stoichiometry topic: mass calculations objectives: day 1 of 4 to learn how to perform mass...

Stoichiometry

Topic: Mass Calculations

Objectives: Day 1 of 4

• To learn how to perform mass calculations in chemical reactions (calculations that involve grams and chemical reactions)

Unit: Stoichiometry

QuickwriteAnswer one of the questions below 1-2

sentences:• In chemistry, we count by weighing moles; We can’t

measure moles in a lab, BUT what unit of measurement can we use in the lab to count atoms or molecules???

• Or what tool of measurement do we use in class to measure moles???

• Review: Using the equation below, What is the mole ratio between water and oxygen gas, in other words, 2 moles of H2O will produce how many moles of O2?

2H2O(l) → 2H2(g) + 1O2(g)

Stoichiometry• Chemistry is really all about reactions• Reactions involve the rearrangement of atoms• The calculation of the quantities of chemical elements or

compounds involved in chemical reactions is called Stoichiometry

• It is the MOLES in the balanced chemical equation that enables us to determine just how much product forms

2H2O(l) → 2H2(g) + 1O2(g) • In other words, if I have 2 moles of water (36 grams of

water), then I can produce 1 mole of oxygen (32 grams of oxygen)

• Steps:1. What units are you solving for?

2. What units are you given?

3. Write the possible conversion factors.

4. Start with what you know.

5. Set up calculation so that units cancel.

6. Solve the problem.

7. Does the answer make sense?

Stoichiometry Involves Dimensional Analysis

Stoichiometry Involves Dimensional Analysis

• You drove 7 kilometers to school today. How many meters did you drive?

1. What units are you solving for?2. What units are you given?6. Solve the problem3. What are the conversion factors?5. Set up the calculation so the units cancel4. Start with what you know7. Does the answer make sense

1000 meters

1 kilometers= meters= 7000 meters

Kilometers 7 Kilometers

What is Stoichiometry?

• The calculation of ________ of chemical elements or compounds involved in chemical ________

Answer BankReactantsconversion

molesQuantitiesReactionsproducts

quantities

reactions

Mass Calculations

• We just saw how to use balanced equations for a reaction to calculate the numbers of moles

• Remember, moles represent numbers of molecules and we cannot count molecules directly

• In chemistry, we count by weighing!!!!!!!!• We don’t have a machine to count atoms, BUT,

we do have a convenient tool for measuring atoms, it is called the GRAM!!!!

• When we weigh we use the gram, therefore we need to learn how to convert moles to mass

Mass Calculations

• Let’s consider an unbalanced combustion reaction in which propane reacts with oxygen to produce carbon dioxide and water

C3H8(g) + O2(g) → CO2(g) + H2O(g)

• What mass of oxygen will be required to react exactly with 44.1 grams of propane?

Mass Calculations

• First, we need to balance the equation:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

• Let’s summarize what we know and what we want to find

• What we know:– The balanced equation for the reaction– The mass or amount of propane availible(44.1g)

• What we want to calculate: – The mass of oxygen required to react exactly

with all the propane

Mass Calculations

Our overall plan of attack is as follows:1. We are given the number of grams of propane,

so we must convert to moles of propane (C3H8), because the balanced equation deals in moles not grams

2. Next, we can use the coefficients in the balanced equation to determine the moles of oxygen(O2) required

3. Finally, we will use the molar mass of O2 to calculate grams of oxygen

44.1 g propane ??? Grams of O2requires

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Our Plan of Attack!

44.1 g propane ? moles of propane ? moles of O2 ? Grams of O2

We are given gramsof propane

We have to convert grams

of propane into moles

Use mole ratio2 convert moles of propane into

moles of O 2

We have to convert moles

into gramsof O 2

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

44.1 g C3H8 requires ?? g O2

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

• We are given the number of grams of propane, so we must convert to moles of propane (C3H8), because the balanced equation deals in moles not grams

1 mol C3H85 mol O244.1 g C3H8

44.09 g C3H81 mol C3H8

32.0 g O2

1 mol O2

• Next, we can use the coefficients in the balanced equation to determine the moles of oxygen(O2) required

• Finally, we will use the molar mass of O2 to calculate grams of oxygen

= 160 g of O2

2Al(s) + 3I2(g) → 2AlI3(s)

• Consider the above reaction• Calculate how many grams of the product aluminum

Iodide (AlI3)would be produced by the complete reaction of 35.0 grams of Aluminum (Al)

Practice:

Our Plan of Attack!

35.0 g of Al ? moles of Al ? moles of AlI3 ? Grams of AlI3

We are given grams

of Al

We have to convert grams

of Al into moles of Al

Use mole ratio2 convert moles

of Al intomoles of AlI 3

We have to convert moles

into gramsof AlI 3

2Al(s) + 3I2(g) → 2AlI3(s)

2Al(s) + 3I2(g) → 2AlI3(s)

• Back to the problem!!!!• Calculate how many grams of the product aluminum

Iodide (AlI3)would be produced by the complete reaction of 35.0 grams of Aluminum (Al)

1 mol Al 2 mol AlI335.0 g Al

26.98 g Al 2 mol Al

407.68 g AlI3

1 mol AlI3

= 528.87 g of AlI3

Mass of Al Mass of AlI3

Br2(l) + 2NaCl(aq) → 2NaBr(aq) + Cl2(g)

• Consider the above balanced reaction• Calculate how many grams of the product chlorine(Cl2)

be produced by the complete reaction of 3.0 Moles of sodium chloride (NaCl)

• Hint: this time you are given MOLES!!!!! So, start with moles and convert moles of NaCl into grams of Cl2

3.0 mol NaCl 1 mol Cl2

2 mol NaCl

70.9 g Cl2

1 mol Cl2

= 106.5 g of Cl2

Practice:

Moles of NaCl Mass of CI2

Practice:• How many grams of chlorine gas are

needed to produce 10.0 g of sodium chloride? Remember to balance first!!!!!

Cl2 + NaI NaCl + I2

1 mol NaCl 1 mol Cl210.0 g NaCl

58.43 g NaCl 2 mol NaCl

70.9 g Cl2

1 mol Cl2

2

= 6.06 g Cl2

2

Mass of NaCl Mass of CI2

4Al+ 2O2 → 2Al2O3

• Calculate how many Molecules of aluminum oxide (Al2O3)would be produced by the complete reaction of 3.4 grams of Aluminum (Al)

1 mol Al 2 mol Al2O33.4 g Al

26.98 g Al 4 mol Al

6.02 x 1023 molecules Al2O3

1 mol Al2O3

= 3.79 x 1022

Molecules of Al2O3

Summarize: (fill in the blank)

Grams of reactant ___ of reactant Moles of product Grams of_____

You are given____ of reactant

Use Molar massOf reactant to Get moles of

______

Use mole _____To get from

Moles of reactantTo moles of

product

Use Molar ___Of product to Get grams of

product

Answer BankProduct

RatioMassmoles

reactantgrams

Topic: Limiting Reactant


• To learn what the limiting reactant is in a chemical reaction

• To learn how to calculate the limiting reactant using moles of reactants

Unit: Stoichiometry


sentences:• A sandwich recipe requires 2 pieces of bread, 3

slices of meat and 1 slice of cheese; you go into the kitchen and realize that you have 2 pieces of bread, 1 slice of cheese, and NO MEAT; what limited affected your ability to make your sandwich???

Limiting Reactants• Earlier, we discussed making sandwiches• Recall, that the sandwich making process

could be described as follows:

2 pieces of bread + 3 slices of meat + 1 slice of cheese →1 sandwich

• In this equation, all the products are used up, nothing was left over

• Now assume you came to work one day and found the following quantities of ingredients

Limiting Reactants• Now assume you came to work one day and

found the following quantities of ingredients

• 20 slices of bread• 24 slices of meat • 12 slices of cheese

• How many sandwiches can you make?• What will be left over

Limiting Reactants• To solve this problem, let’s see how many

sandwiches we can make with each ingredient:

• Bread:

• Meat:

• Cheese:

1 sandwich20 slices of bread

2 slices of bread

=10 sandwiches

1 sandwich24 slices of meat

3 slices of meat = 8 sandwiches

1 sandwich12 slices of cheese

1 slice of cheese=12 sandwiches

How many sandwiches can you make? The answer

is 8! Once you run out of meat, you must stop making sandwiches.The meat is the limiting ingredient!

Limiting Reactants• What do you have left over?• Making 8 sandwiches requires 16 pieces of

bread• You started with 20 slices, so you have 4

slices of bread left over• You also used 8 pieces of cheese for the 8

sandwiches, so you have 4 pieces of cheese left over

• In this example, the meat was the limiting reactant

Limiting Reactants• When molecules react with

each other to form products, considerations very similar to those making sandwiches arise

• Consider the reaction that occurs when Hydrogen Gas reacts with oxygen gas to form water

H2(gas) H2O(liquid) + O2(gas)

Limiting Reactants• The reaction occurs

between 10 H2

molecules and 7 O2

molecules• Remember, each O2

molecule requires 2 H2 molecules

2H2(gas) 2H2O(liquid) + O2(gas)

Limiting Reactants• After the reaction, 10 water

molecules formed and 2 O2 molecules are left over

• That is, the H2 molecules are used up before the water molecules are consumed

• We have excess (extra) O2 and H2 is the limiting reactant because the reaction runs out of H2 first

2H2(gas) 2H2O(liquid) + O2(gas)

What is the Limiting Reactant?

• The _______ that is completely _________ or used up when a reaction is run to completion

• The reactant that is not completely consumed is in _____

reactant consumed

excessAnswer Bank

Molesconsumed

productLimitingreactantexcess

Practice: Consider the reaction: N2 + 3 H2 2 NH3What is the limiting reactant if 2 moles of Nitrogen gas react with 7 moles of hydrogen gas???? Given: 2 mole of N2

7 mol of H2

Take the moles of each reactant given and divide it by the coefficient (or moles) of the balanced equation.

What is the limiting reactant?

2 < 2.3, So N2 is the limiting reactant because it has the lowest ratio!!!!!!!!!!!!!!

7 moles of H2

3 moles of H2

= 2.3 moles of H22 moles of N2

1 moles of N2

= 2.0 moles of N2

Practice:

Consider the reaction: N2 + 3 H2 2 NH3What is the limiting reactan if you have 3 moles of

N2 and 6 moles of H2

Given: 3 mole of N2

6 mol of H2

2 < 3, Sooooo, H2 is the limiting reactant because it has the lowest ratio!!!!!!!!!!!!!!

6 moles of H2

3 moles of H2

3 moles of N2

1 moles of N2

= 2 moles of H2 = 3 moles of N2

Practice:

• Suppose 25 grams of nitrogen reacted with 5 grams of hydrogen gas are mixed and react to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion

• N2(g) + 3H2(g) → 2NH3(g)

This reaction is different from the others we have doneSo far in that we are mixing specified amounts of

Two reactants together. To know how much product Forms we must we must determine which reactant is consumed first. In other words, we must determine

The limiting reactant

Practice:• Suppose 25 grams of nitrogen reacted with 5

grams of hydrogen gas are mixed and react to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion

• N2(g) + 3H2(g) → 2NH3(g)

First we calculate the moles of the two reactants present :

1 mol N2

28.0 g N2

25.0 g N2

1 mol H2

2.016 g H2

5.0 g H2

= .892 moles of N2

= 2.48 moles of H2



• N2(g) + 3H2(g) → 2NH3(g)

Now we must determine which reactant is the limiting reactant.

We have 0.892 moles of nitrogen Let’s determine how many moles of hydrogenAre required to react with this much nitrogen.

Because 1 mol of nitrogen reacts with 3 mol of Hydrogen, the number of moles of hydrogen we

Need to react completely with 0.892 mol of nitrogen is calculated as follows:

3 mol H2

1 mol N2

0.892 mol N2

= 2.68 moles of H2

Is nitrogen or hydrogen the limiting reactant? The answer comes from the comparison:

2.48 moles of H2 available < 2.68 moles of H2 required

This means that the hydrogen will be consumed first before the nitrogenRuns out, so hydrogen is the limiting reactant



• N2(g) + 3H2(g) → 2NH3(g)

25.0 g N2

2 mol NH317.0 g NH3

3 mol H21 mol NH3

2.48 mol H2

= 28.1 g of NH3

Reflect:We see that 0.892 mol of nitrogen require 2.68 mol

Of hydrogen to react completely. However, only 2.48Mol of hydrogen are available. This means that the hydrogen will be consumed before the nitrogen runs out, so Hydrogen

Is the limiting reactant

If the nitrogen is excess, then the hydrogen will run out first. Again we find

that the hydrogen limits the amount of ammonia Formed

Because the moles of hydrogen are limiting, we must useOur quantity of Hydrogen to determine the moles of ammonia

that can form

Summarize:

• When determining the limiting reactant we first convert mass into_____

• The smaller molar value of the reactants is the _____ reactant and the larger molar value of the reactants is in excess

• The limiting reactant will determine how much ______ will form

Answer BankMoles

consumedproductLimitingreactantexcess

Topic: Theoretical & Percent Yield


• To understand what theoretical yield is

• To learn how to calculate percent yield using the theoretical yield

Unit: Stoichiometry


sentences:• Chemists who work for companies like to know

how efficient a reaction is, in other words, they would like to know how much product forms after a reaction runs to completion; why do you think chemists would want to know this???

Percent Yield• If you recall, the amount of product formed is

determined by the limiting reactant• The amount of product calculated using

stoichiometry is called the theoretical yield• The theoretical yield is the amount of product

predicted from the amounts of reactants used up• Think of it as the maximum amount of product that

could be produced from 100% of the reactants being used up

• The theoretical yield is rarely if ever actually obtained

What is the theoretical yield?

• The ________ maximum amount of product that can be formed when the limiting reactant is completely used up or ________

• It is calculated using ___________

Answer Bankactual

Consumedpercentage

stoichiometryCalculatedtheoretical

calculated

consumed

stoichiometry

Percent Yield• Why is the theoretical yield never reached?• One reason for this is the presence of side

reactions that consume some of the reactants• The actual yield of product, which is the amount of

product actually obtained, is compared to the theoretical yield

• This comparison, called the percent yield, is expressed as a percent

• The percent yield is calculated by dividing the actual yield / by the theoretical yield

• Percent yield = Actual yield x 100% Theoretical yield

What is the percent yield?

• The ______ (experimental) yield of a product given as a _________ over of the __________ yield

• Percent yield = Actual yield x 100%

Theoretical yield

actualpercentage

theoretical

Answer Bankactual

Consumedpercentage


Practice:• Consider the balanced reaction below:

• 2H2(g) + CO (g) → CH3OH(l)

• Suppose 68.5 grams of CO is reacted with 8.6 grams of H2

• Calculate the theoretical yield of methanol CH3OH(l)

• Your experiment actually produces 35.7 grams of methanol CH3OH(l)

• What is the percent yield of methanol CH3OH(l) ?

68.5 g CO

8.60 g H2


• 2H2(g) + CO (g) → CH3OH(l)



28.01 g CO

1 mol H2

2.016 g H2

= 4.27 moles of H2

1 mol CO

Step 1: Calculate the moles of reactants

= 2.45 moles of CO

2.45 mol CO


• 2H2(g) + CO (g) → CH3OH(l)



1 mol CO

2 mol H2

Step 2: Now we determinewhich reactant is limiting using

The mole ratio of between CO and H2

= 4.9 moles of H2

Is CO or H2 the limiting reactant?

The answer comes from the comparison:4.27 moles of H2 present < 4.9 moles of H2 needed to react with all the CO

This means that the hydrogen will be consumed first before the CORuns out, so hydrogen is the limiting reactant

4.27 mol H2


• 2H2(g) + CO (g) → CH3OH(l)



2 mol H2

1 mol CH3OH(l)

Step 3: We must use the amount of H2 and

the mole ratio between CH3OH and H2 to determine

the maximum amount of methanol that can be produced

= 2.14 moles of CH3OH

2.14 represents the theoretical yield Which if you recall, is the maximum amount of a given product that can be formed when the limiting reactant is

completely used up or consumed

2.14 mol CH3OH


• 2H2(g) + CO (g) → CH3OH(l)



1 mol CH3OH

32.04 g CH3OH

Step 4: Convert moles to grams

= 68.6 grams of CH3OH


• 2H2(g) + CO (g) → CH3OH(l)




• Percent yield = 35.7 g CH3OH x100 = 52%

68.6 g CH3OH

Step 5: Calculate percent yield by dividing the actual yield (35.7 g) by the theoretical yield (68.6 g)

Summarize:

• The _____ ______ is the calculated maximum amount of product that can be formed when the limiting reactant is completely used up or consumed

• The _____ ______ is the actual (experimental) yield of a product given as a percentage over of the theoretical yield

• % Yield = ___????____ x 100% ????

Topic: Percent Yield Calculations


• To understand what theoretical yield is

• To learn how to calculate percent yield using the theoretical yield

Unit: Stoichiometry


sentences:• Chemists who work for companies like to know

how efficient a reaction is, in other words, they would like to know how much product forms after a reaction runs to completion; why do you think chemists would want to know this???

• According to your recipe, you should have produced 2 liters of lemonade, but something went wrong and you actually produced 1.8 liters of lemonade; What percentage of lemonade did you make????

Percent Yield

• If you recall, the amount of product formed is determined by the limiting reactant

• The amount of product calculated using stoichiometry is called the theoretical yield

• The theoretical yield is the amount of product predicted from the amounts of reactants used up

• Think of it as the maximum amount of product that could be produced from 100% of the reactants being used up

• The theoretical yield is rarely if ever actually obtained

What is the theoretical yield?

• The ________ maximum amount of product that can be formed when the limiting reactant is completely used up or ________

• It is calculated using ___________

Answer Bankactual

Consumedpercentage


calculated

consumed

stoichiometry

Actual Yield What your produce in the lab

• The amount of product actually produced by a chemical reaction is the actual yield

• The actual yield is the amount of product that actually forms when the reaction is carried out in the laboratory

• In other words it is the true amount of product formed from the experiment

What is the Actual yield?

• The amount of product that ______ forms when the reaction is carried out in the ______

actually

laboratory

Answer Bank

Consumedlaboratory

stoichiometryCalculatedTheoretical

actually

Percent Yieldratio of actual/theoretical

• Why is the theoretical yield never reached?• One reason for this is the presence of side reactions that

consume some of the reactants• Other times, the reaction is never run to completion• The actual yield of product, which is the amount of product

actually obtained, is compared to the theoretical yield• This comparison, called the percent yield, is expressed as a

percent• The percent yield is calculated by dividing the actual yield /

by the theoretical yield

• Percent yield = Actual yield x 100% Theoretical yield

What is the percent yield?

• The ______ yield divided by the _______ yield then multiplied by 100 as a _________

• Percent yield = Actual yield x 100%

Theoretical yield

actualpercentage

theoretical

Answer Bankactual

Consumedpercentage


Practice:Consider the balanced reaction below:

2H2(g) + CO (g) → CH3OH(l)

Suppose 68.5 grams of CO is reacted with 8.6 grams of H2

1) Calculate the theoretical yield of methanol CH3OH(l)

2) Your experiment actually produces 35.7 grams of methanol CH3OH(l)

3) What is the percent yield of methanol CH3OH(l) ?

68.5 g CO

8.60 g H2


• 2H2(g) + CO (g) → CH3OH(l)



28.01 g CO

1 mol H2

2.016 g H2

= 4.27 moles of H2

1 mol CO

Step 1: Calculate the moles of reactants

= 2.45 moles of CO

2.45 mol CO


• 2H2(g) + CO (g) → CH3OH(l)



1 mol CO

2 mol H2

Step 2: Now we determinewhich reactant is limiting using

The mole ratio of between CO and H2

= 4.9 moles of H2

Is CO or H2 the limiting reactant?

The answer comes from the comparison:4.27 moles of H2 present < 4.9 moles of H2 needed to react with all the CO

This means that the hydrogen will be consumed first before the CORuns out, so hydrogen is the limiting reactant

4.27 mol H2


• 2H2(g) + CO (g) → CH3OH(l)



2 mol H2

1 mol CH3OH(l)

Step 3: We must use the amount of H2 and

the mole ratio between CH3OH and H2 to determine

the maximum amount of methanol that can be produced

= 2.14 moles of CH3OH

2.14 represents the theoretical yield Which if you recall, is the maximum amount of a given product that can be formed when the limiting reactant is

completely used up or consumed

2.14 mol CH3OH


• 2H2(g) + CO (g) → CH3OH(l)



1 mol CH3OH

32.04 g CH3OH

Step 4: Convert moles to grams

= 68.6 grams of CH3OH


• 2H2(g) + CO (g) → CH3OH(l)




• Percent yield = 35.7 g CH3OH x100 = 52%

68.6 g CH3OH

Step 5: Calculate percent yield by dividing the actual yield (35.7 g) by the theoretical yield (68.6 g)

Summarize:

• The _____ ______ is the calculated maximum amount of product that can be formed when the limiting reactant is completely used up or consumed

• The _____ ______ is the actual (experimental) yield of a product given as a percentage over of the theoretical yield

• % Yield = ___????____ x 100% ????

Topic: Molecular and Empirical Formulas


• To understand the difference between molecular and empirical formulas

• To learn how to calculate empirical formulas and molecular formulas given percent composition and mass

Unit: Stoichiometry

Empirical Formulas

• The formula for a compound that is determined experimentally.

• A formula that represents the Smallest whole-number mole ratio of the different atoms in the compound.

• In other words, it is the simplest formula for a compound.

CHCH22OO CC22HH44OO22

CHCH33OOCHCH33OO

Empirical FormulaEmpirical FormulaMolecular Formula Molecular Formula A A formula formula based on the based on the actual numbers of actual numbers of

atoms atoms of each type in the of each type in the Empirical FormulaEmpirical Formula

A formula that gives the simplest whole-number A formula that gives the simplest whole-number ratioratio of the of the atoms of each element in a compound.atoms of each element in a compound.

Molecular Molecular FormulaFormula

Empirical FormulaEmpirical Formula

HH22OO22 HOHO

CHCH22OOCC66HH1212OO66

What is an Empirical Formula??

• A formula that represents the _____ whole-number ratio of the different atoms in the compound.

• In other words, it is the _____ formula for a compound.

• Example glucose

Answer BankSimplestNumbersoxygen

SmallestCH2O

Molecular Formula

Empirical Formula

C6H12O6??????

Practice: Practice: Write the empirical formula for NWrite the empirical formula for N22OO44

NN22OO44 NONO22

A formula that represents the A formula that represents the Using Using the smallest or lowest whole-the smallest or lowest whole-number ratio of number ratio of NN22OO4 4 we get….we get….

Steps for determining Empirical Formulas

1) Assume a 100 g sample when given percents. This makes 10.3 % = 10.3 g

2) Convert grams into moles for each element.

3) Divide the all the moles by smallest number of moles to get the lowest whole number ratio.

4) Write the empirical formula.

A compound was found to contain 36.11 % calcium and 63.89 % chlorine by mass. What is its empirical formula? What assumption did you make?

36.11 % Ca = 36.11 g Ca

63.89 % Cl = 63.89 g Cl

= 0.9009 mol Ca

= 1.802 mol Cl

0.9009

0.9009

= 1 mol Ca

= 2 mol Cl

Therefore the empirical formula is CaCl2

Step 1 Assume a 100 g sample when given %

Step 2 Convert gramsinto moles for each element.

Step 3 Divide the all the moles by smallest number of moles

Step 4 Write the empirical formula

1 mol Ca

40.08g Ca

1 mol Cl

35.45g Ca

Problem: Write the Empirical Formula for a compound composed of: 72% iron and 27.6% oxygen by mass.

72.% Fe = 72.00 g Ca

27.6 % O = 27.60 g O

= 1.230 mol Fe

= 1.730 mol O

1.230

1.230

= 1 mol Fe

= 1.5 mol O

This gives us the empirical formula is Fe1O1.4

Step 1 Assume a 100 g sample when given %

Step 2 Convert gramsinto moles for each element.

Step 3 Divide all the moles by smallest number of moles

Step 4 Write the empirical formula

1 mol Fe

55.84g Fe

1 mol O

16.00g O

Since 1.4 atoms does not exist, we need to multiply the compound by 2, so we get 2(Fe1O1.5) = Fe2O3

• A molecular formula is based on the actual number of atoms in each type of compound or molecule

• For example, consider glucose or sugar:

• The molecular formula tells us that it contains 2 Carbon atoms, 4 Hydrogen atoms, and 2 Oxygen atoms

Molecular Formulas

C2H4O2

A formula based on the actual _______ of atoms in each type of compound or molecule

Example: glucose C2H4O2 has 2 Carbon atoms, 4 Hydrogen atoms, and 2 ______ atoms

What is a Molecular Formula?

Answer BankSimplestNumbersoxygen

SmallestCH2O

Steps for determining Molecular Formulas

1. Find molar mass of the empirical formula (EF).

2. The molar mass of the molecule will be given.

3. Divide ___molar mass _of molecule______

molar mass of Empirical Formula 4. Multiply your answer from “step e” by the

subscripts given in the empirical formula.

Practice: Find the molecular formula for a compound with an empirical formula of CH4N if the

molar mass of the molecule is 60.12 g/mole. 1. Find molar mass of the empirical formula

Molar mass of Empirical Formula – CH4NC = 1 x12.0 = 12.0 g/moleH = 1.0 x 4 = 4.0 g/moleN = 1 x 14 = 14.0 g/mole

Molar mass of Empirical Formula = 30.0 g/mole

2. The molar mass of the molecule will be given.

Molar mass molecule (given) = 60.12 g/mole

Practice: Find the molecular formula for a compound with an empirical formula of CH4N if the

molar mass of the molecule is 60.12 g/mole.

4. Multiply your answer from the previous step by the subscripts given in the empirical formula.

2(CH4N) = C2H8N2

Therefore the Molecular Formula is C2H8N2

3. Divide _________molar mass _of molecule_________

molar mass of Empirical Formula ____Molar Mass _molecule____ = 60.12 g/mole = 2.00

Molar mass Empirical Formula 30.0 g/mole

Practice: Determine the molecular formula of a compound with an empirical formula of NH2 and Molecular or molar mass of 32.06

g/mole.1. Find molar mass of the empirical formula

Molar mass of Empirical Formula – NH2

N = 1 x14.0 = 14.0 g/moleH = 2.0 x 1 = 2.0 g/mole

Molar mass of Empirical Formula = 16.0 g/mole 2. The molar mass of the molecule will be given.

Molar mass molecule (given) = 32.06 g/mole

Practice: Determine the molecular formula of a compound with an empirical formula of NH2 and Molecular or molar mass of 32.06

g/mole.

4. Multiply your answer from the previous step by the subscripts given in the empirical formula.

2(NH2) = N2H4

Therefore the Molecular Formula is N2H4

3. Divide ______molar mass _of molecule_______________

molar mass of Empirical Formula ____Molar Mass _molecule____ = 32.06 g/mole = 2.00

Molar mass Empirical Formula 16.0 g/mole

Summarize:• Compare and contrast the empirical

formula with the molecular formula:

• Can the empirical formula be the same as the molecular formula????

• What do you do if the subscript is not a whole number such as 1.4????

• Complete the table:Molecular Formula Empirical Formula

P4O6

C6H9

stoichiometry topic: mass calculations objectives: day 1 of 4 to learn how to perform mass...

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