Moles http://www.learner.org/vod/vod_window.html?pid=803.

12 2 500 6.022 x 1023 Do Now – 3/10/17 What is a dozen?
What is a pair? 2 What is a ream? 500 What is a mole? 6.022 x 1023

I can… Define a mole as equal to 6.02 x 1023 particles (atoms or molecules). Explain the quantity of one mole is set by defining one mole of carbon 12 atoms to have a mass of exactly 12 grams. Determine the molar mass of a molecule from its chemical formula and a table of atomic masses and convert the mass of a molecular substance to moles, number of particles, or volume of gas at standard temperature and pressure.

Quick Question What is an isotope?
Where do you get an atoms mass on the periodic table? An atom of a given element that varies in its number of neutrons. The average atomic mass is below the symbol

Isotope Review Isotope: all isotopes of a given element share the same number of protons, each isotope differs from the others in its number of neutrons

Average Atomic Mass

Average Atomic Mass Measured in amu’s (atomic mass unit)
Isotopes are not present in even distribution, so to get the average atomic mass we need to take into account percentage of each isotope. How?

Counting with moles Chemical reactions involve extremely large numbers of particles. Chemists use counting unit called the mole to measure amounts of a substance. Why? We always deal with macroscopic samples containing enormous amounts of atoms It’s convenient to have a special unit that describes a large number of atoms

The Mole Similar to a pair or a dozen. A shoe salesman sells shoes in pairs. Chemists measure atoms and molecules in moles Like all units, a mole has to be based on something reproducible. A mole is the quantity of anything that has the same number of particles found in grams of carbon-12. That number is roughly 6.02x1023. Mole: the amount of a substance that contains as many atoms, molecules, or other particles as there are atoms in exactly 12 g of the carbon - 12 isotope Avogadro's Number (after Amedeo Avogadro) X 1023 Ex.) Mole of iron = __6.02 X 1023___ atoms

Quick Question 1 mole of hydrogen contains how many atoms?
1 mole of CO contains how many molecules?

Where else can you find that mass?

Vocab Review Molar Mass = the mass of one mole of a substance
Can be used to convert between moles and grams in problems The atomic mass of an element in amu’s is equal to its molar mass (g/mol) Ex.) Carbon-12 has an atomic mass of 12 amu. So, Carbons molar mass is 12 grams (12 g/mol) Molar Masses differ with elements. A mole of carbon weighs different than a mole of iron. A dozen rocks weigh different than a dozen feathers

Calculating Molar Mass
You can calculate the molar mass by adding up the atomic masses taking into account how many of each atom there are. Calculate the Molar Mass of carbon dioxide, CO2. 12.01 g + 2(16.00 g) = 44.01 g  One mole of CO2 (6.02 x 1023 molecules) has a mass of grams Now, you do H2O on your marker boards.

Molar Mass H2O 18.02 g 2(1.01 g) g =  One mole of H2O (6.02 x 1023 molecules) has a mass of grams

Marker Board Practice 2(12.01 g) + 6(1.01) + 16.00 g = 46.08 g/mol
What is the Molar Mass of ethanol - C2H5OH? 2(12.01 g) + 6(1.01) g = 46.08 g/mol

What is a mole? 1 mole of particles = the number of atoms in exactly 12g of ______________ 1 mole of atoms has a _____________ the _____________ of the element in grams. 1 mole of molecules has a mass equal to the molecular mass in grams. examples 1 mole H2O is the number of molecules in _____________ g H2O 1 mole H2 is the number of molecules in ___________ g H2. Carbon-12 Mass equal to atomic mass 18.02 2.02 1 mole of particles = __________________ particles ____________________________! 6.02x1023 OF ANY SUBSTANCE

What is a mole? Avogadro’s Number 6.02x1023 molecules atoms ions
__________________________ is the number of molecules in one mole for any substance = ___________________ can be used to convert between particles (______________, __________, ________) and moles in problems So why use moles? many properties depend on the __________________________ in the sample, _______________________ of the sample. Avogadro’s Number 6.02x1023 molecules atoms ions Number of particles not on the mass

So far we have learned that:
1 mol =________________________________ 1 mol =_______________________________ Molar mass (from the periodic table) 6.02 x 1023 particles Each equality can be written as a set of two conversion factors. They are: 1 mole Molar Mass Molar Mass 1 mole 6.022 x 1023 1 mole 1 mole 6.022 x 1023

We can count by weighing
1 jellybean = 2.3 g And we weigh a jar of jellybeans and it weighs 1518 g, we can determine the number of jellybeans using this mass just like we can use the molar mass (the mass of one mole) 1 jellybean 2.3 g 1518 g x = 660 jellybeans We could also convert this to dozen…. 660 jelly beans x 1 dozen 12 jellybeans = 55 dozen jellybeans

So far we have learned that:
1 mol =________________________________ 1 mol =_______________________________ Molar mass (from the periodic table) 6.02 x 1023 particles Each equality can be written as a set of two conversion factors. They are: 1 mole Molar Mass Molar Mass 1 mole 6.022 x 1023 1 mole 1 mole 6.022 x 1023

Coefficient number not exponent 6.02 x 1023
How do I do scientific notation on my calculator??? (YOU MUST HAVE A SCIENTIFIC CALCULATOR!) Coefficient number Punch the __________________________ into your calculator Push the EE or EXP button. Do ______ use the x (times) button!! Enter the ______________ number using the +/- button to change its sign if necessary. not exponent 6.02 x 1023 exponent Coefficient .

Examples 3.01 x 1022 atoms Mg 1 mol Mg 6.02x1023 atoms = mol Mg 0.0500
1. How many moles of magnesium is 3.01 x 1022 atoms of magnesium equal to 3.01 x 1022 atoms Mg 1 mol Mg 6.02x1023 atoms = mol Mg 0.0500

Examples 4.00 moles glucose 6.02x1023 molecules 1 mol
2. How many molecules are there in 4.00 moles of glucose, C6H12O6? 4.00 moles glucose 6.02x1023 molecules 1 mol = molecules glucose 2.41x1024

Your Turn 1 mol 1.20 x 1025 atoms P 6.02x1023 atoms = mol P 19.9
3. How many moles are 1.20 x 1025 atoms of phosphorous equal to? 1 mol 1.20 x 1025 atoms P 6.02x1023 atoms = mol P 19.9

(atoms, molecules, formula units)
g representative particles (atoms, molecules, formula units) mol ratio may be needed L molar mass 1 mol 6.02 x1023 rp 6.02 x 1023 rp 22.4 L *Notice any time you are going to or from grams you need the molar mass

Mole-Mass Conversions
You can use ratios to convert between moles and mass and between moles and number of atoms ___1 mole X___ ___1 mole X___ molar mass of X (g) and X 1023 X atoms If you have 1.34 moles of CO2, how many grams do you have? 1.34 mol CO2 44.01 g CO2 = 59.0 g 1 mol CO2

Next Example: Converting moles to grams
How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li 6.94 g Li = g Li 24.3 1 mol Li

Your turn: Converting grams to moles
How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li 1 mol Li = mol Li 2.62 6.94 g Li

Your Turn 25 g H2O 1 mol H2O = mol H2O 1.39 18.02 g H2O
a. How many moles are in 25 grams of water? 25 g H2O 1 mol H2O = mol H2O 1.39 18.02 g H2O

One Quick Note Atoms – Used to describe a single element
Formula Unit – Used to describe an ionic compound Molecule – Used to describe covalent compound

Marker Board Practice atoms
Please label the following elements or compounds with the correct units for representative particles (atoms, molecules, or formula units) 5.9 x 1017 __________________ Al2O3 1.8 x ______________ Fe 4.5x1024 _______________ H2O2 formula units atoms molecules

Do Now 3/14/17 How many moles of He are in 6.46 g of He?
How many particles is this equal to? 6.46 g He 1 mol = moles 1.62 4.00 g He 1.62 mol He 6.02 x 1023 atoms = atoms 9.75 x 1023 1 mol

One Quick Note Atoms – Used to describe a single element
Formula Unit – Used to describe an ionic compound Molecule – Used to describe covalent compound

Marker Board Practice atoms
Please label the following elements or compounds with the correct units for representative particles (atoms, molecules, or formula units) 5.9 x 1017 __________________ Al2O3 1.8 x ______________ Fe 4.5x1024 _______________ H2O2 formula units atoms molecules

I can… Define a mole as equal to 6.02 x 1023 particles (atoms or molecules). Explain the quantity of one mole is set by defining one mole of carbon 12 atoms to have a mass of exactly 12 grams. Determine the molar mass of a molecule from its chemical formula and a table of atomic masses and convert the mass of a molecular substance to moles, number of particles, or volume of gas at standard temperature and pressure. Any H.W. ?’s

2 Step Problems 11.9 g Na 1 mol 6.02 x 1023 atoms 22.99 g Na 1 mol
i g Na to representative particles 11.9 g Na 1 mol 6.02 x 1023 atoms 22.99 g Na 1 mol = atoms 3.12 x 1023

Another Example What is the molar mass of Fe2O3?
How many grams are in 1.23 x 1024 formula units of Fe2O3? MM – 2(55.85 g) + 3(16.00 g) = 159.7 g/mol 1 mol 1.23  1024 f.u.’s 159.7 g 6.02 x 1023 f.u.’s 1 mol = grams Fe2O3 326.3

Your Turn j. 26.8 g C2H2 to molecules 26.8 g C2H2 1 mol
6.02 x 1023 molecules 26.04 g C2H2 1 mol = molecules 6.20 x 1023

Your Turn r  1021 molecules of C12H22O11 to grams 1 mol g 2.19  1021 molecules 6.02 x 1023 molecules 1 mol = grams C12H22O11 1.25 MM – 12(12.01 g) + 22(1.01 g) + 11(16.00) = g/mol

Standard Molar Volume 1 mole of a gas occupies 22.4 liters of volume
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amedeo Avogadro At STP (Standard Temperature and Pressure): 1 mole of a gas occupies 22.4 liters of volume

The Volume of a Mole of Gas
It’s difficult to put oxygen gas on a scale, so one way we can measure gas is by volume. *Remember in the video Avogadro found that at the same temperature and pressure, equal volumes of different gases contain the same number of molecules or atoms - Avogadro’s Law

Avogadro’s Law At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present. So how can we use this information for conversions? If we figure out how many liters is equal to a mole then…

We’re back to using ratios
representative particles (atoms, molecules, formula units) mol ratio may be needed L molar mass 1 mol 6.02 x1023 rp 6.02 x 1023 rp 22.4 L

Molar Volume of a Gas Standard Temperature & Pressure 273 K 1 atm
Recall that STP stands for _____________ _________________ ____________________and represents the conditions of T = ______________ and P = ________________ At STP, __________________, it has been found that _________ of ______ gas will occupy a ___________ of __________. We can write this as the following conversion factor: & Pressure 273 K 1 atm & only at STP 1 mole any volume 22.4 L Only for a gas at STP, 1 mol = 22.4 L 1 mole L 22.4 L mole or

We can use this to help solve problems such as the following: Example #1 At STP, how many liters will 3.85 moles of ammonia (NH3) gas occupy? 3.85 moles NH3 22.4 L 86.24 = L 1 mol

73.9 L NH3 1 mol 3.30 22.4 L 3.30 moles 6.02 x 1023 molecules 1 mol
Example #2 At STP, how many moles will be in a 73.9 L container of ammonia (NH3) gas? How many molecules of NH3 are in the container described above? 73.9 L NH3 1 mol = moles 3.30 22.4 L 3.30 moles 6.02 x 1023 molecules 1 mol = molecules 1.99x1024

or 3.30 moles 17.04 g NH3 56.2 1 mol 73.9 L NH3 1 mol 17.04 g NH3
Example #2 What is the mass of the NH3 gas in the container described above? 3.30 moles 17.04 g NH3 = g 56.2 1 mol or 73.9 L NH3 1 mol 17.04 g NH3 22.4 L 1 mol = g 56.2

Molar Volume Problems 1.) What volume will 1.27 moles of helium gas occupy at STP? 1.27 moles 22.4 L He = L 28.4 1 mol

Do Now – 3/15/17 Convert 9.3 grams of Ca(OH)2 to its representative units 9.3 g Ca(OH)2 1 mol 6.02 x 1023 formula units 74.1 g 1 mol = formula units Ca(OH)2 7.56 x 1022 MM – g+ 2(16 g) + 2(1.01) = 74.1 g/mol

I can… Define a mole as equal to 6.02 x 1023 particles (atoms or molecules). Explain the quantity of one mole is set by defining one mole of carbon 12 atoms to have a mass of exactly 12 grams. Determine the molar mass of a molecule from its chemical formula and a table of atomic masses and convert the mass of a molecular substance to moles, number of particles, or volume of gas at standard temperature and pressure. Any H.W. ?’s

More Mole Calculations
Substance Mass of One Mole Silver g/mol

Substance # of Moles # of Particles Silver 5.00 3.01x1024 atoms 5.0 moles 6.02 x 1023 atoms = atoms Ag 3.01x1024 1 mol

Substance # of Moles Mass (grams) Silver 5.00 539.4 g Ag 5.0 moles g = g Ag 539.4 1 mol

Relative Mass The calculated relative atomic mass is not the mass of exact atom. It is a ratio of actual mass respect to 1/12th of the mass of carbon-12 atom. the relative atomic mass of an element is now based on the arbitrary value of the carbon-12 isotope being assigned a mass of by international agreement!

Determining Avogadro’s Number
The number of particles in a mole, can be experimentally determined by first "counting" the number of atoms in a smaller space and then scaling up to find the number of particles that would have a mass equal to the atomic or molecular mass in grams. Some real data from which Avogadro's number can be determined: X-Ray diffraction studies show that gold consists of a repeating atomic arrangement where the repeating unit (called a unit cell) is a cube containing 4 gold atoms. Each side of the cube has a length of 4.08x10-8 cm The density of gold is 19.3 g/cm3 and its atomic mass is 197. v= side3= (4.08 x 10-8cm)3= 6.79 x 10-23cm3 197g x 1 cm3 x __4 atoms = x 1023 atoms/mole mol g x 10-23cm3 "...the molecular weight of a substance, expressed in grams, shall henceforth be called mole.”

Do Now - 3/20/17 Write a sentence(s) describing how you would convert volume (liters) to mass (grams) (Be specific about each step) To go from liters to grams you first must convert to moles by dividing by 22.4 L. You then convert to grams by multiplying by the substances molar mass.

Quick Practice A sample of Cl2 gas has a volume of 65.0 L at STP. How many grams is there? 65.0 L Cl2 1 mol 70.9 g = g Cl2 205.7 22.4 L 1 mol

I can… Define a mole as equal to 6.02 x 1023 particles (atoms or molecules). Determine the molar mass of a molecule from its chemical formula and convert the mass of a substance to moles, number of particles, or volume of gas at standard temperature and pressure. Determine the Percent composition of a covalent or ionic compound Any H.W. ?’s

Discussion Question a number or ratio expressed as a fraction of 100
What is the definition of percent? What is the definition of composition? What is Percent Composition? a number or ratio expressed as a fraction of 100 the way in which a whole or mixture is made up. the percent by mass of each element in a compound

Percent Composition, Empirical and Molecular Formulas
Courtesy

Percent Composition The formula of a compound tells us what?
The ratio of atoms in a compound Usefulness of figuring this out experimentally: To compare experimental to actual, which helps up determine purity. How? Most direct and common – Mass Spectrometry

Mass Spectrometry

Percent Composition This is the percent by mass of each element in a compound. How is it found?

Calculating Percentage Composition
Calculate the percentage composition of magnesium carbonate, MgCO3. Formula mass of magnesium carbonate: 24.31 g g + 3(16.00 g) = g 100.00

Percent Composition W.S.
Example 1: Percent composition of potassium chromate (K2CrO4). Molar mass: Subscripts x atomic mass Percent composition K 2(39.10) = g x 100% = Cr = g O 4(16.00) = g Molar mass = g %’s should add up to ~100% 78.20 g 40.27 % g 52.00 g 26.78 % g 64.00 g 32.96 % g %

Percent Composition W.S.
Example 2: Percent composition of magnesium oxide (MgO). Molar mass: Subscripts x atomic mass Percent composition Mg = g x 100% = O = g Molar mass = g 24.31 g 60.31 % 40.31 g 16.00 g 39.69 % 40.31 g % %’s should add up to ~100%

Calculate the percent composition of each of these compounds
Calculate the percent composition of each of these compounds. Name the compound if the name is missing. SHOW AS MUCH WORK, AS THE EXAMPLES ABOVE! Na3PO4 Name: Sodium Phosphate Na – 3(22.99) = g x 100 = 42.07 % g P – = g x 100 = 18.89 % g O – 4(16.00) = g x 100 = 39.04 % g 100% Molar Mass = g

Your Turn 2. (NH4)2C2O4 Name: Ammonium Oxalate N – 2(14.01) = 28.02 g
22.57 % g H – 8(1.01) = g x 100 = 6.51 % g C - 2(12.01) = g x 100 = 19.35 % g O – 4(16.00) = g x 100 = 51.56 % g 100% Molar Mass = g

I can… Define a mole as equal to 6.02 x 1023 particles (atoms or molecules). Determine the molar mass of a molecule from its chemical formula and convert the mass of a substance to moles, number of particles, or volume of gas at standard temperature and pressure. Determine the Percent composition of a covalent or ionic compound

Marker Board Check Calculate the percent composition of (NH4)2S
N – 2( g) = g H – 8(1.01 g) = 8.08 g S – 1( g) = g + g N – g g x 100 = % H – g g x 100 = 11.85% S – g g x 100 = 47.04% 99.99

I can… Define a mole as equal to 6.02 x 1023 particles (atoms or molecules). Determine the molar mass of a molecule from its chemical formula and convert the mass of a substance to moles, number of particles, or volume of gas at standard temperature and pressure. Determine the Percent composition of a covalent or ionic compound

Do Now – 3/26/15 1. Calculate the number of atoms, molecules, or formula units in 26.8 g Ca(OH)2. 2. Now write 3-5 complete sentences describing what you found in question 1 using the correct name of the compound. (Please use the following key terms: formula unit, Avogadro’s number, mole, molar mass)

Answer 26.8 g Ca(OH)2 1 mol 6.02 x 1023 f.u.’s 1. 1 mol
= formula units 2.18x1023 2. I converted 26.8 g of calcium hydroxide to formula units by first converting to moles. I did this by dividing the given mass by the molar mass (74.1 g). I then converted moles of Ca(OH)2 to formula units by multiplying by Avogadro’s number (6.02 x 1023), which is equal to the number of particles in one mole. This gave me an answer of 2.18x1023 formula units of Ca(OH)2.

Do Now – 3/22/17 Calculate percent composition of (NH4)2Cr2O7
N – g g x 100 = % N – 2(14.01 g) = g H – 8(1.01 g) = g Cr – 2(52.00 g) = g O - 7(16.00 g) = g + g H – g g x 100 = 3.20% Cr – g g x 100 = 41.24% O – g g x 100 = 44.42% 99.97%

Any Percent Composition Questions?
I can… Determine the Percent composition of a covalent or ionic compound Determine the Empirical formula being given the percent composition Compare and Contrast empirical and molecular formulas Any Percent Composition Questions?

Empirical Formula What is it? CH2
Empirical – “based only on observation and measurement” Empirical Formula: The formula of a compound that expresses the smallest whole-number ratio of atoms present (simplest formula). (All formulas of ionic compounds are empirical formulas, but are often called formula units.) What is the empirical formula of C2H4? CH2

Empirical Formulas As we saw in the last unit, chemists can often predict the products of reactions, by looking at solubility charts or knowing the solubility rules. Sometimes though, a chemical reaction gives a product that cannot be predicted or has never been produced before. So how do we identify what it is?

Empirical Formula Remember, a chemical formula tells us the ratio of atoms present in a compound. Ex.) In CO2 – there is one carbon atom for every 2 oxygen atoms in each molecule of CO2 Using mass spectrometry, we can find the mass ratio of compounds to help us figure out the ratio of the formula How do we figure out this formula? 4 Steps

Empirical Formula – When given grams
Obtain the mass of each element present (in grams). If given percent of each element base calculation on 100 grams of compound. Determine moles of each element in 100 grams of compound. Turn your mass of each compound to moles Divide the number of moles of each element by the smallest number of moles from step 2, which will convert the smallest number to 1. (If all of the numbers are whole numbers, these are the subscripts for your empirical formula. If not, go to step 4) Multiply the numbers you derived in step 3 by the smallest integer that will convert all of them to whole numbers (This set of whole numbers is your empirical formula subscripts).

Example – Empirical Formula
By analysis, a compound was found to consist of 40.01% carbon: 6.67% hydrogen and 53.2% oxygen. Find the empirical formula of the compound. Step I: Divide the percent (as grams) of each element by atomic mass of this element. The answer (ratio) given by this calculation tells us the moles of one kind of atom as compared to the moles of the other atoms present in the compound.

By analysis, a compound was found to consist of 40.01% carbon: 6.67% hydrogen and 53.2% oxygen. Find the empirical formula of the compound. Step II: Note that the ratio is not always a whole number. Atoms do not come in fractional parts!! Therefore we must divide each of the ratios by the smallest in order to change all the ratios to small whole numbers. 3.33 mol C =1.02 3.325 6.60 mol H =2.03 3.325 3.325 mol O =1.0 3.325

By analysis, a compound was found to consist of 40.01% carbon: 6.67% hydrogen and 53.2% oxygen. Find the empirical formula of the compound. Step III. Sometimes this final ratio still is not a whole number – if this happens, pick out a small number of the ratios by this factor you will arrive at a series of small whole numbers. 3.33 mol C =1.02 3.25 CH2O 6.60 mol H =2.03 3.25 3.25 mol O =1.0 3.25

g Na, 39.7g Cr, 42.7g O Na2Cr2O7 17.6 g Na 1 mol = mol Na =1.00 x 2 22.99 g 0.763 = 2 39.7 g Cr 1 mol = mol Cr =1.00 x 2 52.00 g 0.763 = 2 42.7 g 0 1 mol = mol O =3.50 x 2 16.00 g 0.763 = 7

Scratch Paper Practice
A 10 gram sample of a compound contains 7.22 grams of magnesium and 2.78 grams of nitrogen. What is its empirical formula? 7.22 g Mg 1 mol = mol Mg =1.5 x 2 24.31 g 0.198 = 3 2.78 g N 1 mol = mol N =1.00 x 2 14.01 g 0.198 = 2 Mg3N2

Your Turn - Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? 1)Treat % as mass, and convert grams to moles

Empirical Formula Determination
2. Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

Empirical Formula Determination
3. Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 Empirical formula: C3H5O2

Molecular Formula What is it?
Molecular Formula: A chemical formula indicating the elements of a molecular compound and showing the number of atoms of each element comprising the compound. (Always covalently bonded). CH vs C6H vs C18H72

Formulas Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11 How do you get it?

Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = g

The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = g 2. Divide the molecular mass by the mass given by the emipirical formula.

The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? (C3H5O2) x 2 = C6H10O4 3. Multiply the empirical formula by this number to get the molecular formula.

EXAMPLE:. A compound was found to consist of 30. 51% Carbon, 1
EXAMPLE: A compound was found to consist of 30.51% Carbon, 1.69% Hydrogen, and 67.80% Bromine. The molecular weight is Find the molecular formula of the compound. STEP I: Find the empirical formula

C3H2Br STEP I: Find the empirical formula 2.54 mol C =2.99 0.849
1.67 mol H =1.97 0.849 0.849 mol Br =1.0 0.849

EXAMPLE: A compound was found to consist of 30. 51% Carbon, 1
EXAMPLE: A compound was found to consist of 30.51% Carbon, 1.69% Hydrogen, and 67.80% Bromine. The molecular weight is Find the molecular formula of the compound. STEP II: Find the empirical formula mass and find the molecular formula. C3H2Br C – 3(12.01 g) = g H – 2(1.01 g) = g Br – g = g + g (C3H2Br) x 2 = C6H4Br2

Marker Board Check NO2 =1.00 =2.00
The molecular mass of a compound is The analysis of the compound shows that there are grams of nitrogen and grams of oxygen. What is the empirical formula of the compound? What is the molecular formula of this compound? =1.00 0.0434 =2.00 0.0434 NO2

Marker Board Check NO2 (NO2) x 2 = N2O4
The molecular mass of a compound is The analysis of the compound shows that there are grams of nitrogen and grams of oxygen. What is the empirical formula of the compound? What is the molecular formula of this compound? NO2 N g = g O – 2(16.00 g) = g + 46.01 g (NO2) x 2 = N2O4

Ticket Out the Door C9OH17 76.5 g C 1 mol = 6.37 mol C =9.02 12.01 g
A fat is composed, in part, of long chains of carbon and hydrogen atoms. What is the empirical formula of a fat containing 76.5 % C, 11.3 % O and 12.1 % H? C9OH17 76.5 g C 1 mol = mol C =9.02 12.01 g 0.706 11.3 g O 1 mol = mol O =1.00 16.00 g 0.706 12.1 g H 1 mol = mol H =16.97 1.01 g 0.706

I can… Determine the Percent composition of a covalent or ionic compound Determine the Empirical formula being given the percent composition Compare and Contrast empirical and molecular formulas

Do Now – 4/3/17 CH2Cl C2H4Cl2 C - 12.01 H - (2)1.01 Cl - 35.45
A compound used as an additive for gasoline to help prevent engine knock has a percent composition of 71.65% Cl, 24.27% C, 4.07% H. The molar mass is known to be g. Determine the empirical formula and molecular formula for this compound CH2Cl 71.65 g Cl 1 mol C H - (2)1.01 Cl 49.48 g/mol = 2.02 mol Cl =1.00 35.45 g 2.02 24.27 g C 1 mol = 2.02 mol C =1.00 12.01 g 2.02 4.07 g H 1 mol = 4.03 mol H =2.00 1.01 g 2.02 C2H4Cl2

Ticket Out the Door C9OH17 76.5 g C 1 mol = 6.37 mol C =9.02 12.01 g
A fat is composed, in part, of long chains of carbon and hydrogen atoms. What is the empirical formula of a fat containing 76.5 % C, 11.3 % O and 12.1 % H? C9OH17 76.5 g C 1 mol = mol C =9.02 12.01 g 0.706 11.3 g O 1 mol = mol O =1.00 16.00 g 0.706 12.1 g H 1 mol = mol H =16.97 1.01 g 0.706

Importance of Decimals
6.37 mol C 6 mol C =8.6 =9.02 0.7 0.706 0.7 mol O 0.706 mol O =1.00 =1.0 0.7 0.706 11.98 mol H 12 mol H =16.97 =17.1 0.706 0.7 C?OH17 C9OH17

I can… Determine the Percent composition of a covalent or ionic compound Determine the Empirical formula being given the percent composition Compare and Contrast empirical and molecular formulas

Do Now 4/4/17 C4H5ON2 C8H10O2N4 C - (4)12.01 H - (5)1.01 O - 16.00
Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is g/mol. What is its molecular formula? C4H5ON2 49.48 g C 1 mol = 4.12 mol C = 4 C - (4)12.01 H - (5)1.01 O N - (2)14.01 97.11 g/mol 12.01 g 1.03 5.19 g H 1 mol = mol H = 4.99 1.01 g 1.03 16.48 g O 1 mol = 1.03 mol O = 1 16.00 g 1.03 28.85 g N 1 mol = mol N C8H10O2N4 = 2 14.01 g 1.03

Quick Refresher What is the Percent Composition of Cu2O?
Cu – 2(63.55 g) = g O – g = g g Cu – g g x 100 = % O – g g x 100 = 11.18% 100%

Quick Refresher 2. (NH4)2C2O4 Name: Ammonium Oxalate
N – 2(14.01) = g x 100 = 22.57 % g H – 8(1.01) = g x 100 = 6.51 % g C - 2(12.01) = g x 100 = 19.35 % g O – 4(16.00) = g x 100 = 51.56 % g 100% Molar Mass = g

Quick Refresher A 10 gram sample of a compound contains 7.22 grams of magnesium and 2.78 grams of nitrogen. What is its empirical formula? 7.22 g Mg 1 mol = mol Mg =1.5 x 2 24.31 g 0.198 = 3 2.78 g N 1 mol = mol N =1.00 x 2 14.01 g 0.198 = 2 Mg3N2

Quick Write – Write a quick summary describing similarities and differences between Empirical and Molecular Formulas (remember use Key Terms – Empirical Formula and Molecular Formula) and how they are used together in science. The Empirical Formula is a formula of a compound that expresses the smallest whole-number ratio of atoms present (simplest formula). The Molecular Formula indicates the elements of a covalent compound and shows the actual number of atoms of each element in the compound (Always covalently bonded and some multiple of the empirical formula).

I can… Determine the Percent composition of a covalent or ionic compound Determine the Empirical formula being given the percent composition Compare and Contrast empirical and molecular formulas Organize and write a proper conclusion Any H.W./Lab ?’s

Writing a Conclusion Any Scientific Lab Conclusion, you must…
1. Restate the overall purpose of the experiment 2. What were the major findings? (Summarize your data and/or graph results) 3. Was the hypothesis supported by the data? 4. How could this experiment be improved/sources of error? 5. What could be studied next after this experiment? What new experiment could continue study of this topic?

Example Purpose The purpose of this experiment was to experimentally verify the empirical formula of magnesium chloride. Major findings/was hypothesis supported? I found that my data disagreed with the known formula which is, MgCl2. When determining the mole ratio of Magnesium to Chlorine using my experimental data, I determined a whole number ratio 1:4, which disagrees with the 1:2 ratio expected.

Example Sources of Error The disagreement in my experimental data to the known values shows room for improvement. Sources of error could be from the fact that we could not heat to a constant weight due to the lack of time. This could have caused the mass of Chlorine to be heavier than it actually was. Also, our balances only allowed 2 decimal places, which affected the precision our mole ratios could be. Further Study/Ways to improve To improve this experiment, more precise balances could be used. Also, additional investigations using various compounds would be a good additional experiment to increase our lab skills collecting quantitative data.

Example The purpose of this experiment was to experimentally verify the empirical formula of magnesium chloride. I found that my data disagreed with the known formula which is, MgCl2. When determining the mole ratio of Magnesium to Chlorine using my experimental data, I determined a whole number ratio of 1:4, which disagrees with the 1:2 ratio expected. The disagreement in my experimental data to the known values shows room for improvement. Sources of error could be from the fact that we could not heat to a constant weight due to the lack of time. This could have caused the mass of Chlorine to be heavier than it actually was. Also, our balances only allowed 2 decimal places, which affected the precision our mole ratios could be. To improve this experiment, more precise balances could be used. Also, additional investigations using various compounds would be a good additional experiment to increase our lab skills collecting quantitative data.

I can… Determine the Percent composition of a covalent or ionic compound Determine the Empirical formula being given the percent composition Compare and Contrast empirical and molecular formulas Organize and write a proper conclusion

Do Now – 4/11/16 Convert 12.7 g of CH4 to liters
Convert 7.3 g H2O to representative particles 1 mol 22.4 L 12.7 g CH4 = L 17.7 16.05 g 1 mol 7.3 g H2O 1 mol 6.02x1023 molecules 18.02 g 1 mol = molecules 2.44x1023

More Practice By analysis a compound was found to contain grams of copper and 2.31 g of oxygen. Calculate the empirical formula of the compound. Cu2O 18.32 g Cu 1 mol = mol Cu =2.00 63.55 g 0.144 2.31 g O 1 mol = mol O =1.00 16.00 g 0.144

More Practice What is a mole, in chemistry?
The number of atoms in exactly 12g of Carbon 12 (6.02 x 1023)

Key Terms Law of Conservation of Matter Avogadro’s Law
Empirical Formula Chemical Equation Molecular Formula Mole Percent Error Mole Ratio Percent Composition Relative Mass Know how to convert between moles, grams, liters, and representative particles Balancing Chemical Reaction Reactant Product Coefficient Avogadro’s Number

Any Review/Lab Questions?
I can… Any Review/Lab Questions? Define a mole as equal to 6.02x1023 particles (atoms, molecules, or formula units). Explain the quantity of one mole is set by defining one mole of carbon 12 atoms to have a mass of exactly 12 grams. Determine the molar mass of a molecule from its chemical formula and a table of atomic masses, and convert the mass of a molecular substance to moles, number of particles, or volume of gas at standard temperature and pressure. Determine the percent composition of a covalent or ionic compound Determine the Empirical formula being given the percent composition Compare and Contrast empirical and molecular formulas and be able to determine each

Test Time!!! Please turn in: 1. Review 2. Do Nows 3. Notes

Do Now – 4/13/15 Describe what information you need and how you would solve an unknowns Molecular Formula. Now take a minute and explain it to your neighbor

How do you get molecular formulas?
To obtain the molecular formula, we must know the approximate molar mass and the percent composition. Then, we must compare the empirical formula mass to the molar mass of the molecular formula. *The molecular formula contains a whole number of empirical formula units

Quick Refresher 1. What is the molar mass of AlCl3?
2. Convert 7.90 X 1015 formula units of AlCl3 to grams.

Marker Board Answer Molar Mass – Al – g Cl – ( g) g/mol 1 mol AlCl3 6.02 X 1023 f.u. AlCl3 g AlCl3 1 mol AlCl3 7.90 X 1015 f.u. AlCl3 =1.75 X 10-6 g of AlCl3

Quick Question If I start with 5.2 g of Cu before it becomes oxidized and I end with 10.8 g of CuO, how much Cu and how much O is in the compound in grams? 10.8 – 5.2 = 5.6 g O 5.2 g Cu

Quick Question Why is the empirical formula important?

Percent Error Percent Error: Measured percentage of how far off we are on our measurements. *Gives us an idea of how big our error is compared to the size of what we are measuring (the closer to zero the better) Error

Example A compound was found to consist of 30.51% Carbon, 1.69% Hydrogen, and 67.80% Bromine. The molecular weight is Find the molecular formula of the compound.

Molecular Formula The molecular formula is always some multiple of the empirical formula Molecular formula = n x empirical formula Then, Molar mass = n x empirical formula mass So, n is the number of empirical formula units that make up the molecular formula. So, you multiply all subscripts in the empirical formula by n to get the molecular formula n =

Do Now – 2/19/13 Explain a mole and what it represents in chemistry.
Explain what molar mass (relative mass) is and how to get it for an element and a compound.

Example #1

Example #2 – (#1 on W.S.) At STP, how many liters will 3.85 moles of ammonia (NH3) gas occupy? Your Turn - #2

Chemical Calculations
You can also use ratios to solve the mass or number of moles of a reactant or product by using a balanced chemical equation of the reactants and products. 2 H2 + 1 O H2O “2 molecules of hydrogen react with one molecule of oxygen and form two molecules of water”

Converting Mass to Moles
Balance equation Determine how many moles of water you are trying to make. Question: How much oxygen must be supplied to produce 144 g H2O? Ex.) 144 g H2O X 1 mole H2O =_______ moles H2O 18.0 g H2O Equation 2 H O H2O Amount mol Molar Mass g/mol 18.0 g/mol Mass (moles x molar mass) g 144 g

Using Mole Ratios Using Mole ratios we can convert moles of O2 to H2O.
1 mole of O2 2 mole H2O 8.00 mol H2O X 1 mole O2 = ________ mol O2 2 mole H2O The last step is to covert moles to mass

Converting Moles to Mass
The last step is to covert moles to mass _________ moles O2 x g O2 = ________ g O2 1 mol O2 So, in order to produce 144 grams of H2O, you must supply _______ g of O2.

Percent Yield Percent Yield: Ratio of actual yield to theoretical yield Actual Yield: Amount of product produced in the experiment Theoretical Yield: Max amount of product produced through calculations (The most you should have)

Quick Review So if we get the mass of a solid compound, how can we figure out how many moles of the compound there are? Ex. 2.7 grams of NaCl Then how can use this to figure out how many formula units there are? But what if we are dealing with a gas? It’s hard to put a gas on a scale. How can we calculate moles or number of particles?

1. Calculate the moles in:
Marker Board Practice 1. Calculate the moles in: 4.22  1018 molecules of C8H18

Answer Part 1 X 1018 molecules C8H18 X mole C8H = X 10-6 moles C8H18 6.02 X 1023 molecules C8H18 * Remember written in scientific notation is 7.01 X 10-6

Marker Board – Part 2 2. Now convert to grams.

Answer to Do Now 2. To convert from moles to grams you need the molar mass Molar Mass of C8H18: 8(12.01) + 18(1.01) = g/mol C8H18 Then, 7.01 X 10-6 moles C8H18 X g C8H18 = 8.01 X 10-4 g C8H18 1 mole C8H18

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