Stoichiometry Chapter 3

Avogadro’s Number NA 6.022 X 1023 6.022 X 1023 H atoms = 1.008 g H 6.022 X 1023 O atoms = 16.00 g O

The mole 1 mole = 6.022 X 1023 atoms 1 mole H = 6.022 X 1023 H atoms = 1.008 g H 1 mole Cl2 = 6.022 X 1023 Cl2 molecules = 70.09 g Cl2 1 mole e- = 6.022 X 1023 e-

Molar mass MM,GMM, Is numerically equal to the sum of the masses of the atoms in the formula CaCl2 =110.98 g/mol

Mass-mole conversions Calculate the mass in grams of 13.2 mol CaCl2

Mass-mole conversions Calculate the mass in grams of 13.2 mol CaCl2

Calculate the number of moles in 16.4 g C6H12O6

Calculate the number of moles in 16.4 g C6H12O6

Molarity Concentration of a solute in a solution [ ] M =

What volume of 12 M HCl must be taken to obtain 0.10 mol HCl?

What mass of NaOH is contained in 125mL of a 6.00 M NaOH solution?

What are the molarities of Al 3+ and SO42- in 0.100 M Al2(SO4)3 ?

What is the percent composition of K2CrO4 ?

What is the percent composition of K2CrO4 ? MM = 2(39.10) + 52.00 + 4(16.00) = 194.20 g/mol

What is the percent composition of K2CrO4 ? MM = 2(39.10) + 52.00 + 4(16.00) = 194.20 g/mol

Empirical formula-simplest Assume a 100 g sample unless otherwise stated Convert grams to moles Find the mole ratio

Find the empirical formula of a compound containing 26. 6 % K, 35 Find the empirical formula of a compound containing 26.6 % K, 35.4 % Cr, and 38.0 % O

A sample of acetic acid weighing 1. 000 g burns to give 1 A sample of acetic acid weighing 1.000 g burns to give 1.466 g CO2 and 0.6001 g H2O. Find the empirical formula.

A sample of acetic acid weighing 1. 000 g burns to give 1 A sample of acetic acid weighing 1.000 g burns to give 1.466 g CO2 and 0.6001 g H2O. Find the empirical formula. CxHyOz + O2 CO2 + H2O 1.000 g 1.466 g 0.6001 g Find the mass of C from CO2 Find the mass of H from H2O Find the mass of O by difference

Molecular formulas Must be told the molar mass MM/FM

Find the molecular formula of acetic acid. MM = 60 g/mol

Chemical Equations Balancing The coefficients of a balanced equation represent numbers of moles of reactant and products

Combustion of propane in air C3H8 + O2 CO2 + H2O

Combustion of propane in air C3H8 + O2 3CO2 + H2O

Combustion of propane in air C3H8 + O2 3CO2 + 4H2O

Combustion of propane in air C3H8 + 5O2 3CO2 + 4H2O

Combustion of propane in air C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l)

How many moles of CO2 are produced when 1.65 mol C3H8 burns?

How many moles of CO2 are produced when 1.65 mol C3H8 burns?

What mass of O2 is required to react with 12.0 g C3H8

What mass of O2 is required to react with 12.0 g C3H8

Limiting reactant Determines the amount of product formed It is completely used up Is used to determine the theoretical yield The other reactant is in excess There will be some left over

Calculate the theoretical yield of AgI and determine the limiting reactant of a reaction between 1.00 g Ag and 1.00 g I2

Calculate the theoretical yield of AgI and determine the limiting reactant of a reaction between 1.00 g Ag and 1.00 g I2 2Ag(s) + I2(s) 2AgI(s)

Calculate the theoretical yield of AgI and determine the limiting reactant of a reaction between 1.00 g Ag and 1.00 g I2 2Ag(s) + I2(s) 2AgI(s)

1.85 g AgI is the theoretical yield I2 is the limiting reactant Ag is in excess

How much Ag is in excess?

What is the percent yield if the actual yield was 1.50 g AgI?