Stoichiometry

Chapter 3Chemical Formulas and

Equations

Stoichiometry

Chemical Equations

Concise representations of chemical reactions

Stoichiometry

Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Anatomy of a Chemical Equation

Reactants appear on the left side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Anatomy of a Chemical Equation

Products appear on the right side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Anatomy of a Chemical Equation

The states of the reactants and products are written in parentheses to the right of each compound.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Anatomy of a Chemical Equation

Coefficients are inserted to balance the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule

Stoichiometry

Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule

• Coefficients tell the number of molecules

Stoichiometry

Cations = + (Lose Electrons)Anions = - (Gain Electrons)

Stoichiometry

Trends in the Periodic Table• Group 1 metals lose 1 electron to

give 1+ cations, such as Na+1 or Li+1.• Group 2 metals lose 2 electrons to

give 2+ cations, such as Mg+2.• Mg = 12 electrons• 1s22s22p63s2

Stoichiometry

Binary Ionic Compounds• Compounds composed of two elements• The total charges of both elements must

be equal.• First find each elements charge according

to the periodic table.• Mg = +2 (loses 2 electrons)• N = -3 (gains 3 electron)

Stoichiometry

• Because compounds are electrically neutral:– The charge on the cation becomes the

subscript on the anion.– The charge on the anion becomes the

subscript on the cation.– Now we have 3 Mg with each +2 charge = +6– And we have 2 N with each -3 charge = -6

Stoichiometry

Na1+Cl1- Ca2+Br1- Al3+O2-

Stoichiometry

Using Chemical Formulas

Stoichiometry

Using Formulas in Problem Solving•Correctly written chemical formulas hold a large amount of information for the prepared student to find.•The subscripts tell us the number of atoms of each kind that is present in the compound.•Example: NaCl has one atom of sodium and one atom of chlorine.•H2SO4 has two atoms of hydrogen, one atom of sulfur and four atoms of oxygen

Stoichiometry

• The mass of a formula can be found by adding the masses of the atoms in the formula.

• The formula mass of any molecule is the sum of the average atomic masses of all atoms present in the formula.

• H2O• H = 2 atoms (each with 1.01 amu) = 2.02 amu• O = 1 atom (16 amu) = 16 amu• Total = 16 +2.02 = 18.02 amu

Formula Mass

Stoichiometry

What is the formula mass for H2SO4?

Stoichiometry

Molar Mass = Formula Mass•The sum of the atomic masses times the number of atoms of each kind of element is equal to the mass of one mole of the substance.•Examples: Na = 23g/mol and Cl = 35.5g/mol so NaCl has a molar mass of 58.5g/mol•Nitric acid is HNO3so its molar mass is

H = 1 x 1 = 1N = 14 x 1 = 14O = 16 x 3 = 48

Total = 63g/mol

Stoichiometry

What is the molar mass of Al2S3?

Stoichiometry

Molar Mass as a Conversion Factor• To convert a known amount of a

compound in moles to mass in grams, multiply the amount of moles by the molar mass.

• Amount of moles x molar mass (g/mol) = mass in grams

• Notice the moles cancel out

Stoichiometry

What is the mass in grams of 2.5 moles of oxygen gas?

Stoichiometry

• To convert a known mass of a compound in grams to an amount in moles, the mass is divided by the molar mass.

• Mass in grams x 1 / molar mass (g/mol) = moles

• Notice the grams cancel out.

Stoichiometry

Ibuprofen is C13H18O2. What is the molar mass? If you have 33 g of ibuprofen, how many moles

do you have?

Stoichiometry

Percent Composition

•The percentage composition of each element in a compound can be determined using the formula and the atomic masses.•Mass of element / mass of compound x 100 = % of element

Stoichiometry

• Example: Sodium chloride or NaCl• Find the total mass of the compoundNa = 23.0Cl = 35.5 58.5g/mol• Divide each mass by the total then x 100% Na = 23.0 x 100 = 39.3%Na

58.5% Cl = 35.5 x 100 = 60.7% Cl 58.5• Notice that the total of the percentages is

always equal or very close to 100%.

Stoichiometry

H2O

Stoichiometry

Determining Chemical Formulas

Stoichiometry

Calculating Empirical Formulas• An empirical formula consists of the symbols for

the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound.

• Begin by assuming you have 100g of substance and using the percent composition as the amount of grams.

• 78.1% of B = 78.1 grams of B• 21.9% of H = 21.9 grams of H

Stoichiometry

• Once you have grams, convert to moles by looking up the amount of grams on the periodic chart which equals 1 mole.

• 78.1 grams B x 1 mole/10.81 g = 7.22 mol B• 21.9 grams H x 1 mole/1.01 g = 21.7 mol H• These values give us a mole ratio of 7.22 mol

of B to 21.7 mol of H. However, we need this reduced to the smallest numbers.

• To find such ratio, divide each number of moles by the smallest number. 7.22/21.7 = 3.01

• The ratio then is 1 mol B to 3 mol H• Final Answer = BH3

Stoichiometry

A compound has 40% Fe and 60% S. Find the empirical formula

Stoichiometry

• Sometimes mass composition is known instead of percent composition.

• To find the empirical formula, convert mass composition to moles.

• Then calculate the smallest whole number mole ratio by dividing the smaller number over the larger.

• This removes 1 step!

Stoichiometry

A 10.15 g sample of P and O has 4.433 g of P. What is the empirical formula?

• O = 4.433 g• P = 10.15 – 4.433 = 5.717 g• 4.433 g x 1 mole/16 g = .3573 mol Oxygen• 5.717 g x 1 mole/30.97g = .1431 mol P• .1431 /.3573 = 2.497• 1 mol P = 2.5 mol O• Must have whole numbers so multiply by 2• Final answer = P2O5

Stoichiometry

A 20 g compound has 4 g of Ca and some Br. What is the empirical formula?

Stoichiometry

Calculating the molecular formula from the empirical formula

• The empirical formula is the lowest whole number ratio for a compound.

• It is not always the actual formula or the molecular formula.

• For example the empirical formula is CH but the molecular formula is C2H4.

Stoichiometry

• To calculate the molecular formula from the empirical formula:

• x(empirical formula)=molecular formula• x = experimental formula mass/empirical

formula mass• For example, the empirical formula is P2O5 and

it’s experimental molar mass is 283.89 g/mol.• To find the empirical formula mass go to the

periodic chart. P = 30.874 and O = 16• (30.874 x 2) + (16 x 5) = 141.94 amu• 141.94 / 283.89 = 2.0001• 2(P2O5) = P4O10

Stoichiometry

What is the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.11 amu?