Stoichiometry © 2009, Prentice-Hall, Inc. Unit 10: Stoichiometry 1 Calculations with Chemical Formulas

Stoichiometry

© 2009, Prentice-Hall, Inc.

Unit 10: Stoichiometry 1Unit 10: Stoichiometry 1

Calculations with Chemical Calculations with Chemical FormulasFormulas

Stoichiometry


Mass in Elements and Compounds

Stoichiometry

Atomic Mass

Atoms are so small, it is difficult to discuss how much they weigh in grams.

• Use atomic mass unitsatomic mass units or amuamus– 1 amu is 1/12 the mass of a carbon-12 atom.

This gives us a basis for comparison.• The decimal numbers on the periodic table

are atomic masses in amu.


Stoichiometry

Gram Atomic Mass

• Atomic mass in gramsgrams instead of amu’s.– Represents an amount that we can actually

measure in lab.

• Also known as a molemole, or molar mass, which we will come back to later…


Stoichiometry


Gram Formula Mass • The gram formula mass is the sum of the

atomic masses for the atoms in a chemical formula.

• So, the gram formula mass of calcium chloride, CaCl2, would be

Ca: 1 x 40.1 = 40.1

+ Cl: 2 x 35.5 = 71.0

111.1 amu

• Gram formula mass is generally used for either molecular or ionic compounds.

Stoichiometry


Gram Formula Mass

• For the molecule, ethane (C2H6), the formula mass would be:

• For the ionic compound, (NH4)2CO3

N: 2 x 14.0 amu = 28.0

H: 8 x 1.0 amu = 8.0

C: 1 x 12.0 amu = 12.0

+O: 3 x 16.0 amu = 48.0

= 96.0 amu96.0 amu

C: 2 x 12.0 = 24.0

30.0 amu30.0 amu

+ H: 6 x 1.0 = 6.0

Stoichiometry


Percent Composition

You can find the percentage (%) of the mass of a compound that comes from each of the elements in the compound by using these steps:

1.Calculate the formula mass of the compound.

2.Divide the mass of each element by the formula mass and multiply that fraction x 100.

Stoichiometry


Percent Composition

So the percentage of carbon in ethane,C2H6, is…

C is 2 x 12.0 = 24.0

H is 6 x 1.0 = 6.0

30.030.024.0 amu

30.0 amu= x 100

= 80.0%

% Carbon

amu

Stoichiometry

(stop)


Stoichiometry


Moles

Stoichiometry


Avogadro’s Number

• 6.02 x 1023

• 1 mole of 12C has a mass of 12 g.

Stoichiometry


Molar Mass

• By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol).– The molar mass of an element is the mass

number for the element that we find on the periodic table.

– The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).

Stoichiometry


Using Moles

Moles provide a bridge from the molecular scale to the real-world scale.

Stoichiometry


Mole Relationships

• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.

• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.

Stoichiometry


Finding Empirical Formulas

Stoichiometry


Calculating Empirical FormulasCalculating Empirical Formulas

• Empirical formula: Empirical formula: smallest, whole-number ratio of atoms of elements in a compound.

• You can calculate the empirical formula from the percent composition.

Stoichiometry

Steps for Empirical FormulaSteps for Empirical Formula

1. For each element, convert mass to moles. – If you have percents, use the percent as the

number of grams/100 grams of compound. Example: 67% would be 67 grams.

2. Find the mole ratio: Divide all the numbers by the smallest number of moles

3. Use the smallest, whole number ratio as the subscripts for the formula.


Stoichiometry



Example:The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of:

carbon (61.31%)hydrogen (5.14%) nitrogen (10.21%)oxygen (23.33%)

Find the empirical formula of PABA.

Stoichiometry



Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol12.01 g

1 mol14.01 g

1 mol1.01 g

1 mol16.00 g

Sunscreen PABAcarbon (61.31%)

hydrogen (5.14%) nitrogen (10.21%)oxygen (23.33%)

Stoichiometry


Calculating Empirical FormulasCalculating Empirical FormulasCalculate the mole ratio by dividing by the smallest number of moles:

C: = 7.005 = 77

H: = 6.984 = 77

N: = 1.000 = 11

O: = 2.001 = 22

5.105 mol0.7288 mol

5.09 mol0.7288 mol

0.7288 mol0.7288 mol

1.458 mol0.7288 mol

Stoichiometry



These numbers are the subscripts for the empirical formula: C7H7NO2

The molecule is shown here.

Stoichiometry

Let’s work some examples in your notes packet…


Stoichiometry

Molecular FormulaMolecular Formula


Empirical Formula is the smallest whole number ratio of atoms of elements in a compound.

Molecular Formula is the real formula of a compound. It is a multiple of the Empirical Formula. They may be the same!

1.1. CalculateCalculate the mass of the empirical formula.

2.2. DivideDivide the molecular mass given in the problem by the empirical formula mass.

3.3. MultiplyMultiply the subscripts in the empirical formula by the number you get to make new subscripts.

Molecular formula is a multiple Molecular formula is a multiple of the empirical formula! of the empirical formula!

Stoichiometry

Molecular Formula ProblemMolecular Formula Problem

• Analysis of a chemical used in photographic developing fluid indicates a chemical composition of:

65.4% C

5.45% H

29.09% O

• The molar mass is found to be 110.0 g/mol. Determine the empirical and molecular formulas.


Stoichiometry

Find Empirical Formula First!Find Empirical Formula First!

65.4g C 1 mole C = 5.45 ÷ 1.82 = 33

12g C

5.45g H 1 mole H = 5.45 ÷1.82 = 33

1g H

29.09g O 1 mole O = 1.82 ÷ 1.82 = 11

16g O

CC33HH33OO = Empirical Formula


Stoichiometry

Molecular FormulaMolecular Formula 1. Find molar mass C3H3O

C – 3 x 12 = 36

H – 3 x 1 = 3

O – 1 x 16 = +16

55 g/mole is Empirical Formula Mass

2. Divide the given Molecular Formula Mass (110g/mole) by the Empirical Formula Mass

110/55 = 2, so C3x2H3x2O1x2

3. Molecular Formula = CMolecular Formula = C66HH66OO22 © 2009, Prentice-Hall, Inc.

Stoichiometry

Molecular formula Molecular formula is some multiple of the empirical formulaempirical formula. It may be the same as the empirical formula!

Two samples of a compound must have the same percent composition to be the same compound or they would have different empirical formulas.


Stoichiometry

HydratesHydrates

A hydratehydrate is an ionic compound that has a specific number of water molecules bound to the

atoms in its crystals. (This is NOT the same as being dissolved in water)


Many compounds are found in nature as hydrates, such as protein crystals

Stoichiometry

Naming HydratesNaming Hydrates

To name a hydrate, give the name of the compound, and add the prefix for the number of waters + hydrate:

CuSOCuSO44·5H·5H22O O

copper(II) sulfate pentahydrate


Stoichiometry

Formula of a HydrateFormula of a Hydrate

1.1. Weigh Weigh the hydrate, then heatheat it in a partly covered crucible to drive off the water.

2.2. Weigh it againWeigh it again to determine the amount of water lost from the anhydrous (no water) compound. Repeat until mass stops changing.

3.3. Calculate the empirical formula Calculate the empirical formula with the anhydrous (no water) compound as one unit and the water as the second.


Stoichiometry

Hydrate ProblemHydrate ProblemProblem:

If 11.75 g of cobalt(II) chloride is heated, 9.25 g of anhydrous cobalt chloride, CoCl2, remains. What is the formula and name for this hydrate?

11.75 – 9.25 = 2.5 g water removed

9.25 g anhydrous CoCl2

Molar mass of CoCl2

Co – 1 x 59 = 59

Cl – 2 x 35.5 = +71

130 g/mol

9.25 g CoCl2 1 mole = .0712 ÷.0712 = 11

130 g

2.5 g H20 1 mole = .139 ÷ .0712 = 22

18 g

CoCl2·2H2O is cobalt(II) chloride dihydrate


Stoichiometry

Hydrate ProblemHydrate ProblemA hydrate is found to have the following percent

composition: 48.8% MgSO4 and 51.22% H2O. What is the formula and name of this hydrate?

Molar mass of MgSO4 = 24.3 + 32 + 64 =120.3g/mol

Molar mass of H2O = 2 + 16 = 18 g/mol

48.8g 1 mole = .406 ÷ .406 = 11 120.3 g

51.22 g 1 mole = 2.85 ÷ .406 = 77 18 g

Magnesium sulfate heptahydrate: Magnesium sulfate heptahydrate: MgSOMgSO44· 7H· 7H22OO© 2009, Prentice-Hall, Inc.

Stoichiometry

HydratesHydrates

Sometimes hydration can have nifty color changes!


Stoichiometry

(end)


Stoichiometry

Empirical Formula

• A blue solid is found to contain 36.84% nitrogen and 63.16% oxygen. What is the empirical formula for this solid?

• Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur.


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Stoichiometry © 2009, Prentice-Hall, Inc. Unit 10: Stoichiometry 1 Calculations with Chemical Formulas