The Mole and Stoichiometry

The Mole and Stoichiometry
Avogadro’s Number Formula Mass % Composition

Atomic Mass: The Average Mass of an Element’s Atoms
Atomic mass is sometimes called atomic weight or standard atomic weight. The atomic mass of each element is directly beneath the element’s symbol in the periodic table. It represents the average mass of the isotopes that compose that element, weighted according to the natural abundance of each isotope.

Molar Mass: Counting Atoms by Weighing Them
As chemists, we often need to know the number of atoms in a sample of a given mass. Why? Because chemical processes happen between particles. Therefore, if we want to know the number of atoms in anything of ordinary size, we count them by weighing.

The Mole: A Chemist’s “Dozen”
When we count large numbers of objects, we often use units such as 1 dozen objects = 12 objects. 1 gross objects = 144 objects. The chemist’s “dozen” is the mole (abbreviated mol). A mole is the measure of material containing × 1023 particles: 1 mole = × 1023 particles This number is Avogadro’s number. We like to know the number of atoms or moles in a sample because when we react one substance with another they combine as particles in a particular stoichiometric relationship. 1 atom may react with 1 atom or 1 mole may react with 1 mole. Being able to calculate the number of moles of a substance (which contains Avogadro’s number of molecules) is therefore very important as we plan chemical reactions.

The Mole First thing to understand about the mole is that it can specify Avogadro’s number of anything. For example, 1 mole of marbles corresponds to × 1023 marbles. 1 mole of sand grains corresponds to × 1023 sand grains. One mole of anything is × 1023 units of that thing.

The Mole The second, and more fundamental, thing to understand about the mole is how it gets its specific value. The value of the mole is equal to the number of atoms in exactly 12 grams of pure C-12. 12 g C = 1 mol C atoms = × 1023 C atoms The definition of the mole gives us a relationship between mass and the number of atoms. Thus, we can count atoms by weighing them.

Converting between Number of Moles and Number of Atoms
Converting between number of moles and number of atoms is similar to converting between dozens of eggs and number of eggs. For atoms, you use the conversion factor 1 mol atoms = × 1023 atoms. The conversion factors take the following forms:

Converting between Mass and Amount (Number of Moles)
To count atoms by weighing them, we need one other conversion factor—the mass of 1 mol of atoms. The mass of 1 mol of atoms of an element is the molar mass. An element’s molar mass in grams per mole is numerically equal to the element’s atomic mass in atomic mass units (amu).

Converting between Mass and Moles
The lighter the atom, the less mass in 1 mol of atoms. The lighter the atom, the more atoms there are in 1 g. Note how the number of grams per mole of a substance is equal to its atomic weight in the periodic table. The atomic weight of carbon is – therefore one mole of carbon weights grams and contains Avogadro’s number of atoms.

Relationship Between Moles and Mass
Which weighs more: a dozen apples or a dozen eggs Which weighs more: a mole of copper (Cu) or a mole of gold (Au)?

Relationship Between Moles and Mass
Which has more units: 5 kg of bricks or 5 kg of feathers? Which has more atoms: 100 g of copper (Cu) or 100 g of gold (Au)?

Mole and Mass Relationships
sulfur 32.06 g 1 mole carbon 12.01 g

Converting between Mass and Moles
The molar mass of any element is the conversion factor between the mass (in grams) of that element and the amount (in moles) of that element. For carbon, We use molar mass as a conversion factor

Conceptual Plan We now have all the tools to count the number of atoms in a sample of an element by weighing it. First, we obtain the mass of the sample by weighing. Then, we convert it to the amount in moles using the element’s molar mass. Finally, we convert it to the number of atoms using Avogadro’s number. The conceptual plan for these kinds of calculations takes the following form:

Calculate the moles of carbon in 0.0265 g of pencil lead
Given: Find: g C mol C Conceptual Plan: Relationships: 1 mol C = g g C mol C Solution: Check: because the given amount is much less than 1 mol C, the number makes sense

How many copper atoms are in a penny weighing 3.10 g?
Given: Find: 3.10 g Cu atoms Cu Conceptual Plan: Relationships: g Cu mol Cu atoms Cu 1 mol Cu = g, 1 mol = x 1023 Solution: Check: because the given amount is much less than 1 mol Cu, the number makes sense

Practice — How many aluminum atoms are in a can weighing 16.2 g?
Do this at home for practice. 3.62 X 1023 atoms

Which has more atoms, 10.0 g Mg or 10.0 g Ca?
Magnesium Calcium Both have the same number of atoms.

Which has more atoms, 10.0 g Mg or 10.0 g Ca?
Magnesium Calcium Both have the same number of atoms. Answer: a

Which of the following has the largest mass?
10.0 g Li 10.0 moles of Li 100 g Na 10.0 moles of K 100 g Rb

Which of the following has the largest mass?
10.0 g Li 10.0 moles of Li 100 g Na 10.0 moles of K 100 g Rb Answer: d

If a pure copper penny has 2.94 × 1022 atoms, what will be its mass?
1.30 g 0.323 g 2.79 g 3.10 g 1.12 g

If a pure copper penny has 2.94 × 1022 atoms, what will be its mass?
1.30 g 0.323 g 2.79 g 3.10 g 1.12 g Answer: d

Formula Mass mass of 1 molecule of H2O
The mass of an individual molecule or formula unit also known as molecular mass or molecular weight Sum of the masses of the atoms in a single molecule or formula unit whole = sum of the parts! mass of 1 molecule of H2O = 2(1.01 amu H) amu O = amu

Molar Mass of Compounds
The relative masses of molecules can be calculated from atomic masses formula mass = 1 molecule of H2O = 2(1.01 amu H) amu O = amu 1 mole of H2O contains 2 moles of H and 1 mole of O so… molar mass = 1 mole H2O = 2(1.01 g H) g O = g so the Molar Mass of H2O is g/mole Molar mass = formula mass (in g/mole)

Your Turn! Formula is Mg3(PO4)2
What is the formula mass of magnesium phosphate? 262.9 g/mol 150.9 g/mol 333.6 g/mol 119.3 g/mol 166.9 g/mol Formula is Mg3(PO4)2

What is the molar mass of calcium phosphate?
g/mole g/mole g/mole g/mole g/mole Ca3(PO4)2

What is the molar mass of calcium phosphate?
g/mole g/mole g/mole g/mole g/mole Answer: a The formula for calcium phosphate is Ca3(PO4)2

How many moles are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00)
Given: Find: 50.0 g mol PbO2 moles PbO2 Conceptual Plan: Relationships: 1 mol PbO2 = g g PbO2 mol PbO2 Solution: Check: because the given amount is less than g, the moles being < 1 makes sense

Find the number of CO2 molecules in 10.8 g of dry ice
Given: Find: 10.8 g CO2 molecules CO2 g CO2 mol CO2 molec CO2 Conceptual Plan: Relationships: 1 mol CO2 = g, 1 mol = x 1023 Solution: Check: because the given amount is much less than 1 mol CO2, the number makes sense

Which of the following contains the largest number of molecules?
10.0 g CH4 10.0 g C2H6 10.0 g SO2 10.0 g Xe

Which of the following contains the largest number of molecules?
10.0 g CH4 10.0 g C2H6 10.0 g SO2 10.0 g Xe Answer: a

Which sample represents the greatest number of moles?
44.01 g CO2 1.0 moles C3H8 6.022 x 1023 molecules C4H10 18.02 g H2O All of the samples have the same number of moles.

Which sample represents the greatest number of moles?
44.01 g CO2 1.0 moles C3H8 6.022 x 1023 molecules C4H10 18.02 g H2O All of the samples have the same number of moles. Answer: e

Learning Check: Using Molar Mass
Example: If we need mole Ca3(PO4)2 for an experiment, how many grams do we need to weigh out? What do we want to determine? 0.168 mole Ca3(PO4)2 = ? g Ca3(PO4)2 Calculate MM of Ca3(PO4)2 3 × mass Ca = 3 × g = g 2 × mass P = 2 × g = g 8 × mass O = 8 × g = g 1 mole Ca3(PO4)2 = g Ca3(PO4)2 Start End

Set up ratio so that what you want is on the top and what you start with is on the bottom = g Ca3(PO4)2

Learning Check: Mole Conversions
Calculate the number of formula units of Na2CO3 in 1.29 moles of Na2CO3. How many moles of Na2CO3 are there in 1.15 × 105 formula units of Na2CO3 ? = 7.77 × 1023 particles Na2CO3 = 1.91×10–19 mol Na2CO3

Using Avogadro’s Number
What is the mass, in grams, of one molecule of octane, C8H18? Molecules octane  mol octane  g octane 1. Calculate molar mass of octane Mass C = 8 × g = g Mass H = 18 × g = g 1 mol octane = g octane 2. Convert 1 molecule of octane to grams = × 10–22 g octane

Mole-to-Mole Conversion Factors
Chemical formula may be used to relate the amount of each atom to the amount of the compound In N2O5 there are three relationships: 2 mol N ⇔ 1 mol N2O5 5 mol O ⇔ 1 mol N2O5 2 mol N ⇔ 5 mol O Can also use these on atomic scale: 2 atom N ⇔ 1 molecule N2O5 5 atom O ⇔ 1 molecule N2O5 2 atom N ⇔ 5 molecule O

Stoichiometric Equivalencies
These ratios constitute conversion factors Equivalency Mole Ratio 2 mol N ⇔ 1 mol N2O5 2 mol N 1 mol N2O5 5 mol O ⇔ 1 mol N2O5 5 mol O 1 mol N2O5 2 mol N ⇔ 5 mol O 5 mol O 2 mol N

Calculating the Amount of a Compound by Analyzing One Element
Calcium phosphate is widely found in natural minerals, bones, and some kidney stones. A sample is found to contain moles of phosphorus. How many moles of Ca3(PO4)2 are in that sample? What do we want to find? 0.864 mol P = ? mol Ca3(PO4)2 What do we know? 2 mol P ⇔ 1 mol Ca3(PO4)2 Solution = mol Ca3(PO4)2

Your Turn! Calculate the number of moles of calcium in moles of Ca3(PO4)2 2.53 mol Ca 0.432 mol Ca 3.00 mol Ca D mol Ca E mol Ca 2.53 moles of Ca3(PO4)2 = ? mol Ca 3 mol Ca  1 mol Ca3(PO4)2 = 7.59 mol Ca

Your Turn! A sample of sodium carbonate, Na2CO3, is found to contain 10.8 moles of sodium. How many moles of oxygen atoms (O) are present in the sample? 10.8 mol O 7.20 mol O 5.40 mol O D mol O E mol O 10.8 moles of Na = ? mol O 2 mol Na  3 mol O = 16.2 mol O

Mass-to-Mass Calculations
Common laboratory calculation Need to know what mass of reagent B is necessary to completely react given mass of reagent A to form a compound Stoichiometry comes from chemical formula of compounds Use the subscripts Summary of steps mass A → moles A → moles B → mass B

Mass-to-Mass Calculations
Chlorophyll, the green pigment in leaves, has the formula C55H72MgN4O5. If g of Mg is available to a plant for chlorophyll synthesis, how many grams of carbon will be required to completely use up the magnesium? Analysis g Mg ⇔ ? g C g Mg → mol Mg → mol C → g C Assembling the Tools g Mg = 1 mol Mg 1 mol Mg ⇔ 55 mol C 1 mol C = g C

Ex. Mass-to-Mass Conversion
1 mol C ⇔ g C g Mg ⇔ 1 mol Mg g Mg → mol Mg → mol C → g C 1 mol Mg ⇔ 55 mol C = g C

Your Turn! How many g of iron are required to use up all of g of oxygen atoms (O) to form Fe2O3? A g B g C g D. 134 g E g = 59.6 g Fe mass O  mol O  mol Fe  mass Fe 25.6 g O  ? g Fe 3 mol O  2 mol Fe

Your Turn! Silver is often found in nature as the ore, argentite (Ag2S). How many grams of pure silver can be obtained from a 836 g rock of argentite? A g B. 728 g C. 364 g D. 775 g E. 418 g mass Ag2S  mol Ag2S  mol Ag  mass Ag 836 g Ag2S  ? g Ag 1 mol Ag2S  2 mol Ag = 728 g Ag

Percentage Composition of Compounds
Percentage of each element in a compound by mass Can be determined from the formula of the compound and the experimental mass analysis of the compound. The percentages may not always total to 100% due to rounding. Mass of 1 mol of a compound is its molar mass.

(Mass) Percent Composition of an Element
Consider the chlorofluorocarbon CCl2F2. What is the mass percent of Cl? We know: 1) molar mass CCl2F2 = (35.45) + 2(19.00) = g/mol; molar mass Cl = g/mol. 2) 1 mol CCl2F2 contains 2 mol Cl. Mass percent Cl = 2 X g/mol ÷ g/mol X 100% = 58.64%

Mass Percent as a Conversion Factor
The mass percent tells you the mass of a constituent element in 100 g of the compound the fact that CCl2F2 is 58.64% Cl by mass means that 100 g of CCl2F2 contains g Cl This can be used as a conversion factor 100 g CCl2F2 : g Cl

Determine the mass % composition of CaCl2 (Ca = 40.08, Cl = 35.45)

Benzaldehyde is 79. 2% carbon. What mass of benzaldehyde contains 19
Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C? Given: Find: 19.8 g C, 79.2% C g benzaldehyde Conceptual Plan: Relationships: 100 g benzaldehyde : 79.2 g C g C g benzaldehyde Solution: Check: because the mass of benzaldehyde is more than the mass of C, the number makes sense

Conversion Factors from Chemical Formula
We can also use relationships derived from chemical formulas as conversion factors. How many moles of Cl are in 38.5 mol of CCl2F2? We know: 1 mol CCl2F2: 2 mol Cl Therefore, 38.5 mol CCl2F2 X 2 mol Cl / 1 mol CCl2F2 = 77.0 mol Cl The chemical formula gives us a relationship between the amounts (in moles) of substances, not between the masses (in grams) of them.

How many grams of sodium are in 6. 2 g of NaCl. (Na = 22. 99; Cl = 35
Given: Find: 6.2 g NaCl g Na Conceptual Plan: Relationships: 1 mol NaCl = g, 1 mol Na = g, 1 mol Na : 1 mol NaCl g NaCl mol NaCl mol Na g Na Solution: Check: because the amount of Na is less than the amount of NaCl, the answer makes sense

Which of the following contains the largest mass percent hydrogen?
10.0 g CH4 10.0 g C2H6 10.0 g H2O 10.0 g H2S

Which of the following contains the largest mass percent hydrogen?
10.0 g CH4 10.0 g C2H6 10.0 g H2O 10.0 g H2S Answer: a

Calculate the number of carbon atoms in 25
Calculate the number of carbon atoms in 25.0 grams of isopropyl alcohol (C3H8O). 1.25 C atoms 15.0 C atoms 2.51 x 1023 C atoms 7.52 x 1023 C atoms 2.07 x 10–24 C atoms

Calculate the number of carbon atoms in 25
Calculate the number of carbon atoms in 25.0 grams of isopropyl alcohol (C3H8O). 1.25 C atoms 15.0 C atoms 2.51 x 1023 C atoms 7.52 x 1023 C atoms 2.07 x 10–24 C atoms Answer: d

Percent Composition One can determine percentage composition based on the chemical analysis of a substance Example: A sample of a liquid with a mass of g was decomposed into its elements and gave g of carbon, g of hydrogen, and g of oxygen. What is the percentage composition of this compound? Analysis: Calculate percentage by mass of each element in sample Tools: Equation for percentage by mass Total mass = g Mass of each element

% Composition of Compound
For C: For H: For O: Percentage composition tells us mass of each element in g of substance In g of our liquid 60.26 g C, g H, and g O = 60.26% C = 11.11% H = 28.62% O Sum of percentages: 99.99%

Your Turn! A sample was analyzed and found to contain g nitrogen and g oxygen. What is the percentage composition of this compound? 1. Calculate total mass of sample Total sample mass = g g = g 2. Calculate % Composition of N 3. Calculate % Composition of O = 25.94% N = 74.06% O

Because the rest must be sulfur,
Your Turn! In a previous “Your Turn!” slide, you found that an 836 g rock of argentite (Ag2S) contained 728 g of silver. What are the percentages of silver and sulfur, by mass, in argentite? 50.0% Ag, 50.0% S 66.6% Ag, 33.4% S 12.9% Ag, 87.1% S D. 115% Ag, 108% S E. 87.1% Ag, 12.9% S Calculate % comp. of Ag: = 87.1% Ag Because the rest must be sulfur, % S = 100 – 87.1 = 12.9% S

Percent Compositions and Chemical Identity
Theoretical or Calculated Percentage Composition Calculated from molecular or ionic formula. Lets you distinguish between multiple compounds formed from the same two elements If experimental percent composition is known Calculate theoretical percentage composition from proposed chemical formula Compare with experimental composition e.g., N and O form multiple compounds N2O, NO, NO2, N2O3, N2O4, and N2O5

Using Percent Composition
Are the mass percentages 30.54% N and 69.46% O consistent with the formula N2O4? Procedure: Assume one mole of compound Subscripts tell how many moles of each element are present 2 mol N and 4 mol O Use molar masses of elements to determine mass of each element in 1 mole Molar Mass of N2O4 = g N2O4 / 1 mol Calculate % by mass of each element

Using Percent Composition (cont)
= g N = g O = 30.54% N in N2O4 = 69.46% O in N2O4 The experimental values match the theoretical percentages for the formula N2O4.

Your Turn! If a sample containing only phosphorous and oxygen has percent composition 56.34% P and 43.66% O, is this P4O10? Yes B. No 4 mol P  1 mol P4O10 10 mol O  1 mol P4O10 4 mol P = 4  g/mol P = g P 10 mol O = 10 16.00 g/mol O = g O 1 mol P4O10 = g P4O10 This compound is not P4O10. It is P2O3. = 43.64% P = 56.36% O

Empirical & Molecular Formulas Balancing Chemical Equations

Determining Empirical and Molecular Formulas
When making or isolating new compounds one must characterize them to determine structure Empirical Formula Simplest ratio of atoms of each element in compound Obtained from experimental analysis of compound Molecular Formula Exact composition of one molecule Exact whole number ratio of atoms of each element in molecule glucose Empirical formula CH2O Molecular formula C6H12O6

Three Ways to Calculate Empirical Formulas
From Masses of Elements e.g., g sample of which g is Fe and g is O. From Percentage Composition e.g., 43.64% P and 56.36% O From Combustion Data Given masses of combustion products e.g., The combustion of a g sample of a compound of C, H, and O in pure oxygen gave g CO2 and g of H2O

Strategy for Determining Empirical Formulas
Determine mass in g of each element Convert mass in g to moles Divide all quantities by smallest number of moles to get smallest ratio of moles Convert any non-integers into integer numbers. If number ends in decimal equivalent of fraction, multiply all quantities by the denominator of the fraction Otherwise, round numbers to nearest integers

1. Empirical Formula from Mass Data
When a g sample of a compound was analyzed, it was found to contain g of C, g of H, and g of N. Calculate the empirical formula of this compound. Step 1: Calculate moles of each substance 3.722  10–3 mol C 1.860  10–2 mol H 3.723  10–3 mol N

Step 2: Select the smallest number of moles Smallest is × 10–3 mole C = H = Step 3: Divide all number of moles by the smallest one Mole ratio Integer ratio 1.000 = 1 4.997 = 5 N = 1.000 = 1 Empirical formula = CH5N

One of the compounds of iron and oxygen, “black iron oxide,” occurs naturally in the mineral magnetite. When a g sample was analyzed it was found to have g of Fe and g of O. Calculate the empirical formula of this compound. Assembling the tools: 1 mol Fe = g Fe 1 mol O = g O Find the moles of each element, then find the ratio of the moles of the elements. 1. Calculate moles of each substance mol Fe mol O

2. Divide both by smallest number mol to get smallest whole number ratio. =1.000 Fe × 3 = Fe =1.33 O × 3 = 3.99 O Or Empirical Formula = Fe3O4

2. Empirical Formula from Percentage Composition
New compounds are characterized by elemental analysis, from which the percentage composition can be obtained Use percentage composition data to calculate empirical formula Must convert percentage composition to grams Assume g sample Convenient Sum of percentage composition = 100% Sum of masses of each element = 100 g

Calculate the empirical formula of a compound whose percentage composition data is 43.64% P and 56.36% O. If the molar mass is determined to be g/mol, what is the molecular formula? Step 1: Assume 100 g of compound. Calculate moles. 43.64 g P 56.36 g O 1 mol P = g 1 mol O = g = mol P = mol P

Step 2: Divide by smallest number of moles  2 = 2  2 = 5 Step 3: Multiply to get integers 1.000  2 = 2 2.500  2 = 5 Empirical formula = P2O5

3. Empirical Formulas from Indirect Analysis:
In practice, compounds are not broken down into elements, but are changed into other compounds whose formula is known Combustion Analysis Compounds containing carbon, hydrogen, and oxygen, can be burned completely in pure oxygen gas Only carbon dioxide and water are produced e.g., Combustion of methanol (CH3OH) 2CH3OH + 3O2  2CO2 + 4H2O

Combustion Analysis Classic Modern CHN analysis

3. Empirical Formulas from Indirect Analysis:
Carbon dioxide and water are separated and weighed separately All C ends up as CO2 All H ends up as H2O Mass of C can be derived from amount of CO2 mass CO2  mol CO2  mol C  mass C Mass of H can be derived from amount of H2O mass H2O  mol H2O  mol H  mass H Mass of oxygen is obtained by difference mass O = mass sample – (mass C + mass H)

Ex. Indirect or Combustion Analysis
The combustion of a g sample of a compound of C, H, and O in pure oxygen gave g CO2 and g of H2O. Calculate the empirical formula of the compound. C H H2O CO2 MM (g/mol) 12.011 1.008 18.015 44.01 1. Calculate mass of C from mass of CO2. mass CO2  mole CO2  mole C  mass C = g C

The combustion of a g sample of a compound of C, H, and O gave g CO2 and g of H2O. Calculate the empirical formula of the compound. 2. Calculate mass of H from mass of H2O. mass H2O  mol H2O  mol H  mass H = g H 3. Calculate mass of O from difference. 5.217 g sample – g C – g H = g O

H O At. mass 12.011 1.008 15.999 g 2.021 0.5049 2.691 4. Calculate mol of each element = mol C = mol H = mol O

Preliminary empirical formula C0.1683H0.5009O0.1682 5. Calculate mol ratio of each element Because all values are close to integers, round to = C1.00H2.97O1.00 Empirical Formula = CH3O

Your Turn! The combustion of a g sample of a compound of C, H, and S in pure oxygen gave g CO2 and g of H2O. Calculate the empirical formula of the compound. C4H12S CH3S C2H6S D. C2H6S3 E. CH3S2

Your Turn! Solution 1. Calculate mass of C from mass of CO2.
mass CO2  mole CO2  mole C  mass C = g C 2. Calculate mass of H from mass of H2O. mass H2O  mol H2O  mol H  mass H = g H

Your Turn! Solution Continued
3. Calculate mass of S from difference. 4. Convert individual elements to moles to obtain a preliminary empirical formula. 13.66 g sample – g C – g H = g S = mol C = mol H = mol S

Your Turn! Solution Continued
Preliminary empirical formula C0.4497H1.319S0.2198 5. Calculate mol ratio of each element Because all values are close to integers, round to = C2.03H6.00S1.00 Empirical Formula = C2H6S

Answer on following slides.
Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53% Try this at home. Answer on following slides.

Example: Find the empirical formula of aspirin with the given mass percent composition
Information Given:C = 60.00%, H = 4.48 %, O =35.53% \ g C, 4.48 g H, g O Find: empirical formula, CxHyOz g C mol C conceptual plan whole number ratio mole ratio empirical formula pseudoformula g H mol H g O mol O C4.996H4.44O2.220 {C2.25H2O1} x 4 C9H8O4

Determining Molecular Formulas
Empirical formula Accepted formula unit for ionic compounds Molecular formula Preferred for molecular compounds In some cases molecular and empirical formulas are the same When they are different, the subscripts of molecular formula are integer multiples of those in empirical formula If empirical formula is AxBy Molecular formula will be An × xBn × y

Determining Molecular Formula
Need molecular mass and empirical formula Calculate ratio of molecular mass to mass predicted by empirical formula and round to nearest integer Example: Glucose Molecular mass is g/mol Empirical formula = CH2O Empirical formula mass = g/mol Molecular formula = C6H12O6

Learning Check The empirical formula of a compound containing phosphorous and oxygen was found to be P2O5. If the molar mass is determined to be g/mol, what is the molecular formula? Step 1: Calculate empirical mass Step 2: Calculate ratio of molecular to empirical mass = 2 Molecular formula = P4O10

Your Turn! The empirical formula of hydrazine is NH2, and its molecular mass is What is its molecular formula? NH2 N2H4 N3H6 N4H8 N1.5H3 Atomic Mass: N = ; H = 1.008; O = Molar mass of NH2 = (1 × 14.01) g + (2 × 1.008) g = g n = (32.0/16.02) = 2

Molecular formula = {C5H3} x 4 = C20H12
Benzopyrene has a molar mass of 252 g/mol and an empirical formula of C5H3. What is its molecular formula? (C = 12.01, H=1.01) C5 = 5(12.01 g) = g H3 = 3(1.01 g) = g C5H3 = g molar mass empirical formula Molecular formula = {C5H3} x 4 = C20H12

Chemical Reactions Reactants  Products
Reactions involve chemical changes in matter resulting in new substances. Reactions involve rearrangement and exchange of atoms to produce new molecules. Elements are not transmuted during a reaction. Reactants  Products

Chemical Reactions Shorthand way of describing a reaction
Provides information about the reaction formulas of reactants and products states of reactants and products relative # of reactant/products that are required/made used to determine weights of reactants used and products that can be made

CH4(g) + O2(g) ® CO2(g) + H2O(g)
Combustion of Methane Methane gas burns to produce carbon dioxide gas and gaseous water. Whenever something burns it combines with O2(g). CH4(g) + O2(g) ® CO2(g) + H2O(g) If you look closely, you should immediately spot a problem.

Combustion of Methane Notice also that the left side has four hydrogen atoms while the right side has only two. To correct these problems, we must balance the equation by changing the coefficients, not the subscripts.

Combustion of Methane, Balanced
To show the reaction obeys the Law of Conservation of Mass the equation must be balanced. We adjust the numbers of molecules so there are equal numbers of atoms of each element on both sides of the arrow. 1 C H O

Rules for Balancing Chemical Equations
Write a skeletal equation. Co2O3 + C  Co + CO2 Balance atoms in more complex substances first: 2 Co2O3 + C  Co + 3 CO2 (since 3O  2O) Balance free elements last: 2 Co2O3 + 3 C  4 Co + 3 CO2 Clear fractions by dividing through. NA Count atoms to ensure equation is balanced.

What are the coefficients for the decomposition of nitroglycerin?
__ C3H5N3O9  __ N2 + __ CO2 + __ H2O + __ O2 2,3,6,2,1 2,3,6,5,1 4,6,12,10,12 4,3,12,10,1 4,6,12,10,1

What are the coefficients for the decomposition of nitroglycerin?
__ C3H5N3O9  __ N2 + __ CO2 + __ H2O + __ O2 2,3,6,2,1 2,3,6,5,1 4,6,12,10,12 4,3,12,10,1 4,6,12,10,1 Answer: e

Which of the following statements is true about a balanced chemical reaction?
The mass of the reactants is equal to the mass of the products. The number of each type of atom is the same in the reactants and in the products. The moles of each type of atom is the same in the reactants and in the products. All of the above.

Which of the following statements is true about a balanced chemical reaction?
The mass of the reactants is equal to the mass of the products. The number of each type of atom is the same in the reactants and in the products. The moles of each type of atom is the same in the reactants and in the products. All of the above. Answer: d

Write a balanced equation for the combustion of octane, C8H18
Write a skeletal equation C8H18(l) + O2(g)  CO2(g) + H2O(g) Balance atoms in complex substances first 8  C  1 x 8 C8H18(l) + O2(g)  8 CO2(g) + H2O(g) 18  H  2 x 9 C8H18(l) + O2(g)  8 CO2(g) + 9 H2O(g) Balance free elements by adjusting coefficient in front of free element 25/2 x 2  O  25 C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9 H2O(g) If fractional coefficients, multiply thru by denominator {C8H18(l) + 25/2 O2(g)  8 CO2(g) + 9 H2O(g)}x 2 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) Check 16  C  16; 36  H  36; 50  O  50

Your Turn! Balance each of the following equations.
What are the coefficients in front of each compound? __ Ba(OH)2(aq) +__ Na2SO4(aq) → __ BaSO4(s) + __ NaOH(aq) 1 1 1 2 ___KClO3(s) → ___KCl(s) +___ O2(g) 2 2 3 __H3PO4(aq) + __ Ba(OH)2(aq) → __Ba3(PO4)2(s) + __H2O(l ) 2 3 1 6

Quantities in Chemical Reactions
The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction. Law of conservation of mass Balancing equations by balancing atoms The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry.

Reaction Stoichiometry
The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) 2 molecules of C8H18 react with 25 molecules of O2 to form 16 molecules of CO2 and 18 molecules of H2O 2 moles of C8H18 react with 25 moles of O2 to form 16 moles of CO2 and 18 moles of H2O 2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O

Stoichiometry of Making Pizza: Ingredient Relationships
The number of pizzas you can make depends on the amount of the ingredients you use 1 crust + 5 oz. tomato sauce + 2 cups cheese  1 pizza This relationship can be expressed mathematically 1 crust : 5 oz. sauce : 2 cups cheese : 1 pizza If you have 10 cups of cheese how many pizzas can you make (assuming you have enough crusts and tomato sauce)? Since 2 cups cheese : 1 pizza, then, If you want to make 3 pizzas, how much cheese do you need?

Making Molecules: Mole-to-Mole Conversions
We use the ratio from the balanced chemical equation in the same way that we used the ratio from the pizza recipe. The ratio of the coefficients acts as a conversion factor between the amount in moles of the reactants and products. The ratio acts as a conversion factor between the amount in moles of the reactant C8H18 and the amount in moles of the product CO2. 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) stoichiometric ratio: 2 moles C8H18 : 16 moles CO2

Suppose That We Burn 22.0 Moles of C8H18; How Many Moles of CO2 Form?
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) stoichiometric ratio: 2 moles C8H18 : 16 moles CO2 The combustion of 22 moles of C8H18 adds 176 moles of CO2 to the atmosphere.

Your Turn! How many moles of hydrochloric acid are required to completely react with 1.43 moles of iron(II) chloride according to the reaction below? 14HCl + Na2Cr2O7 + 6FeCl2 → 2CrCl3 + 7H2O + 6FeCl3 + 2NaCl 20.0 mole 1.43 mole 0.613 mole 3.34 mole 8.58 mole Assembling the tools 1.43 mole FeCl2 = ? moles HCl 6 moles FeCl2 = 14 mole HCl = 3.34 mol HCl

The rapid decomposition of sodium azide, NaN3, to its elements is one of the reactions used to inflate airbags: 2 NaN3 (s)  2 Na (s) + 3 N2 (g) How many grams of N2 are produced from 6.00 g of NaN3? 3.88 g 1.72 g 0.138 g 2.59 g

The rapid decomposition of sodium azide, NaN3, to its elements is one of the reactions used to inflate airbags: 2 NaN3 (s)  2 Na (s) + 3 N2 (g) How many grams of N2 are produced from 6.00 g of NaN3? 3.88 g 1.72 g 0.138 g 2.59 g Answer: a

The overall equation involved in photosynthesis is
The overall equation involved in photosynthesis is 6 CO2 + 6 H2O  C6H12O6 + 6 O2. How many grams of glucose (C6H12O6, g/mol) form when 4.40 g of CO2 react? 18.0 g 3.00 g 108 g g

The overall equation involved in photosynthesis is 6 CO2 + 6 H2O  C6H12O6 + 6 O2. How many grams of glucose (C6H12O6, g/mol) form when 4.40 g of CO2 react? 18.0 g 3.00 g 108 g g Answer: b

Your Turn! How many grams of sodium dichromate are required to produce 24.7 g iron(III) chloride from the following reaction? 14HCl + Na2Cr2O7 + 6FeCl2 → 2CrCl3 + 7H2O + 6FeCl3 + 2NaCl 6.64 g Na2Cr2O7 0.152 g Na2Cr2O7 8.51 g Na2Cr2O7 39.9 g Na2Cr2O7 8.04 g Na2Cr2O7 = 6.64 g Na2Cr2O7

Work this problem at home.
The overall equation involved in photosynthesis is 6 CO2 + 6 H2O  C6H12O6 + 6 O2. How many grams of carbon dioxide (CO2) are required to produce 18.0 g of C6H12O6? 18.0 g 3.00 g 54.0 g 26.4 g 4.40 g Work this problem at home. Answer on the next slide.

The overall equation involved in photosynthesis is 6 CO2 + 6 H2O  C6H12O6 + 6 O2. How many grams of carbon dioxide (CO2) are required to produce 18.0 g of C6H12O6? 18.0 g 3.00 g 54.0 g 26.4 g 4.40 g Answer: d

Predicting Amounts from Stoichiometry
How much CO2 is pumped into the atmosphere by combustion of a tankful of gasoline? 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) 2 moles C8H18 : 16 moles CO2 Given: 57.0 L of gasoline, d = g/mL, octane = g/mol, 1 mol CO2 = g Find: Amount carbon dioxide from combustion of 15 gallons of octane Plan: L → mL → g → moles octane → moles CO2 → g CO2

2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
How much CO2 is pumped into the atmosphere by combustion of a tankful of gasoline? 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) Given: Find: 15.0 gallon tank of gas, 1 gallon = L, d = g/mL, 1 mol octane = g, 1 mol CO2 = g g CO2 gallons octane → L → mL → g → moles octane → moles CO2 → g CO2 Plan: Relation-ships: 3.785 L/1 gal 1000 mL/L 0.703 g/mL 1 mol C8H18 = g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2 Soltn: 15.0 gal x 3.785L/gal x 1000 mL/L x g/mL x 1 mol oct/ g oct x 16 mol CO2/2 mol oct x g CO2/ mol CO2 = 123,030 g kg Check: 15.0 gal of octane is 39.9 kg; because 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense

Limiting Reactants Theoretical & Percentage Yields

Limiting Reactant, Theoretical Yield
Recall our pizza recipe: 1 crust + 5 oz tomato sauce + 2 cups cheese  1 pizza If we have 4 crusts, 10 cups of cheese, and 15 oz tomato sauce. How many pizzas can we make? We have enough crusts to make We have enough cheese to make We have enough tomato sauce to make

Limiting Reactant We have enough crusts for four pizzas, enough cheese for five pizzas, but enough tomato sauce for only three pizzas. We can make only three pizzas. The tomato sauce limits how many pizzas we can make.

Theoretical Yield Tomato sauce is the limiting reactant, the reactant that makes the least amount of product. The limiting reactant is also known as the limiting reagent. The maximum number of pizzas we can make depends on this ingredient. In chemical reactions, we call this the theoretical yield. This is the amount of product that can be made in a chemical reaction based on the amount of limiting reactant. The ingredient that makes the least amount of pizza determines how many pizzas you can make (theoretical yield).

The Limiting Reactant For reactions with multiple reactants, one of the reactants may be completely used up before the others When this reactant is used up, the reaction stops and no more product is made The reactant that limits the amount of product is called the limiting reactant or limiting reagent the limiting reactant gets completely consumed The amount of product that can be made from the limiting reactant is called the theoretical yield Reactants not completely consumed are called excess reactants

Pizza Making Efficiency
Besides amount of ingredients, what other factors effect the actual number of pizzas produced? burn a pizza, drop one on the floor, or other unexpected events may occur actual number of pizzas may be fewer than expected: actual yield The efficiency of making pizzas is the ratio of pizzas actually made versus maximum number of pizzas. This ratio can be expressed as a %. In chemical reactions, this is referred to as the percent yield. If we only made 2 pizzas but we had the ingredients to make 3 pizzas (perhaps we dropped one), our percent yield would be 2/3 X 100% = 67%.

Summarizing Limiting Reactant and Yield
The limiting reactant (or limiting reagent) is the reactant that is completely consumed in a chemical reaction and limits the amount of product. The reactant in excess is any reactant that occurs in a quantity greater than is required to completely react with the limiting reactant.

Summarizing Limiting Reactant and Yield
The theoretical yield is the amount of product that can be made in a chemical reaction based on the amount of limiting reactant. The actual yield is the amount of product actually produced by a chemical reaction. The percent yield is calculated as follows:

Limiting and Excess Reactants in the Combustion of Methane
CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) Balanced equation for the combustion of methane requires that one molecule of CH4 reacts with two molecules of O2

Limiting and Excess Reactants in the Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) If there are five molecules of CH4 and eight molecules of O2, which is the limiting reactant? The reactant that limits the amount of product is called the limiting reactant or limiting reagent since less CO2 can be made from available O2 than CH4, O2 is the limiting reactant

Stoichiometry Problem
When 28.6 kg of C are allowed to react with 88.2 kg of TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. conceptual plan identify limiting reactant calculate theoretical yield of Ti from TiO2 and C (g → mols) calculate theoretical yield from limiting reagent (mols → g) calculate percent yield (actual yield/theoretical yield)

Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. Write a conceptual plan } smallest amount is from limiting reactant kg TiO2 C smallest mol Ti

Collect needed relationships
Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Collect needed relationships 1000 g = 1 kg Molar Mass TiO2 = g/mol Molar Mass Ti = g/mol Molar Mass C = g/mol 1 mole TiO2 : 1 mol Ti (from the chem. equation) 2 mole C : 1 mol Ti (from the chem. equation)

Apply the conceptual plan
Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti Apply the conceptual plan limiting reactant smallest moles of Ti

Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti Apply the conceptual plan theoretical yield

Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti Apply the conceptual plan

limiting reactant = TiO2 theoretical yield = 52.9 kg
Find the limiting reactant, theoretical yield, and percent yield: TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti Check the solutions limiting reactant = TiO2 theoretical yield = 52.9 kg percent yield = 80.9% Because Ti has lower molar mass than TiO2, the T.Y. makes sense and the percent yield makes sense as it is less than 100%

Ammonia is produced using the Haber process:
Ammonia is produced using the Haber process: H2 + N2  2 NH3 Calculate the mass of ammonia produced when 35.0 g of nitrogen react with 12.5 g of hydrogen. 47.5 g 42.6 g 35.0 g 63.8 g 70.5 g

Ammonia is produced using the Haber process: H2 + N2  2 NH3 Calculate the mass of ammonia produced when 35.0 g of nitrogen react with 12.5 g of hydrogen. 47.5 g 42.6 g 35.0 g 63.8 g 70.5 g Answer: b

Ammonia is produced using the Haber process: 3 H2 + N2  2 NH3 What percent yield of ammonia is produced from 15.0 kg each of H2 and N2, if 13.7 kg of product are recovered? Assume the reaction goes to completion. 7.53 × 10–2 % 1.50 × 10–1 % 75.3% 15.0% 16.2%

Ammonia is produced using the Haber process: 3 H2 + N2  2 NH3 What percent yield of ammonia is produced from 15.0 kg each of H2 and N2, if 13.7 kg of product are recovered? Assume the reaction goes to completion. 7.53 × 10–2 % 1.50 × 10–1 % 75.3% 15.0% 16.2% Answer: c

What mass of TiCl4 is needed to produce 25
What mass of TiCl4 is needed to produce 25.0 g of Ti if the reaction proceeds with an 82% yield? TiCl4 + 2Mg  Ti + 2 MgCl2 30.5 g 121 g 99.1 g 81.2 g

What mass of TiCl4 is needed to produce 25
What mass of TiCl4 is needed to produce 25.0 g of Ti if the reaction proceeds with an 82% yield? TiCl4 + 2Mg  Ti + 2 MgCl2 30.5 g 121 g 99.1 g 81.2 g Answer: b

Learning Check: Percentage Yield
A chemist set up a synthesis of solid phosphorus trichloride by mixing 12.0 g of solid phosphorus with g chlorine gas and obtained 42.4 g of solid phosphorus trichloride. Calculate the percentage yield of this compound. Analysis: Write balanced equation P(s) + Cl2(g)  PCl3(s) Is this balanced? Determine Limiting Reagent Determine Theoretical Yield Calculate Percentage Yield

Assembling the Tools: 1 mol P = g P 1 mol Cl2 = g Cl2 3 mol Cl2  2 mol P Solution Determine Limiting Reactant But you only have 35.0 g Cl2, so Cl2 is limiting reactant = 41.2 g Cl2

Solution Determine Theoretical Yield Determine Percentage Yield Actual yield = 42.4 g = 45.2 g PCl3 = 93.8 %

Practice Problems

Calculate the mass in grams of one mole of footballs if one football has a mass of 0.43 kg.

Calculate the mass in grams of one mole of footballs if one football has a mass of 0.43 kg.
Answer: c

Your Turn! How many moles of CO2 are there in 10.0 g? 1.00 mol
E mol Molar mass of CO2 1 × g = g C 2 × g = g O 1 mol CO2 = g CO2 = mol CO2

Macroscopic to Microscopic
How many silver atoms are in a 85.0 g silver bracelet? What do we want to determine? 85.0 g silver = ? atoms silver What do we know? g Ag = 1 mol Ag 1 mol Ag = 6.022×1023 Ag atoms g Ag  mol Ag  atoms Ag = 4.75 × 1023 Ag atoms

Your Turn! How many atoms are in 1.00 × 10–9 g of U?
Molar mass U = g/mole. 6.02 × 1014 atoms 4.20 × 1011 atoms C × 1012 atoms D × 10–31 atoms E × 1021 atoms = 2.53 × 1012 atoms U

Example: How many moles of iron (Fe) are in g Fe? What do we want to determine? 15.34 g Fe = ? mol Fe What do we know? 1 mol Fe = g Fe Set up ratio so that what you want is on top and what you start with is on the bottom Start End = mol Fe

Your Turn! How many grams of platinum (Pt) are in 0.475 mole Pt? 195 g
Molar mass of Pt = g/mol = 92.7 g Pt

Your Turn! Calculate the mass in grams of FeCl3 in 1.53 × 1023 formula units. (molar mass = g/mol) 162.2 g 0.254 g 1.661 ×10–22 g 41.2 g 2.37 × 10–22 = 41.2 g FeCl3

Your Turn! How much, in grams, do 8.85 × 1024 atoms of zinc weigh?
= 961 g Zn

Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) Given: 75.7% Sn, (100 – 75.3) = 24.3% F  in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g F Find: SnxFy Rel: 1 mol Sn = g; 1 mol F = g Conceptual plan: whole number ratio g Sn mol Sn mole ratio pseudoformula empirical formula g F mol F

Determine the empirical formula of stannous fluoride, which contains 75.7% Sn (118.70) and the rest fluorine (19.00) Apply the conceptual plan Sn0.638F1.28 SnF2

Determine the empirical formula of magnetite, which contains 72
Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Given: 72.4% Fe, (100 – 72.4) = 27.6% O  in 100 g magnetite there are 72.4 g Fe and 27.6 g O Find: FexOy Rel: 1 mol Fe = g; 1 mol O = g Conceptual plan: whole number ratio g Fe mol Fe mole ratio pseudoformula empirical formula g O mol O

Determine the empirical formula of magnetite, which contains 72
Determine the empirical formula of magnetite, which contains 72.4% Fe (55.85) and the rest oxygen (16.00) Apply the conceptual plan Fe1.30O1.73

Example: Find the empirical formula of compound with the given amounts of combustion products
Information Given: g compound, 2.445 g CO2, g H Find: empirical formula, CxHyOz conceptual plan g CO2, H2O mol CO2, H2O mol C, H g C, H g O mol O mol C, H, O pseudo formula mol ratio empirical formula

Information Given: g compound, 2.445 g CO2, g H2O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula needed relationships 1 mole CO2 = g CO2 1 mole H2O = g H2O 1 mole C = g C 1 mole H = g H 1 mole O = g O 1 mole CO2 = 1 mole C 1 mole H2O = 2 mole H

Apply the conceptual plan calculate the moles of C and H
Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, 2.445 g CO2, g H2O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan calculate the moles of C and H

Apply the conceptual plan calculate the grams of C and H
Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, 2.445 g CO2, g H2O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan calculate the grams of C and H

Apply the conceptual plan calculate the grams and moles of O
Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, 2.445 g CO2, g H2O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan calculate the grams and moles of O

Information Given: g compound, g CO2, g H2O mol C, g C, mol H, g H, g O, mol O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan write a pseudoformula C H O

Example: Find the empirical formula of compound with the given amounts of combustion products Information Given: g compound, g CO2, g H2O mol C, g C, mol H, g H, g O, mol O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan find the mole ratio by dividing by the smallest number of moles

Information Given: g compound, g CO2, g H2O mol C, g C, mol H, g H, g O, mol O Find: empirical formula, CxHyOz CP: g CO2 & H2O  mol CO2 & H2O  mol C & H  g C & H  g O  mol O  mol ratio  empirical formula Rel: MM of CO2, H2O, C, H, O; mol element : 1 mol compound Apply the conceptual plan multiply subscripts by factor to give whole number, if necessary write the empirical formula 1 12 10 O H C

An unknown compound contains the following percents by mass: C: 60
An unknown compound contains the following percents by mass: C: 60.86%, H: 5.83%, O: 23.16%, and N: 10.14%. Find the empirical formula. C6H8O2N2 C7H8O2N C6H8O2N C8H8O2N C8H8ON

An unknown compound contains the following percents by mass: C: 60
An unknown compound contains the following percents by mass: C: 60.86%, H: 5.83%, O: 23.16%, and N: 10.14%. Find the empirical formula. C6H8O2N2 C7H8O2N C6H8O2N C8H8O2N C8H8ON Answer: b

What is the empirical formula of a compound containing C, H, and O if combustion of 3.69 g of the compound yields 5.40 g of CO2 and g of H2O? CH3O C2H3O CHO C2H3O2 CH2O

What is the empirical formula of a compound containing C, H, and O if combustion of 3.69 g of the compound yields 5.40 g of CO2 and g of H2O? CH3O C2H3O CHO C2H3O2 CH2O Answer: e

Your Turn! 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l )
How many grams of Al2O3 are produced when 41.5 g Al react? 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l ) 78.4 g 157 g 314 g 22.0 g 11.0 g = 78.4 g Al2O3

Ex. Limiting Reactant Calculation
How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to: 4 NH3 + 5 O2  4 NO + 6 H2O Solution: Step 1 mass NH3  mole NH3  mole O2  mass O2 Assembling the tools 1 mol NH3 = g 1 mol O2 = g 4 mol NH3  5 mol O2 Only have 40.0 g O2, O2 limiting reactant = 70.5 g O2 needed

Ex. Limiting Reactant Calculation
How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to: 4 NH3 + 5 O2  4 NO + 6 H2O Solution: Step 2 mass O2  mole O2  mole NO  mass NO Assembling the tools 1 mol O2 = g 1 mol NO = g 5 mol O2  4 mol NO Can only form 30.0 g NO = 30.0 g NO formed

Your Turn! If 18.1 g NH3 is reacted with 90.4 g CuO, what is the maximum amount of Cu metal that can be formed? 2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g) (MM) (17.03) (79.55) (28.01) ( (18.02) (g/mol) 127 g 103 g 72.2 g 108 g 56.5 g 127 g CuO needed Only have 90.4 g so CuO limiting = 72.2 g Cu can be formed

Your Turn! If 8.51 g C2H5SH reacts with 22.4 g O2, what is the maximum amount of sulfur dioxide that can form? 2C2H5SH(l) + 9O2(g)  4CO2(g) + 6H2O(g) + 2SO2(g) (62.13) (32.00) (44.01) (18.02) (64.06) 19.7 g 4.38 g 39.5 g 9.97 g 8.77 g 19.7 g O2 needed Have 22.4 g so C2H5SH is limiting = 8.77 g SO2 can form

Ex. Percentage Yield Calculation
When 18.1 g NH3 and 90.4 g CuO are reacted, the theoretical yield is 72.2 g Cu. The actual yield is 58.3 g Cu. What is the percent yield? 2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g) = 80.7%

Your Turn! When 6.40 g of CH3OH was mixed with 10.2 g of O2 and ignited, 6.12 g of CO2 was obtained. What was the percentage yield of CO2? 2CH3OH + 3O2  2CO2 + 4H2O MM(g/mol) (32.04) (32.00) (44.01) (18.02) 6.12% 8.79% 100% 142% 69.6% = 9.59 g O2 needed; CH3OH limiting = 8.79 g CO2 in theory

The Mole and Stoichiometry

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