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Stoichiometry: Percent Stoichiometry: Percent CompositionCompositionAt the conclusion of our time At the conclusion of our time together, you should be able to:together, you should be able to:

1. Determine the percent composition of each of the elements in a compound

2. Use percent compositions of a compound to determine the empirical and/or molecular formula

Chemical Formulas of CompoundsChemical Formulas of Compounds

Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions).

NO2 molecule: 2 atoms of O for every 1 atom of N

1 mole of NO2 : 2 moles of O atoms to every 1 mole of N atoms

If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

Types of FormulasTypes of Formulas

Empirical Formula

The formula of a compound that expresses the smallest whole number ratio of the atoms present.

Ionic formulas are always empirical formulas

Molecular Formula

The formula that states the actual number of each kind of atom found in one molecule of the compound.

What is the percent composition of hydrogen in water?

2.02 g H2

18.02 g H2O= 11.21 % H2

Determine the grams in 2 moles of hydrogen and divide that by the grams in 1 mole of H2O,

then multiply by 100.

x 100

To Obtain an To Obtain an Empirical FormulaEmpirical Formula

1. Determine the mass in grams of each element present, if necessary.

2. Calculate the number of moles of each element.

3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.

4. If whole numbers are not obtained in step 3, multiply through by the smallest number that will give all whole numbers

A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine the formula for this substance.

This requires mole ratios, so convert grams to moles

moles of N = 2.34 g of N

14.01 g/mole

= 0.167 moles of N

moles of O = 5.34 g

16.00 g/mole

= 0.334 moles of O

Formula:

0.334 0.167ON 0.167 0.334 2

0.167 0.167

N O NO

Empirical Formula from % Empirical Formula from % CompositionComposition

A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H

What is the empirical formula of the substance?

Consider a sample size of 100 gramsThis would contain 60.80 g of Na,

28.60 grams of B and 10.60 grams HDetermine the number of moles of eachDetermine the simplest whole number ratio

Empirical Formula from % Empirical Formula from % CompositionComposition

Determine the number of moles of each 2.64 mol Na 2.65 mol B10.50 mol H

Determine the simplest whole number ratio

NaBH4

Calculation of the Molecular Calculation of the Molecular FormulaFormula

A compound has an empirical formula of NO2. The colorless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?

Formula mass = 14.01 + 32.00 =

46.01 g/mol

92.0 g/mol 46.01 g/mol = ~2

N2O4

Stoichiometry: Percent Stoichiometry: Percent CompositionCompositionLet’s see if you can:Let’s see if you can:

1. Determine the percent composition of each of the elements in a compound

2. Use percent compositions of a compound to determine the empirical and/or molecular formula

Percent CompositionPercent Composition

What is the percent carbon in C5H8NO4 (the

glutamic acid used to make MSG monosodium

glutamate), a compound used to flavor foods

and tenderize meats?

a) 8.22 % C

b) 24.3 % C

c) 41.1 % C

d) not listed

Empirical Formula from % Empirical Formula from % CompositionComposition

A substance has the following composition by mass: 12.8% C, 2.1% H, and 85.1% Br (by mass). Calculate the empirical formula and the molecular formula of this compound given that the molar mass is 188 g/mol.

 Consider a sample size of 100 grams

Determine the number of moles of eachDetermine the simplest whole number ratio

Empirical Formula from % Empirical Formula from % CompositionComposition

Determine the number of moles of each 12.8 g/12.01 g/mol =

1.07 mol C2.1 g/1.01 g/mol =

2.08 mol H85.1 g/79.90 g/mol =

1.07 mol Br

CH2Br

Molecular Formula from Molecular Formula from Empirical FormulaEmpirical Formula

CH2BrEmpirical Molar Mass =

93.93 g/molMolecular Molar Mass = 188 g/mol

93.93 g/mol~2

C2H4Br2