stoichiometry chapter 3 chemical formulas and equations
DESCRIPTION
Stoichiometry Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)TRANSCRIPT
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?