molar mass, empirical & molecular formulas

Molar Mass, Molar Mass, Empirical & Empirical & Molecular Molecular FormulasFormulas

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Molecular Mass & Formula Molecular Mass & Formula MassMass

•A A chemical formulachemical formula is a is a convenient way of showing convenient way of showing how many and what type of how many and what type of atoms make up a particular atoms make up a particular molecule of formula unitmolecule of formula unit


•While the mass of an atom While the mass of an atom can be found by looking at can be found by looking at the periodic table, the the periodic table, the mass of a molecule or mass of a molecule or formula unit must be formula unit must be calculated.calculated.


• The The molecular massmolecular mass of a of a substance can be found by substance can be found by finding the sum of the finding the sum of the atomic masses that make up atomic masses that make up a particular moleculea particular molecule

•CHCH22ClCl2 2 = 12 + 2(1) + 2(35.45)= 12 + 2(1) + 2(35.45)

= 84.9 amu= 84.9 amu


• The The formula massformula mass, which is , which is used for ionic compounds, is used for ionic compounds, is found in the same fashion as found in the same fashion as molecular compounds.molecular compounds.

•NaCl = 22.99 + 35.45 NaCl = 22.99 + 35.45 = 58.44 amu= 58.44 amu

Calculating Formula MassCalculating Formula MassCalculate the formula mass of magnesium Calculate the formula mass of magnesium carbonate, MgCOcarbonate, MgCO33..

24.31 g + 12.01 g + 3(16.00 g) 24.31 g + 12.01 g + 3(16.00 g) ==

84.32 g84.32 g

The Avogadro ConstantThe Avogadro Constant

•Atoms are much too small to Atoms are much too small to count individuallycount individually

• The mass of a single molecule The mass of a single molecule is so small that it is impossible is so small that it is impossible to measure by ordinary means to measure by ordinary means in the laboratory.in the laboratory.


• For this reason, scientists For this reason, scientists had to come up with a way had to come up with a way to convert atomic mass to convert atomic mass units to something that units to something that could be used in the could be used in the laboratory, gramslaboratory, grams


• Chemists have Chemists have found that 6.02 found that 6.02 x 10x 102323 atoms of atoms of an element has an element has a mass in gram a mass in gram equivalent to equivalent to the mass of one the mass of one atom in atomic atom in atomic mass unitsmass units


• This number, This number, 6.02 x 106.02 x 102323 is is known as the known as the Avogadro Avogadro ConstantConstant

The MoleThe Mole

• Just like a dozen donuts is Just like a dozen donuts is equivalent to twelve donuts, equivalent to twelve donuts, one one molemole of donuts would be of donuts would be 6.02 x 106.02 x 102323 donuts. Unlike donuts. Unlike the term dozen, however, the term dozen, however, moles are used to quantify moles are used to quantify very small objects, like very small objects, like atoms.atoms.

The MoleThe Mole• A A molemole is just a is just a

particular particular number of number of atoms, ions, atoms, ions, molecules, or molecules, or formula unitsformula units

• The The molemole is the is the SI base unit SI base unit representing the representing the chemical chemical quantity of a quantity of a substance, and substance, and has been given has been given the symbol Nthe symbol NAA

ReviewReview: The Mole: The Mole

The number equal to the The number equal to the numbernumber of of carbon atoms in exactly 12 grams of carbon atoms in exactly 12 grams of pure pure 1212C.C.

1 mole1 mole of anything = of anything = 6.022 6.022 10 102323 units of that thingunits of that thing

The MoleThe Mole

• The mass of The mass of one mole of one mole of molecules, molecules, atoms, ions, atoms, ions, or formula or formula units is called units is called the the molar molar massmass of that of that speciesspecies

ReviewReview: Molar : Molar MassMass

A substance’s A substance’s molar mass molar mass (molecular weight) is the mass in (molecular weight) is the mass in grams of one mole of the compound.grams of one mole of the compound.

COCO22 = 44.01 grams per mole = 44.01 grams per mole

HH22O = 18.02 grams per moleO = 18.02 grams per mole

Ca(OH)Ca(OH)22 = 74.10 grams per mole = 74.10 grams per mole

Mass-Mole and Mass-Mole and Mole-Mass Mole-Mass

ConversionsConversions

• In order to convert the mass of an object In order to convert the mass of an object to the number of moles, you divide by to the number of moles, you divide by the molar mass of the substancethe molar mass of the substance

EXAMPLE:EXAMPLE:

4.56 g CO4.56 g CO22 x x 1 mole CO1 mole CO22 = 0.104 mole CO = 0.104 mole CO22

44 g CO44 g CO22

Mass-Mole and Mole-Mass Mass-Mole and Mole-Mass ConversionsConversions

• In order to convert moles to mass, In order to convert moles to mass, you multiply by the molar mass of you multiply by the molar mass of the substancethe substance

EXAMPLE:EXAMPLE: 0.58 moles NH0.58 moles NH44NONO33 x x 80.g NH80.g NH44NONO33 = 46 g = 46 g

NHNH44NONO33

1 mole NH1 mole NH44NONO33

Volume-Mole and Volume-Mole and Mole-Volume Mole-Volume ConversionsConversions

• In order to do any conversions In order to do any conversions with moles and volume, the with moles and volume, the substance you are dealing substance you are dealing with must be a gas and exist with must be a gas and exist at STP which is 0°C and 1 atmat STP which is 0°C and 1 atm

• To convert the volume of an To convert the volume of an object to the number of object to the number of moles, you divide by 22.4 Lmoles, you divide by 22.4 L

EXAMPLE:

54.2 L of N2 x 1 mole N2 = 2.42 moles N2

22.4 L N2

Volume-Mole and Volume-Mole and Mole-Volume ConversionsMole-Volume Conversions

• To convert the number of moles To convert the number of moles to volume, you multiply by 22.4 Lto volume, you multiply by 22.4 L

EXAMPLE:EXAMPLE:0.78 moles of He x 0.78 moles of He x 22.4 L He22.4 L He = 17 L He = 17 L He

1 mole He1 mole He

Molecules/Atoms/Formula Units - Molecules/Atoms/Formula Units - Moles and Vice VersaMoles and Vice Versa

• To convert the number of M/A/F to To convert the number of M/A/F to moles, divide by 6.02 x 10moles, divide by 6.02 x 102323 M/A/F M/A/F

• To convert moles to M/A/F multiply by To convert moles to M/A/F multiply by 6.02 x 106.02 x 102323 M/A/F M/A/F

• Remember . . .Remember . . . – moleculesmolecules are for are for covalent covalent

compoundscompounds– formula unitsformula units are for are for ionic compoundsionic compounds

Molecules/Atoms/Molecules/Atoms/Formula Units - Moles Formula Units - Moles

and Vice Versaand Vice Versa

EXAMPLE:

5.21 X 1024 molecules SO2 x 1 mole SO2 = 8.65 moles SO2

6.02 x 1023 molecules SO2

1.25 moles Cu(NO3)2 x 6.02 x 1023 f.u. = 7.53 x1023 f.u. Cu(NO3)2

1 mole Cu(NO3)2

Converting from Mass-Volume, Mass-Converting from Mass-Volume, Mass-M/A/F, Volume-M/A/F, or Anything in M/A/F, Volume-M/A/F, or Anything in

BetweenBetween

• Convert the first unit of moles. Convert the first unit of moles. Then convert from moles to the Then convert from moles to the unit you’re looking for.unit you’re looking for.

MOLES ARE THE MIDDLE MEN!MOLES ARE THE MIDDLE MEN!

The MoleThe Mole

Percentage CompositionPercentage Composition

The The percentage compositionpercentage composition of a of a

compound is a statement of the compound is a statement of the

relative mass each element relative mass each element contributes to the mass of the contributes to the mass of the

compound as a wholecompound as a whole

““PART TO WHOLE!”PART TO WHOLE!”


• To determine percent composition of a To determine percent composition of a particular element in a compound, particular element in a compound, divide the mass of the element by the divide the mass of the element by the molecular mass of the compound, and molecular mass of the compound, and multiply by 100multiply by 100

• If the percent composition were If the percent composition were calculated for all elements of the calculated for all elements of the compound, and added together, the compound, and added together, the sum of the percents should equal 100%sum of the percents should equal 100%


FOR EXAMPLE:FOR EXAMPLE:Calculate the % composition of carbon in Calculate the % composition of carbon in

ethanol (Cethanol (C22HH55OH)OH)1.1. Calculate the molecular mass of the Calculate the molecular mass of the

compound:compound:2(12) + 5(1) + 1(16) + 1 = 46 amu2(12) + 5(1) + 1(16) + 1 = 46 amu

2.2. Calculate the total mass of the element in Calculate the total mass of the element in question: 2(12) = 24 amuquestion: 2(12) = 24 amu

3.3. Divide the mass of the element by the Divide the mass of the element by the mass of the compound and multiply by 100:mass of the compound and multiply by 100:24 amu24 amu x 100 = 52 % carbon x 100 = 52 % carbon46 amu46 amu

Calculating Percentage Calculating Percentage CompositionComposition

Calculate the percentage composition of Calculate the percentage composition of magnesium carbonate, MgCOmagnesium carbonate, MgCO33..

24.31 g + 12.01 g + 3(16.00 g) = 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g84.32 g 24.31

100 28.83%84.32

Mg 12.01

100 14.24%84.32

C 48.00

100 56.93%84.32

O

100.00

Empirical FormulasEmpirical Formulas

The The empirical formulaempirical formula of a of a substance is the simplest, whole-substance is the simplest, whole-number ratio between the atoms number ratio between the atoms

of the elements present in a of the elements present in a compoundcompound


A step-by-step method for determining A step-by-step method for determining empirical formulas is found below:empirical formulas is found below:

1.1. Convert all quantities to moles (if the Convert all quantities to moles (if the quantity is a percent, remove the quantity is a percent, remove the percent sign, and stick in a grams sign!)percent sign, and stick in a grams sign!)

2.2. Divide all numbers by the SMALLEST Divide all numbers by the SMALLEST number of molesnumber of moles

3.3. If the quotient is NOT a whole number, If the quotient is NOT a whole number, convert the decimal to a fraction, then convert the decimal to a fraction, then multiply by the denominator of the multiply by the denominator of the fraction to make it a whole numberfraction to make it a whole number

4.4. Write each element with its Write each element with its corresponding subscriptcorresponding subscript

Empirical FormulasEmpirical FormulasFind the empirical formula for a Find the empirical formula for a

compound containing 43.9% C, 7.32 % compound containing 43.9% C, 7.32 % H, 48.78 % OH, 48.78 % O

Step One:Step One:

43.9 g C x 43.9 g C x 1 mole1 mole = 3.66 moles = 3.66 moles

12 g12 g

7.32 g H x 7.32 g H x 1 mole1 mole = 7.32 moles = 7.32 moles

1 g1 g

48.78 g O x 48.78 g O x 1 mole1 mole = 3.05 moles = 3.05 moles

16 g16 g


Step 2Step 2::

C: 3.66/3.05 =1.2C: 3.66/3.05 =1.2

H: 7.32/3.05 =2.4H: 7.32/3.05 =2.4

O: 3.05/3.05 =1O: 3.05/3.05 =1

Step 3Step 3::C: 1.2 = 6/5 x 5 = 6C: 1.2 = 6/5 x 5 = 6

H: 2.4 = 12/5 x 5 = 12H: 2.4 = 12/5 x 5 = 12

O: 1 x 5 = 5O: 1 x 5 = 5

Step 4Step 4::

CC66HH1212OO55

Molecular FormulaMolecular Formula

• Molecular formulasMolecular formulas indicate the actual indicate the actual number of atoms of each number of atoms of each element making up a element making up a moleculemolecule

• To determine a molecular To determine a molecular formula, one must know formula, one must know both the empirical both the empirical formula and the molecular formula and the molecular mass of the substancemass of the substance

http://www.jozie.net/JF/HS_Chem/images/moleguy.gif

Molecular FormulaMolecular Formula

What is the molecular formula of a substance What is the molecular formula of a substance that has an empirical formula of AgCOthat has an empirical formula of AgCO22 and a and a formula mass of 304 g?formula mass of 304 g?

Step 1Step 1: Determine the formula mass of the : Determine the formula mass of the empirical formula: 108+12+2(16) = 152 gempirical formula: 108+12+2(16) = 152 g

Step 2Step 2: Divide the formula mass by the mass : Divide the formula mass by the mass of the empirical formula: 304 g/152 g = 2of the empirical formula: 304 g/152 g = 2

Step 3Step 3: Multiply the subscripts of the : Multiply the subscripts of the empirical formula by the ratio found in step empirical formula by the ratio found in step 2:2:

AgAg22CC22OO44

FormulasFormulas

molecular formula = (empirical molecular formula = (empirical formula)formula)nn [ [nn = integer] = integer]

molecular formula = Cmolecular formula = C66HH66 = (CH) = (CH)66

empirical formula = CHempirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.

Molecular formula: the true number of atoms of each element in the formula of a compound.

FormulasFormulas (continued)(continued)

Formulas for Formulas for ionic compoundsionic compounds are are ALWAYSALWAYS empirical (lowest whole empirical (lowest whole number ratio).number ratio).Examples:Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

FormulasFormulas (continued)(continued)

Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be empirical (lowest whole be empirical (lowest whole number ratio).number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Empirical Formula Empirical Formula DeterminationDetermination

1.1. Base calculation on 100 grams of Base calculation on 100 grams of compound. compound.

2.2. Determine moles of each element Determine moles of each element in 100 grams of compound.in 100 grams of compound.

3.3. Divide each value of moles by the Divide each value of moles by the smallest of the values.smallest of the values.

4.4. Multiply each number by an integer Multiply each number by an integer to obtain all whole numbers.to obtain all whole numbers.


Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

49.32 14.107

12.01

g C mol Cmol C

g C

6.85 16.78

1.01

g H mol Hmol H

g H

43.84 12.74

16.00

g O mol Omol O

g O


(part 2)(part 2)

4.1071.50

2.74

mol C

mol O

6.782.47

2.74

mol H

mol O

2.741.00

2.74

mol O

mol O

Divide each value of moles by the smallest Divide each value of moles by the smallest of the values.of the values.

Carbon:Carbon:

Hydrogen:Hydrogen:

Oxygen:Oxygen:


(part 3)(part 3)Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.

Carbon: 1.50Carbon: 1.50 Hydrogen: 2.50Hydrogen: 2.50 Oxygen: 1.00Oxygen: 1.00x 2 x 2 x 2

33 55 22

Empirical formula:C3H5O

2

Finding the Molecular Finding the Molecular FormulaFormula

The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

1. Find the formula mass of 1. Find the formula mass of CC33HH55OO22

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g



3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

2. Divide the molecular mass by 2. Divide the molecular mass by the mass given by the emipirical the mass given by the emipirical formula.formula.

1462

73



3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g146

273

3. Multiply the empirical formula by 3. Multiply the empirical formula by this number to get the molecular this number to get the molecular formula.formula.

(C(C33HH55OO22) x 2 ) x 2 ==

CC66HH1010OO44

HydratesHydrates

• HydratesHydrates are are compounds that compounds that have crystallized have crystallized from a water from a water solution. In these solution. In these compounds, the compounds, the water molecules water molecules adhere to the ions adhere to the ions in the compound, in the compound, and become part and become part of the crystalof the crystal

HydratesHydrates

• The formula for a hydrate indicates The formula for a hydrate indicates the number of water molecules the number of water molecules attached with a coefficient attached with a coefficient

CuSOCuSO44•5H•5H220 would have 5 water 0 would have 5 water molecules attached to itmolecules attached to it

• To name hydrates, name the regular To name hydrates, name the regular compound and use a Greek prefix to compound and use a Greek prefix to indicate the number of waters presentindicate the number of waters present

CuSOCuSO44•5H•5H220 would be named copper 0 would be named copper (II) sulfate pentahydrate(II) sulfate pentahydrate

HydratesHydrates• When calculating the formula mass for When calculating the formula mass for

a hydrate, the mass of the water is a hydrate, the mass of the water is added to the mass of the formula unitadded to the mass of the formula unit

CuSOCuSO44•5H•5H220 = 63.5 + 32 + 4(16) + 5(18)0 = 63.5 + 32 + 4(16) + 5(18) = 249.5 amu= 249.5 amu

• To determine the ratio of compound to To determine the ratio of compound to water in a hydrate, the compounds can water in a hydrate, the compounds can be heated to drive off the water. By be heated to drive off the water. By comparing the mass of the sample to comparing the mass of the sample to the mass of the water, one can the mass of the water, one can determine the formula of the hydratedetermine the formula of the hydrate

HydratesHydratesEXAMPLEEXAMPLE: : A 10.407 g sample of hydrated barium A 10.407 g sample of hydrated barium

iodide is heated to drive off the water. The mass of iodide is heated to drive off the water. The mass of the dry sample is 9.520 g. What is the formula of the dry sample is 9.520 g. What is the formula of the hydrate?the hydrate?

Step 1Step 1: Find the mass of the water in the : Find the mass of the water in the compound: 10.407 g – 9.520 g = 0.887 g watercompound: 10.407 g – 9.520 g = 0.887 g water

Step 2Step 2: Convert mass of both compounds to : Convert mass of both compounds to moles:moles:

9.520 g BaI9.520 g BaI22 x x 1 mole BaI1 mole BaI22 = 0.0243 moles BaI = 0.0243 moles BaI22

391 g BaI391 g BaI22

0.887 g H0.887 g H22O x O x 1 mole H1 mole H22OO = 0.0493 moles H = 0.0493 moles H22O O

18 g H18 g H22O O

HydratesHydrates

Step 3Step 3: Divide both results by the smaller : Divide both results by the smaller number of moles:number of moles:

0.0243 moles / 0.0243 moles = 10.0243 moles / 0.0243 moles = 1

0.0493 moles / 0.0243 moles = 20.0493 moles / 0.0243 moles = 2

Step 4Step 4: Write the formula for the hydrate:: Write the formula for the hydrate:

BaIBaI22•2H•2H2200

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