” and the other one which is not in excess is “limiting reagent”.

One good example is a recipe. A recipe is a balance equation. In hamburger-making process, we always purchased the ingredients in the correct ratios so that when we used all the components, with nothing left over.

1 burger buns + 2 pieces of patties + 2 slices of cheese ------> 1 hamburger

When you came to home hungry after going to school, you checked the refrigerator and found the following quantities of ingredients:

10 pieces of burger buns

14 pieces of patties

16 slices of cheese

Then, how many hamburgers can you make? What will be the left over?

To solve this problem, let’s see how many hamburgers we can make with each component.

Burger Buns:

__1 hamburger__

**10 burger buns X**1 burger buns

**= 10 hamburgers**

Patties:

__1 hamburger__

**14 pieces patties X**2 pieces of patties

**= 7 hamburgers**

Cheese:

__1 hamburger__

**16 slices cheese X**2 slices of cheese

**= 8 hamburgers**

How many hamburgers can you make? The answer is seven. When you run out of patties, you must stop making hamburgers. The patty is the limiting ingredient.

What do you have left over? Making 7 hamburgers requires 7 pieces of burger buns. You started 10 pieces, so you have 3 pieces of buns left. You also used 14 slices of cheese for the 7 hamburgers, so you have 16 – 14 = 2 slices of cheese left.

In this example, the ingredient present in the largest number (the patties) was actually the component that limited the number of hamburgers you could make. This situation arose because each hamburger required 2 pieces of patties- more than the quantity required of any other ingredient.

Another one, suppose a calamansi juice call for 1 cup of sugar for every 15 calamansi. You have 45 calamansi and 9 cups of sugar. Try to find out which ingredient is limiting, the calamansi or the sugar?

Consider our laundry habits. Most people try to put off doing laundry as long as possible but cannot avoid doing so any longer when the supply of clean underwear runs out. This implies that underwear is the limiting reagent with respect to avoiding laundry.

You probably have been dealing with limiting reactant problems for most of your life

Sometimes in a chemical reaction there is not enough of one reactant to go around. In such cases, this limiting reagent will determine how much product can be formed. During the course of reaction, the limiting reagent is the reactant that is completely consumed.

Camille R. Villania

IV-Zara

## Limiting Reagent and Excess Reagent Concept

Chemical reaction Equations give the ideal stoichiometric relationship among reactants and products. However, the reactants for a reaction in an experiment are not necessarily a stoichiometric mixture. In a chemical reaction, reactants that are not use up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limit the amount of products formed.

Let us consider the reaction between sodium and chlorine. The reaction can be represented by the equation:

2 Na + Cl2->2 NaCl

It represents a reaction of a metal and a diatomic gas chlorine. This balanced reaction equation indicates that two Na atoms would react with two Cl atoms or one Cl2 molecule. Thus, if you have 6 Na atoms, 3 Cl2 molecules will be required. If there is an excess number of Cl2 molecules, they will remain unreacted. We can also state that 6 moles of sodium will require 3 moles of Cl2 gas. If there are more than 3 moles of Cl2 gas, some will remain as an excess reagent, and the sodium is a limiting reagent. It limits the amount of the product that can be formed.

Example:

Ferrous Oxide is reduced to molten iron by heating it with aluminium metal. The reaction is :

FeO(s) + Al (s) -> Fe (l) + Al2 O3

Since all stoichiometry problems require a chemical Equation, we must balance the reactants and the products. Thus,

3FeO(s) + 2Al (s) ->3Fe (l) + Al2 O3

Suppose we need to find out how much molten iron is produced from the equation of 50.0g of FeO with 25.0g Of Al. Since the masses of both reactants are given, it is not obvious whether FeO or Al limits the amount of product. There are several ways to solve limiting reactant problems. We will use the following approach.

1) Calculate the maximum mass of the product that could be produced from the first reactant.

(a) Calculate the moles of reactant

(b) Calculate the moles of product

(c) Calculate the mass of the product

2) Calculate the maximum mass of the product that could be produced from the second reactant.

(a) Calculate the moles of reactant

(b) Calculate the moles of product

(c) Calculate the mass of the product

3) Determine the Limiting Reagent. The limiting reagent is the reactant that produces the lesser amount of product. Determine the actual mass of product that can be obtained from the two reactants. It is the lesser of the masses obtained from steps 1 and 2.

Let’s apply the three-step process to the problem. According to step 1, we calculate the mass of the product obtained from the first reactant, FeO. Using the balanced equation for the reaction, we can calculate the amount of iron produced from 50.0 g of FeO.

From the periodic table, we find the molar mass of Fe is 55.8 g/mol and FeO is 71.8 g/mol. Performing a unit analysis solution to the problem, we have

1molFeO 3 mol Fe 55.8g Fe

50.0g FeOX--------------X --------------- X -------------- = 38.9 g Fe

71.8molFeO 3 mol FeO 1 mol Fe

Next using step 2, we calculate the mass of the product obtained from the second reactant. Using the balanced equation, we can calculate the amount of iron produced from 25.0 g of Al.

From the periodic table, we find that the molar mass of Al is 27.0 g/mol. The unit analysis solution to the problem is

1molAl 3 mol Fe 55.8g Fe

25.0g Al X-------------- X --------------- X -------------- = 77.5g Fe

27.0 gAl 2 mol Al 1 mol Fe

Using Step 3, we now compare the mass of product from each of the reactants. This gives us

50.0g FeO 38.9 g Fe

25.9g Al 77.5 g Fe

We see that FeO yields a smaller mass of product and therefore is the limiting reagent. Al is the excess reagent, and not all the aluminium metal is used in the reaction. The maximum yield of molten iron form 50.0g of FeO and 25.0g of Al is therefore 38.9g of Fe.

Crissa Chirene P. Bug-os

IV-Zara

## STOICHIOMETRY: MOLE TO MOLE CONCEPT

**Terminologies, constant or people that you are going to encounter:**

Mole -is one of the seven fundamental units in the International System of Units (SI).

-is the SI measure of quantity of a “chemical entity” which can be an atom, molecule, formula units, electrons or photons.

Molar mass- is the mass in grams of a substance that is numerically equal to the substances FORMULA MASS.

Atomic mass (Atomic Weight)-is the mass of the atom in atomic masses in the masses.

Amadeo Avogadro-Italian physicist and chemist who proposed a hypothesis that later became known as Avogadro's law.

23

Avogadro’s number = 6 x 10

*1 mole of any compound is equal to Atomic mass.

23

1mole= 6 x 10 atoms (elements)

23

= 6 x 10 molecules (compound)

23

= 6 x 10 formula units (ionic compound)

STEPS IN MOLE TO MOLE CONCEPT:

1. WRITE THE

**CORRECT**CHEMICAL FORMULA/EQUATION.

2.

**BALANCE**THE EQUATION(IF NEEDED).

3.

**IDENTIFY**THE UNKNOWN.

4.

**SOLVE**FOR THE UNKNOWN.

EXAMPLES:

22

· HOW MANY MOLES OF MAGNESIUM ARE 3.01 x 10 ATOMS OF MAGNESIUM?

**RATIO AND PROPORTION:**

**22**

__3.01 x 10 atoms Mg__=

__x mole Mg__x mole Mg = 0.05 mole Mg

23

6.02 x 10 atoms Mg 1 mole Mg

**UNIT FACTOR ANALYSIS:**

**22**

3.01 x 10 atoms Mg x

__1 mole Mg__= 0.05 mole Mg

23

6.02 x 10 atoms Mg

· C6H12O6 + 6O2 6CO2 + 6H2

Mole of glucose: 2.5

Mole of carbon dioxide:?

2.5 mol C6H12O6

· HOW MANY MOLECULES OF H2O CAN BE PODUCED BY REACTING 0.010 MOLE OF OXYGEN?

2H2+ O2 2H2O

Jean Ellice P. Roy

IV-Zara

## FOUR STEPS TO SOLVE MOLE PROBLEMS

**STEP 1:**You must figure out how many parts of your calculationwill haveby using diagrams.

**STEP 2:**You should make a T- chart, then put the given information about the problem. After that, put the units of the given to the bottom right of the T- chart, and the units of what you want to find out in the top right of the T- chart.

**STEP 3:**You must put the conversion factors into the T-chart before the units.

**STEP 4:**You must cancel out the units from the top left and the bottom right then find the answer by multiplying all the numbers on the top together and divide it by the numbers on the bottom.

**For Example:**

Howmany moles of Potassium are present in 69 grams of Potassium?

In all problems like this,you need to go through four steps to find a solution.

**Step 1:**Looking at the diagram, we can see that we are going to use grams to moles, which is a one-step conversion. Furthermore, we can see that we need to use the atomic mass of Potassium as our conversion factor.

**Atomic Mass of Potassium**: 39. 10 g/mol

**Step 2:**The problem gave you "69 grams of potassium" as the starting information. Because this is what you were given, put "69 grams of potassium" in the top left of the T- chart. Since "grams of potassium" is the unit of what you were given, put this in the bottom right of the T. Since you want to find out how many moles of potassium are going to be made, put "moles of potassium" as your unit in the top right. When you've done this, your calculation should look like this:

STEP 3:The conversion factor between grams and moles is the atomic mass of potassium. Because we measure atomic mass in grams, you need to put the atomic mass in front of the unit "grams of potassium". What do you put in front of moles? Whenever you do a calculation of this kind, you need to put "1" in front of moles, like you see here:

STEP 4: In this case, you'd multiply 69 by one and divide the result by 39.10. Your answer, 1.76 moles of potassium:

CHRISTINE JOY G. JIMENEZ

IV-Zara