How you measure how much? You can measure mass, or volume, or you can count pieces. We measure mass in grams. We measure volume in liters. We count pieces.

How you measure how much?

You can measure mass,

or volume,

or you can count pieces.

We measure mass in grams.

We measure volume in liters.

We count pieces in MOLES.

For modern systems; 12C is taken as a standard. So; 12C is assumed to be 12 a.m.u.

1 12C = 12 a.m.u. 1 a.m.u.=1/12 of the mass of a single 12C atom =1/12 (12/6.02 X 1023) =1/6.02 X 1023 grams

=mass of 1 1H atomThen by comparing the masses of each atom with the mass of 12C atomic masses of other atoms can be easily calculated “RELATIVE ATOMIC MASSES”. Relative atomic masses easily calculated by MASS SPECTROMETER.1 16O = 16 a.m.u1 40Ca = 40a.m.u.1 1H = 1 a.m.u.

Moles

Defined as the number of carbon atoms in exactly 12 grams of carbon-12.

1 mole contains 6.02 x 1023 particles.

Treat it like a very large dozen

6.02 x 1023 is called Avogadro’s number.

Representative particles

The smallest pieces of a substance.

For a molecular compound it is a molecule.

For an ionic compound it is a formula unit.

For an element it is an atom.

Units of measurement used with equations.We use a mole concept to bring together the concepts of counting numbers and atomic weights of elements.The mole is derived from the following information.Atomic weights are an average of the relative masses of all of the isotopes of the given element.The number of C-12 atoms in exactly 12.00 g of C-12 is 6.02 X 1023.This called Avogadro’s number.

An amount of a substance that contains Avogadro’s number of atoms, ions, molecules, or any other chemical unit is called A MOLE.

A mole of C-12 atoms is defined as having a mass of exactly 12.00 g, a mass that is equal to its atomic weight.

Mole Calculations1 mole (mol):

Amount of matter that contains as many objects (atoms, molecules) as the number of atoms in

isotopically pure 12C.6.022 x 1023

Also known as Avogadro’s number.1 mol 12C atoms = 6.02 x 1023 12C atoms

1 mol H2O molecules = 6.02 x 1023 molecules

1 mole NO3- ions = 6.02 x 1023 NO3

- ions.

The mole concept for (A) elements, (B) compounds, and (C) molecular substances.

A mole contains 6.02 X 1023 particles.

Since every mole contains the same number of particles, the ratio of the mass of any two moles is the same as the ratio of the masses of individual particles making up the two moles.

Molar Mass

The atomic mass of any substance expressed in

grams corresponds to 1 mol of the substance.

Atomic mass of a substance expressed in grams is the molar mass.

The molar mass of a diatomic substance is equal to twice its atomic mass.

Molar Mass

The generic term for the mass of one mole.

The same as gram molecular mass, gram formula mass, and gram atomic mass.

Gram Atomic Mass

The mass of 1 mole of an element in grams.

12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and

55.85 grams of iron.

We can right this as 12.01 g C = 1 mole

We can count things by weighing them.

Mole Calculations

Calculate the number of sodium atoms in 0.120 mol Na?

0.120 mol Na x 6.02 x 1023 atoms Na = 7.22 x 1022 1 mol Na

Mole Calculations

Calculate the number of moles of potassium in 1.25 x 1021 atoms of K.

1.25 x 1021 atoms K x 1 mol K = 2.08 x 10-3 mol K

6.02 x 1023 atoms K

Mole Calculations

What is the mass in grams of 2.01 x 1022 atoms of sulfur?

2.01 x 1022 atoms S x 1 mol S x 32.07 g S

6.02 x 1023 atoms S 1 mol S

= 1.07g S

Mole Calculations

How many O2 molecules are present in 0.470g of oxygen gas?

0.470 g O2 x 1mol O2 x 6.02 x 1023 molecules O2

32.00 g O2 1 mol O2

= 8.84 x 1021 molecules O2

What about compounds?

In 1 mole of H2O molecules there are two

moles of H atoms and 1 mole of O atoms

To find the mass of one mole of a compound determine the moles of the elements they have

Find out how much they would weigh add them up.

Molar Mass

The Mole

Calculate the number of Magnesium and Chlorine ions present in 0.450 mol of MgCl2.

# Mg2+ ions: 0.450 mol MgCl2 x 6.02 x 1023 formula units x 1 Mg2+

1 mole 1 formula unit

# Mg2+ ions = 2.71 x 1023

# Cl- ions:0.450 mol MgCl2 x 6.02 x 1023 formula units x 2 Cl-

1 mole 1 formula unit # Cl- ions = 5.42 x 1023

The mole concept applied to compounds

The formula weight of a species is the sum of

atomic masses (amu) of the atoms in a species.

Molecular weight of NH3 =

For an ionic cmpd formula weight MgF2 =

Mass of one mole of MgF2 is

Mass of one formula unit of MgF2 is

Mass of 6.022 x 1023 formula units of MgF2 is

Gram Formula MassThe mass of one mole of an ionic compound.

Calculated the same way.

What is the GFM of Fe2O3?

2 moles of Fe x 55.85 g = 111.70 g

3 moles of O x 16.00 g = 48.00 g

The GFM = 111.70 g + 48.00 g = 159.70g

Molar Mass

What about compounds?What is the mass of one mole of CH4?

1 mole of C = 12.01 g

4 mole of H x 1.01 g = 4.04g

1 mole CH4 = 12.01 + 4.04 = 16.05g

The Gram Molecular mass of CH4 is 16.05g

The mass of one mole of a molecular compound.

The gram atomic weight of an element is the mass in grams of one mole of an element that is numerically equal to its atomic weight.

The gram formula weight of a compound is the mass in grams of one mole of the compound that is numerically equal to its formula weight.

The gram formula weight of a compound is the sum total of all the individual atomic weight in the formula.The gram molecular weight is the gram formula weight of a molecular compound.

Molar Mass

Calculate the molar mass of Ag and of magnesium nitrate, Mg(NO3)2.

Ag is an element so its molar mass equals its atomic mass = 107.87g/mol.

Mg(NO3)2 = 24.31 + 2(14.01 + 3 x 16.00)

= 148.33 g/mol

For example

How many moles is 5.69 g of NaOH?

5 69. g 1 mole

40.00 = 0.142 mol NaOH

g

need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g

1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

Mole Calculations

Calculate the mass in grams of a single molecule of carbon dioxide, CO2.

44.01 g CO2 x 1mol CO2 = 7.31 x 10-23 g/molecule

1 mol CO2 6.02 x 1023 molecules

Molar Mass

The number of grams of 1 mole of atoms,ions,or molecules.

We can make conversion factors from these.

To change grams of a compound to moles of a compound.

Examples

How many moles is 4.56 g of CO2 ?

How many grams is 9.87 moles of H2O?

How many molecules in 6.8 g of CH4?

49 molecules of C6H12O6 weighs how much?

Calc the molar mass of Ca(NO3)2.

Calc the molar mass of a compound if 0.372 mol of it has a mass of 152g.

How many grams of each are required to have 0.100 mol ofA. NaOH

B. H2SO4

C. C2H5OH

D. Ca3(PO4)2

How many moles are in 50.0 g of

A. CS2

B. Al2(CO3)3

C. Sr(OH)2

D. LiNO3

Calc the no. of C, H, and O atoms in 1.50 g of glucose (C6H12O6).

What is the average mass of one C3H8 molecule?

What is the mass of 5.00 x 1024 molecules of NH3?

Types of questionsHow many molecules of CO2 are the in 4.56

moles of CO2 ?

How many moles of water is 5.87 x 1022 molecules?

How many atoms of carbon are there in 1.23

moles of C6H12O6 ?

How many moles is 7.78 x 1024 formula units

of MgCl2?

GasesMany of the chemicals we deal with are gases.

They are difficult to weigh.

Need to know how many moles of gas we have.

Two things effect the volume of a gas

Temperature and pressure

Compare at the same temp. and pressure.

Standard Temperature and Pressure

0ºC and 1 atm pressureabbreviated STP

At STP 1 mole of gas occupies 22.4 LCalled the molar volume

Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

Molar Volume

Avogadro's Theory: Two gases containing equal numbers of molecules occupy equal volumes under similar conditions.

Standard Temperature and Pressure: 0OC and 1 atm.

Molar Volume at STP = 22.4 L

Examples

What is the volume of 4.59 mole of CO2 gas

at STP?

How many moles is 5.67 L of O2 at STP?

What is the volume of 8.8g of CH4 gas at

STP?

Molar Volume

Density of gas = molar mass in grams

molar volume in liters

Calculate the density of ammonia gas, NH3, at STP.

Density of NH3 = molar mass NH3 = 17.04g

molar volume NH3 22.4 L

= 0.761 g/L

Density of a gas

D = m /Vfor a gas the units will be g / L

• We can determine the density of any gas at STP if we know its formula.

• To find the density we need the mass and the volume.If you assume you have 1 mole then the mass is

the molar mass (PT)At STP the volume is 22.4 L.

Examples

Find the density of CO2 at STP.

Find the density of CH4 at STP.

Moles

MassVolume

Representative Particles

6.02 x 1023

PT

Atoms Ions

22.4 L

Mole Calculations

What is the mass of 3.36 L of ozone gas, O3, at STP?

3.36 L O3 x 1 mol O3 x 48.00g O3 = 7.20 g O3

22.4 L O3 1 mol O3

Mole Calculations

How many molecules of hydrogen gas, H2, occupy 0.500 L at STP?

0.500 L H2 x 1 mol x 6.02 x 1023 molecules H2

22.4 L 1 mol

= 1.34 x 1022 molecules H2

Mole Calculations

Percent Composition

Percentage Composition: Percentage by mass contributed by each element in a substance.

% Element (E) = (# atoms of Element)(AW of E) x 100%

Formula Weight of Compound

According to the law of definite composition the elements in a compound are always present in

the same proportion by mass.

Percentage composition by mass of a compound

It shows us how many grams of each element exist in 100 gram of a compound.

Percent Composition

Like all percents

Part x 100 %

whole

Find the mass of each component,

divide by the total mass.

Percent CompositionCalculate the percentage of Nitrogen, by mass in

Ca(NO3)2:

Formula Weight = 1 x Ca + 2 x N + 6 x O = 1 x 40.1 + 2 x 14.0 + 6 x 16.0

= 164.1 amu

% N = (2)(14.0) = 17.1% 164.1

Percent CompositionCalculate the percent composition of

trinitrotoluene (TNT), C7H5(NO2)3.

Assume that there is 1 mol of TNT.

1 mol of TNT contains:

7 mol carbon atoms = 84.07 g

5 mol hydrogen atoms = 5.05 g

3 mol of nitrogen atoms = 42.03 g

6 mol oxygen atoms = 96.00 g

Percent Composition

Calculate the molar mass of TNT:

Molar mass = 227.15 g/mol.

Compare the mass of each element to the molar mass of the compound:

84.07 g C x 100 = 37.01% C

227.15 g C7H5(NO2)3

Percent Composition

5.05g H x 100 = 2.22% H

227.15 g C7H5(NO2)3

42.03 g N x 100 = 18.50% N

227.15 g C7H5(NO2)3

96.00 g 0 x 100 = 42.26% O

227.15 g C7H5(NO2)3

Empirical Formula

Empirical formula gives:the ratio of atoms in a molecule: H2O.

the ratio of moles of atoms on the molar level.

The molar ratio of the elements in a compound allows the calculation of the subscripts in the empirical formula of a compound.

Empirical Formula

Percent composition - the mass percent of each element present in a compound.

1. Calculate the chemical formula from percent composition.2.Find the relative numbers of moles of each element in the compound.3.Use molar masses of the elements as conversion factors.

Empirical Formula

Find the ratio of the numbers of moles by dividing the larger number of moles by the smaller number.

Multiply the subscripts by small integers in a trial-and-error procedure until whole numbers are found. 

Empirical Formula• The flow diagram

summarizes the• relationship among

mass percent, moles, mole ratios and subscripts in the formula of a compound used in the determination of empirical formula from percent composition.

Calculating EmpiricalJust find the lowest whole number ratio

C6H12O6

CH2O

It is not just the ratio of atoms, it is also the ratio of moles of atoms.

In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.

In one molecule of CO2 there is 1 atom of C and 2 atoms of O.

Calculating Empirical

Means we can get ratio from percent composition.

Assume you have a 100 g.

The percentages become grams.

Can turn grams to moles.

Find lowest whole number ratio by dividing by the smallest.

ExampleCalculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.

Assume 100 g so38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 1.01 gH45.11 g N x 1mol N = 3.219 mole N 14.01 gN

ExampleThe ratio is 3.220 mol C = 1 mol C

3.219 mol N 1 mol NThe ratio is 16.09 mol H = 5 mol H

3.219 mol N 1 mol N

C1H5N1

• A compound is 43.64 % P and 56.36 % O. What is the empirical formula?• Caffeine is 49.48% C, 5.15% H, 28.87% N and

16.49% O. What is its empirical formula?

Empirical Formula

Q. The percent composition of a solid is known to be 68.4% Ba, 10.3% P, and 21.3% O. What is the empirical formula of the compound?

A. Assuming a 100 g sample gives us 68.4 g Ba, 10.3 g P, and 21.3 g O.

Convert these masses to numbers of moles: 0.498 moles Ba, 0.332 mol P, 1.33 mol O

Empirical Formula

Knowing the relative numbers of moles, find the ratio by dividing the two larger numbers by the smaller number. Answer = 1.5

The O:Ba:P ratio of 4:1.5:1 gives an empirical formula of Ba1.5PO4. However, we must multiply

the subscripts by a small integer (in this case 2) to find whole numbers for the formula. Ba3P2O8, or

Ba3(PO4)2.

Empirical Formula

Q. What is the percent composition of sodium hydrogen carbonate?

A. The formula for sodium hydrogen carbonate is NaHCO3. The Na:H:C:O mole ratio is 1:1:1:3.

Convert this mole ratio into a mass ratio by assuming there is a 1 mole sample present. Answer = 23g Na, 1.0g H, 12g C, 48 g O

Empirical Formula

To determine the percent composition, divide the mass of each element present by the total mass of the compound and multiply by 100.

Total mass of 1 mole of NaHCO3 = 84 g

Answer = 27% Na, 1.2% H, 14% C, 57% O

Empirical Formula

0.500 g of scandium was heated and allowed to react with oxygen. The resulting product oxide had a mass of 0.767g. What is the empirical formula for scandium oxide?Calculate the number of moles of scandium: 0.500 g Sc x 1 mol Sc = 0.0111 mol Sc 44.96 g Sc

Empirical Formula

Calculate the moles of oxygen:

Mass of oxygen = 0.767g– 0.500g O = 0.267 g O

Number of moles of oxygen:

0.267g x 1 mol 0 = 0.0167 mol O

16.00 g

Mole ratio of Sc:O is 0.0111:0.0167

Divide by smallest number: 1:1.50

Empirical Formula

Sc1.00O1.50

Using the law of multiple proportions we need a whole number ratio:

Sc2.00O3.00

Empirical FormulaCalculate the empirical formula of glycine given that it contains 32.0% carbon, 6.7% hydrogen, 18.7% nitrogen and 42.6% oxygen.

Assume a 100g sample. Then the percent composition equals number of grams.

Empirical Formula

32.0 g C, 6.7g H, 18.7g N, 42.6g O.

Convert grams into moles:

32.0 g C x 1 mol = 2.66 mol C

12.01 g C

6.7 g H x 1 mol H = 6.6 mol H

1.01 g H

Empirical Formula

18.7 g N x 1 mol N = 1.33 mol

14.01 g N

42.6 g O x 1 mol O = 2.66 mol O

16.00 g O

Divide the mole ratios by the smallest number: 1.33

C2H5N1O2

The Empirical Formula

The lowest whole number ratio of elements in a compound.

The molecular formula the actual ration of elements in a compound.

The two can be the same.

CH2 empirical formula

C2H4 molecular formula

C3H6 molecular formula

H2O both

Molecular Formula

The formula obtained from percentage compositions is always the empirical formula.

Molecular formula - gives the actual numbers of atoms in a molecule.

1. May be the same as the empirical formula.2. May be a multiple of the empirical formula.

a. Multiple = Molecular mass Empirical Formula

Empirical to molecular

Since the empirical formula is the lowest ratio the actual molecule would weigh more.

By a whole number multiple.

Divide the actual molar mass by the the mass of one mole of the empirical formula.

Molecular Formula

The empirical formula for fructose is CH2O. If the molar mass of fructose is 180 g/mol, what is the molecular formula?

Molar mass of the empirical formula = 30 g/mol.

Molecular Formula

Multiple = molar mass of fructose

molar mass of empirical formula

Multiple = 180 g/mol = 6

30 g/mol

Multiply the empirical formula by 6:

C6H12O6

Molecular FormulaQ. Determine the molecular formula of ethyl butyrate. The molecular mass of the compound is 116 g/mol and

the empirical formula is C3H6O.

A. The molecular formula of the compound may be the same as the empirical formula or it may be a multiple of

the empirical formula.For ethyl butyrate: Multiple = 116 =2

58 The subscripts in the empirical formula are multiplied by 2 giving a molecular formula of C6H12O2.

Some problems

How many atoms are there in 5.10 moles of sulfur? What’s the mass of 5.10 moles of S?

How many moles of calcium atoms are in 1.16 x 1024 atoms of Ca? How many grams?

Which of the following has more atoms: 1.10g of hydrogen atoms or 14.7 g of

chromium atoms?

How many moles are in 0.040 kg Na?

Example

A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

STRUCTURAL FORMULAThe atoms in a molecule are connected or chemically bonded in a precise way. A SF. Shows how the atoms in a molecule are arranged.For ex: H2O H-O-H

C2H6 CH3 H-C- C- H

H H

Empirical formula•The simplest whole number ratio of atoms of elements in a compound, described with the use of subscripts.•Ionic compounds are always shown as empirical formulas.

Molecular FormulaThe actual numbers of atoms in a

molecule.

Structural FormulaShow the relative arrangements of atoms in a molecule

•If you know the name of an ingredient, you can write a chemical formula, and the percent composition of a particular substance can be calculated from the formula. This can be useful information for consumer decisions.

HYDRATESSolids which are found in combined form with water in definite proportion are called as HYDRATES. When hydrates are heated, H2O evaporates, and only solid is obtained in amorphous .(w/o a certain geometric structure, generally in powdered form. H2O molecules surround ionic substances with certain amounts.

WATER OF HYDRATION : Water molecules of a hydrate.

Na2CO3.10H2O Na2CO3(s) + 10H2O(g)

DEHYDRATION: Evaporation of water of hydration.

Na2CO3.10H2O,

CaSO4.2H2O,

CuSO4.5H2O