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Molar Mass, Empirical &amp; Molecular Formulas

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Cartoon courtesy of NearingZero.net. Molar Mass, Empirical &amp; Molecular Formulas. Molecular Mass &amp; Formula Mass. A chemical formula is a convenient way of showing how many and what type of atoms make up a particular molecule of formula unit. Molecular Mass &amp; Formula Mass.

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Molecular Mass & Formula Mass
• A chemical formula is a convenient way of showing how many and what type of atoms make up a particular molecule of formula unit
Molecular Mass & Formula Mass
• While the mass of an atom can be found by looking at the periodic table, the mass of a molecule or formula unit must be calculated.
Molecular Mass & Formula Mass
• The molecular mass of a substance can be found by finding the sum of the atomic masses that make up a particular molecule
• CH2Cl2 = 12 + 2(1) + 2(35.45)

= 84.9 amu

Molecular Mass & Formula Mass
• The formula mass, which is used for ionic compounds, is found in the same fashion as molecular compounds.
• NaCl = 22.99 + 35.45

= 58.44 amu

Calculating Formula Mass

Calculate the formula mass of magnesium carbonate, MgCO3.

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

84.32 g

The Avogadro Constant
• Atoms are much too small to count individually
• The mass of a single molecule is so small that it is impossible to measure by ordinary means in the laboratory.
The Avogadro Constant
• For this reason, scientists had to come up with a way to convert atomic mass units to something that could be used in the laboratory, grams
The Avogadro Constant
• Chemists have found that 6.02 x 1023 atoms of an element has a mass in gram equivalent to the mass of one atom in atomic mass units
The Avogadro Constant
• This number, 6.02 x 1023 is known as the Avogadro Constant
The Mole
• Just like a dozen donuts is equivalent to twelve donuts, one mole of donuts would be 6.02 x 1023 donuts. Unlike the term dozen, however, moles are used to quantify very small objects, like atoms.
The Mole
• A mole is just a particular number of atoms, ions, molecules, or formula units
• The mole is the SI base unit representing the chemical quantity of a substance, and has been given the symbol NA
Review: The Mole
• The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.
• 1 moleof anything = 6.022 ´ 1023units of that thing
The Mole
• The mass of one mole of molecules, atoms, ions, or formula units is called the molar mass of that species
Review: Molar Mass

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

CO2 = 44.01 grams per mole

H2O = 18.02 grams per mole

Ca(OH)2 = 74.10 grams per mole

Mass-Mole and Mole-Mass Conversions
• In order to convert the mass of an object to the number of moles, you divide by the molar mass of the substance

EXAMPLE:

4.56 g CO2 x 1 mole CO2 = 0.104 mole CO2

44 g CO2

Mass-Mole and Mole-Mass Conversions
• In order to convert moles to mass, you multiply by the molar mass of the substance

EXAMPLE:

0.58 moles NH4NO3 x 80.g NH4NO3 = 46 g NH4NO3

1 mole NH4NO3

Volume-Mole and Mole-Volume Conversions
• In order to do any conversions with moles and volume, the substance you are dealing with must be a gas and exist at STP which is 0°C and 1 atm
• To convert the volume of an object to the number of moles, 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 Mole-Volume Conversions
• To convert the number of moles to volume, you multiply by 22.4 L

EXAMPLE:

0.78 moles of He x 22.4 L He = 17 L He

1 mole He

Molecules/Atoms/Formula Units - Moles and Vice Versa
• To convert the number of M/A/F to moles, divide by 6.02 x 1023 M/A/F
• To convert moles to M/A/F multiply by 6.02 x 1023 M/A/F
• Remember . . .
• molecules are for covalent compounds
• formula units are for ionic compounds
Molecules/Atoms/Formula Units - Moles and 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-M/A/F, Volume-M/A/F, or Anything in Between
• Convert the first unit of moles. Then convert from moles to the unit you’re looking for.

MOLES ARE THE MIDDLE MEN!

Percentage Composition

The percentage composition of a

compound is a statement of the

relative mass each element contributes to the mass of the compound as a whole

“PART TO WHOLE!”

Percentage Composition
• To determine percent composition of a particular element in a compound, divide the mass of the element by the molecular mass of the compound, and multiply by 100
• If the percent composition were calculated for all elements of the compound, and added together, the sum of the percents should equal 100%
Percentage Composition

FOR EXAMPLE:

Calculate the % composition of carbon in ethanol (C2H5OH)

• Calculate the molecular mass of the compound:

2(12) + 5(1) + 1(16) + 1 = 46 amu

• Calculate the total mass of the element in question: 2(12) = 24 amu
• Divide the mass of the element by the mass of the compound and multiply by 100:

24 amu x 100 = 52 % carbon

46 amu

Calculating Percentage Composition

Calculate the percentage composition of magnesium carbonate, MgCO3.

24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

100.00

Empirical Formulas

The empirical formula of a substance is the simplest, whole-number ratio between the atoms of the elements present in a compound

Empirical Formulas

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

• Convert all quantities to moles (if the quantity is a percent, remove the percent sign, and stick in a grams sign!)
• Divide all numbers by the SMALLEST number of moles
• If the quotient is NOT a whole number, convert the decimal to a fraction, then multiply by the denominator of the fraction to make it a whole number
• Write each element with its corresponding subscript
Empirical Formulas

Find the empirical formula for a compound containing 43.9% C, 7.32 % H, 48.78 % O

Step One:

43.9 g C x 1 mole = 3.66 moles

12 g

7.32 g H x 1 mole = 7.32 moles

1 g

48.78 g O x 1 mole = 3.05 moles

16 g

Empirical Formulas

Step 2:

C: 3.66/3.05 =1.2

H: 7.32/3.05 =2.4

O: 3.05/3.05 =1

Step 3:

C: 1.2 = 6/5 x 5 = 6

H: 2.4 = 12/5 x 5 = 12

O: 1 x 5 = 5

Step 4:

C6H12O5

Molecular Formula
• Molecular formulas indicate the actual number of atoms of each element making up a molecule
• To determine a molecular formula, one must know both the empirical formula and the molecular mass of the substance
Molecular Formula

What is the molecular formula of a substance that has an empirical formula of AgCO2 and a formula mass of 304 g?

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

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

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

Ag2C2O4

Formulas

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.

• molecular formula = (empirical formula)n [n = integer]
• molecular formula = C6H6 = (CH)6
• empirical formula = CH
Formulas(continued)

Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).

Examples:

NaCl

MgCl2

Al2(SO4)3

K2CO3

Formulas(continued)

Formulas for molecular compoundsMIGHT be empirical (lowest whole number ratio).

Molecular:

C6H12O6

H2O

C12H22O11

Empirical:

H2O

CH2O

C12H22O11

Empirical Formula Determination
• Base calculation on 100 grams of compound.
• Determine moles of each element in 100 grams of compound.
• Divide each value of moles by the smallest of the values.
• Multiply each number by an integer to obtain all whole numbers.
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?

Empirical Formula Determination(part 2)

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

Carbon:

Hydrogen:

Oxygen:

Empirical Formula Determination(part 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

C3H5O2

Empirical formula:

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) = 73.08 g

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?

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

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

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?

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

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

(C3H5O2) x 2 =

C6H10O4

Hydrates
• Hydrates are compounds that have crystallized from a water solution. In these compounds, the water molecules adhere to the ions in the compound, and become part of the crystal
Hydrates
• The formula for a hydrate indicates the number of water molecules attached with a coefficient

CuSO4•5H20 would have 5 water molecules attached to it

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

CuSO4•5H20 would be named copper (II) sulfate pentahydrate

Hydrates
• When calculating the formula mass for a hydrate, the mass of the water is added to the mass of the formula unit

CuSO4•5H20 = 63.5 + 32 + 4(16) + 5(18)

= 249.5 amu

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

EXAMPLE: A 10.407 g sample of hydrated barium iodide is heated to drive off the water. The mass of the dry sample is 9.520 g. What is the formula of the hydrate?

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

Step 2: Convert mass of both compounds to moles:

9.520 g BaI2 x 1 mole BaI2 = 0.0243 moles BaI2

391 g BaI2

0.887 g H2O x 1 mole H2O = 0.0493 moles H2O

18 g H2O

Hydrates

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

0.0243 moles / 0.0243 moles = 1

0.0493 moles / 0.0243 moles = 2

Step 4: Write the formula for the hydrate:

BaI2•2H20