Stoichiometry.

Stoichiometry

Topic: Mass Calculations
Unit: 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)

Quickwrite Answer 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)

Stoichiometry Involves Dimensional Analysis
Steps: What units are you solving for? What units are you given? Write the possible conversion factors. Start with what you know. Set up calculation so that units cancel. Solve the problem. Does the answer make sense?

Stoichiometry Involves Dimensional Analysis
You drove 7 kilometers to school today. How many meters did you drive? 6. Solve the problem 4. Start with what you know 2. What units are you given? 7. Does the answer make sense 1. What units are you solving for? 5. Set up the calculation so the units cancel 3. What are the conversion factors? = meters 7 Kilometers Kilometers 1000 meters = 7000 meters 1 kilometers

What is Stoichiometry? The calculation of ________ of chemical elements or compounds involved in chemical ________ Answer Bank Reactants conversion moles Quantities Reactions products

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

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Mass Calculations 44.1 g propane requires ??? Grams of O2 C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Our overall plan of attack is as follows: 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 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

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Our Plan of Attack! We are given grams of propane We have to convert grams of propane into moles Use mole ratio 2 convert moles of propane into moles of O2 We have to convert moles into grams of O2 44.1 g propane ? moles of propane ? moles of O2 ? Grams of O2 44.1 g C3H8 requires ?? g O2 C3H8(g) O2(g) → 3CO2(g) H2O(g)

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
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 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 44.1 g C3H8 1 mol C3H8 32.0 g O2 5 mol O2 = 160 g of O2 44.09 g C3H8 1 mol O2 1 mol C3H8

Practice: 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)

Our Plan of Attack! 2Al(s) + 3I2(g) → 2AlI3(s) 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 ratio 2 convert moles moles of AlI3 convert moles into grams of AlI3 2Al(s) + 3I2(g) → 2AlI3(s)

2Al(s) + 3I2(g) → 2AlI3(s) = 528.87 g 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) 35.0 g Al 1 mol Al g AlI3 2 mol AlI3 = g of AlI3 26.98 g Al 1 mol AlI3 2 mol Al Mass of Al  Mass of AlI3

2Al(s) + 3I2(g) → 2AlI3(s) = 528.87 g 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 AlI3 35.0 g Al 26.98 g Al 2 mol Al g AlI3 1 mol AlI3 = g of AlI3 Mass of Al  Mass of AlI3

Br2(l) + 2NaCl(aq) → 2NaBr(aq) + Cl2(g)
Practice: Br2(l) + 2NaCl(aq) → 2NaBr(aq) + Cl2(g) 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 70.9 g Cl2 1 mol Cl2 = g of Cl2 2 mol NaCl 1 mol Cl2 Moles of NaCl  Mass of CI2

F2(l) + 2KCl(aq) → 2KF(aq) + Cl2(g)
Practice: F2(l) + 2KCl(aq) → 2KF(aq) + Cl2(g) Calculate how many grams of the product Chlorine gas (Cl2) be produced by the complete reaction of 6.0 Moles of potassium chloride (KCl) Hint: this time you are given MOLES!!!!! So, start with moles and convert moles of NaCl into grams of Cl2 6.0 mol KCl 70.9 g Cl2 1 mol Cl2 = g of Cl2 2 mol KCl 1 mol Cl2 Moles of KCl  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 2 2 10.0 g NaCl 1 mol NaCl 70.9 g Cl2 1 mol Cl2 = 6.06 g Cl2 58.43 g NaCl 2 mol NaCl 1 mol Cl2 Mass of NaCl  Mass of CI2

2Mg+ O2 → 2MgO Calculate how many grams of oxygen gas are required to react completely with 8.9 grams of magnesium (Mg) 8.9 g Mg 1 mol Mg 1 mol O2 32 g O2 = 5.86 grams of O2 24.31 g Mg 1 mol O2 2 mol Mg

4Al+ 3O2 → 2Al2O3 Calculate how many grams of aluminum oxide (Al2O3)would be produced by the complete reaction of 5.8 grams of Aluminum (Al) 5.8 g Al 1 mol Al 2 mol Al2O3 grams Al2O3 = 10.9grams of Al2O3 26.98 g Al 1 mol Al2O3 4 mol Al

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) 3.4 g Al 1 mol Al 2 mol Al2O3 6.02 x 1023 molecules Al2O3 = 3.79 x 1022 Molecules of Al2O3 26.98 g Al 1 mol Al2O3 4 mol Al

__NaN3  __Na + __N2 2 2 3 Automotive airbags inflate when sodium azide (NaN3) rapidly decomposes to sodium and nitrogen. How many molecules of nitrogen gas (N2) will be produced when 38 grams of NaN3 decomposes? 38.0 g NaN3 1 mol NaN3 3 mol N2 6.02 x 1023 molecules N2 = 5.28 x 1023 Molecules of N2 2 mol NaN3 65.02 g NaN3 1 mol N2

Summarize: (fill in the blank)
Grams of reactant ___ of reactant Moles of product Grams of_____ You are given ____ of reactant Use Molar mass Of reactant to Get moles of ______ Use mole _____ To get from Moles of reactant To moles of product Use Molar ___ Of product to Get grams of Answer Bank Product Ratio Mass moles reactant grams

Topic: Limiting Reactant
Unit: Stoichiometry Topic: Limiting Reactant Objectives: Day 2 of 4 To learn what the limiting reactant is in a chemical reaction To learn how to calculate the limiting reactant using moles of reactants

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 a recipe that dealt with making cakes Recall, that the recipe is: 2 eggs + 2 cups of flour + 1 cup of sugar → 1 cake In this equation, all the products are used up, nothing was left over Now assume you visit your kitchen one day and find the following quantities of ingredients

Limiting Reactants Now assume you came to work one day and found the following quantities of ingredients 8 eggs 6 cups of flour 7 cups of sugar How many cakes can you make? What will be left over

Limiting Reactants Eggs: Flour: Sugar:
To solve this problem, let’s see how many sandwiches we can make with each ingredient: Eggs: Flour: Sugar: How many CAKES can you make? The answer is 2! Once you run out of EGGS, you must stop making CAKE, The EGGS are the limiting ingredient or reactant! 4 eggs = 2 cakes 1 cake 2 eggs 6 cups of flour 1 cake = 3 cakes 2 cups of flour 7 cups of sugar = 7 cakes 1 cake 1 cup of sugar

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) H2O(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) H2O(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 _____ Answer Bank Moles consumed product Limiting reactant excess

Practice: Consider the reaction: N2 + 3 H2  2 NH3
What 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 H2 2 moles of N2 1 moles of N2 = 2.0 moles of N2

Practice: Consider the reaction: N2 + 3 H2  2 NH3
What is the limiting reactant 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 = 2 moles of H2 3 moles of N2 1 moles of N2 = 3 moles of N2

Calculations that involve a Limiting Reactant
Consider the following reaction CH4(g) + H2O(g) → 3H2(g) + CO(g) What mass of water is required to react exactly with 249 grams of methane? In other words, how much water will use up 249 grams of methane?

Calculations that involve a Limiting Reactant
This result means that if 249 grams of methane is mixed with 279 grams of water, Both reactants will “run out” at the same time On the other hand, if 249 grams of methane is With 300 grams of water, the methane will be Used up before the water; therefore, the methane Is the limiting reactant and the water is in excess Consider the following reaction CH4(g) + H2O(g) → 3H2(g) + CO(g) What mass of water is required to react exactly with 249 grams of methane? In other words, how much water will use up 249 grams of methane? 249 g CH4 1 mol CH4 18.02 g H2O 1 mol H2O = 279 g of H2O 16.04 g CH4 1 mol H2O 1 mol CH4

Practice: This reaction is different from the others we have done So 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 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)

Practice: First we calculate the moles of the two reactants present : 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) 25.0 g N2 1 mol N2 = .892 moles of N2 28.0 g N2 5.0 g H2 1 mol H2 = 2.48 moles of H2 2.016 g H2

Practice: Now we must determine which
reactant is the limiting reactant. We have moles of nitrogen Let’s determine how many moles of hydrogen Are 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 mol of nitrogen is calculated as follows: 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) 0.892 mol N2 3 mol H2 = 2.68 moles of H2 1 mol N2 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 nitrogen Runs out, so hydrogen is the limiting reactant

Reflect: We see that mol of nitrogen require 2.68 mol Of hydrogen to react completely. However, only 2.48 Mol 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 use Our quantity of Hydrogen to determine the moles of ammonia that can form 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) 25.0 g N2 2.48 mol H2 2 mol NH3 = 28.1 g of NH3 17.0 g NH3 3 mol H2 1 mol NH3

Practice: Homework #7 = 0.12 = 0.30 = 0.15 = 14.0g of NaCl
Consider the reaction: CuCl2+2NaNO3Cu(NO3)2+2NaCl Determine the limiting reactant if 15.6 grams of CuCl2 react with 25.2 grams of NaNO3. How many grams of sodium chloride, NaCl, can be formed? 15.6 g CuCl2 25.2 g NaNO3 STEP 1: Convert Grams to Moles for each reactant 1 mol CuCl2 = 0.12 moles of CuCl2 1 mol NaNO3 = 0.30 moles of NaNO3 134.5 g CuCl2 85.0 g NaNO3 STEP 2: Use one of my mole values from my previous step = 0.15 moles of CuCl2 1 mol CuCl2 0.3 mol NaNO3 2 mol NaNO3 STEP 3: Use moles of limiting reactant to determine how many grams of NaCl will form: 2 mol NaCl = 14.0g of NaCl 0.12 moles CuCl2 58.44 g NaCl 1 mole CuCl2 1 mol NaCl

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 Bank Moles consumed product Limiting reactant excess

Topic: Theoretical & Percent Yield
Unit: Stoichiometry Topic: Theoretical & Percent Yield Objectives: Day 3 of 4 To understand what theoretical yield is To learn how to calculate percent yield using the theoretical yield

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 ___________ calculated Answer Bank actual Consumed percentage stoichiometry Calculated theoretical 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 the __________ yield Percent yield = Actual yield x 100% Theoretical yield actual percentage theoretical Answer Bank actual Consumed percentage stoichiometry Calculated theoretical

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) ?

Step 1: Calculate the moles of reactants
Practice: Step 1: Calculate the moles of reactants 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) 68.5 g CO = 2.45 moles of CO 1 mol CO 28.01 g CO 8.60 g H2 = 4.27 moles of H2 1 mol H2 2.016 g H2

The answer comes from the comparison:
Practice: Step 2: Now we determine which reactant is limiting using The mole ratio of between CO and H2 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) 2.45 mol CO = 4.9 moles of H2 2 mol H2 1 mol CO 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 CO Runs out, so hydrogen is the limiting reactant

Step 3: Now that we have determined the Limiting reactant, we can determine the theoretical yield of methanol 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) 4.27 mol H2 = grams of CH3OH 1 mol CH3OH(l) 32.04 g CH3OH 2 mol H2 1 mol CH3OH 68.14 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

Step 4: Calculate percent yield by dividing the actual yield (35.7 g) by the theoretical yield (68.6 g) 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) Percent yield = 35.7 g CH3OH x100 = 52% 68.6 g CH3OH

Practice: Homework #13 = 1.30 = 3.1 = 1.03 = 247.2g of PbO2
Consider the reaction: PbS2 + 3O2 → PbO SO2 Consider the following unbalanced reaction, where lead (IV) sulfide reacts with oxygen to produce lead (IV) oxide and sulfur dioxide. If 350 grams of lead (IV) sulfide reacts with 100 grams of oxygen gas, how many grams of lead (IV) oxide are produced? (theoretical yield) What are the limiting and excess reactants? What is the percent yield if your experimental procedure produces grams of lead (IV) oxide? chloride? 350 g PbS2 Because 1.3 > 1.0, (O2) oxygen gas is your limiting reactant, In other words, In order to be completely used up, 3.1 mole of (O2) requires 1.0 moles of PbS2 and we have 1.3 moles of PbS2 so clearly we have enough PbS2 so (O2) oxygen gas is the limiting reactant 100 g O2 STEP 1: Convert Grams to Moles for each reactant 1 mol PbS2 1 mol O2 = 1.30 moles of PbS2 = 3.1 moles of O2 271.3g PbS2 32.0 g O2 STEP 2: Use one of the mole values from my previous step 1 mol PbS2 3.1 mol O2 = 1.03 moles of PbS2 3 mol O2 247.2g of PbO2 is your Theoretical Yield!!!!! STEP 3: Use moles of limiting reactant to determine Theoretical Yield of PbO2 : 1 mol PbO2 = 247.2g of PbO2 3.1 moles O2 239.2g PbO2 1 mol PbO2 3 moles O2

Practice: Homework #13 Percent Yield = _205.3 g PbO2 x 100 = 83.05%
Consider the reaction: PbS2 + 3O2 → PbO SO2 Consider the following unbalanced reaction, where lead (IV) sulfide reacts with oxygen to produce lead (IV) oxide and sulfur dioxide. If 350 grams of lead (IV) sulfide reacts with 100 grams of oxygen gas, how many grams of lead (IV) oxide are produced? (theoretical yield) What are the limiting and excess reactants? What is the percent yield if your experimental procedure produces grams of lead (IV) oxide? chloride? STEP 4:Calculate % yield by dividing the actual yield (205.3 g) by the theoretical yield (247.2g) : Percent Yield = _205.3 g PbO2 x 100 = 83.05% 247.2g PbO2

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
Unit: Stoichiometry Topic: Molecular and Empirical Formulas Objectives: Day 4 of 4 To understand the difference between molecular and empirical formulas To learn how to calculate empirical formulas and molecular formulas given percent composition and mass

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.

Empirical Formula Molecular Formula Empirical Formula H2O2 HO CH2O
Molecular Formula A formula based on the actual numbers of atoms of each type in the Empirical Formula A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Molecular Formula Empirical Formula H2O2 HO CH2O C6H12O6 CH3O CH3O C2H4O2 CH2O

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 Bank Simplest Numbers oxygen Smallest CH2O Molecular Formula Empirical Formula C6H12O6 ??????

Practice: Write the empirical formula for N2O4
A formula that represents the Using the smallest or lowest whole-number ratio of N2O4 we get…. N2O4 NO2

Steps for determining Empirical Formulas
Assume a 100 g sample when given percents. This makes 10.3 % = 10.3 g Convert grams into moles for each element. Divide the all the moles by smallest number of moles to get the lowest whole number ratio. Write the empirical formula.

Therefore the empirical formula is CaCl2
A compound was found to contain % calcium and % chlorine by mass. What is its empirical formula? What assumption did you make? 36.11 % Ca = g Ca % Cl = g Cl Step 1 Assume a 100 g sample when given % Step 2 Convert grams into moles for each element. Step 3 Divide the all the moles by smallest number of moles 1 mol Ca = mol Ca = 1 mol Ca 40.08g Ca 0.9009 1 mol Cl = mol Cl = 2 mol Cl 35.45g Ca 0.9009 Step 4 Write the empirical formula Therefore the empirical formula is CaCl2

This gives us the empirical formula is Fe1O1.4
Problem: Write the Empirical Formula for a compound composed of: 72% iron and 27.6% oxygen by mass. 72.% Fe = g Ca 27.6 % O = g O Step 1 Assume a 100 g sample when given % Step 2 Convert grams into moles for each element. Step 3 Divide all the moles by smallest number of moles 1 mol Fe = mol Fe = 1 mol Fe 55.84g Fe 1.230 1 mol O = mol O = 1.5 mol O 16.00g O 1.230 Step 4 Write the empirical formula This gives us the empirical formula is Fe1O1.4 Since 1.4 atoms does not exist, we need to multiply the compound by 2, so we get 2(Fe1O1.5) = Fe2O3

Molecular Formulas C2H4O2 For example, consider glucose or sugar:
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 C2H4O2 68

What is a Molecular Formula?
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 Answer Bank Simplest Numbers oxygen Smallest CH2O 69

Steps for determining Molecular Formulas
1. Find molar mass of the empirical formula 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 3” by the subscripts given in the empirical formula.

1. Find molar mass of 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 g/mole. 1. Find molar mass of the empirical formula Molar mass of Empirical Formula – CH4N C = 1 x12.0 = 12.0 g/mole H = 1.0 x 4 = 4.0 g/mole N = 1 x 14 = 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) = g/mole

Practice: Find the molecular formula for a compound with an empirical formula of CH4N if the molar mass of the molecule is g/mole. 3. Divide __molar mass _of molecule___ molar mass of Empirical Formula ____Molar Mass _molecule____ = g/mole = 2.00 Molar mass Empirical Formula 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

1. Find molar mass of the empirical formula
Practice: Determine the molecular formula of a compound with an empirical formula of NH2 and Molecular or molar mass of g/mole. 1. Find molar mass of the empirical formula Molar mass of Empirical Formula – NH2 N = 1 x14.0 = 14.0 g/mole H = 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) = g/mole

Practice: Determine the molecular formula of a compound with an empirical formula of NH2 and Molecular or molar mass of g/mole. 3. Divide ______molar mass _of molecule_______________ molar mass of Empirical Formula ____Molar Mass _molecule____ = g/mole = 2.00 Molar mass Empirical Formula 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

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

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