Mole Concept PBL
Megan Allen, Conner Mann, and Gabby Rouke
Just in case you missed it!
The total atomic mass for the compound. You find it by adding up the atomic masses for all the elements within the compound.
***Note: subscripts tell you how many atoms of that specific element are in the compound, and when calculating the molar mass for these elements you need to multiply it by the subscript
H( 2 atoms) = 1g x 2
2g + 16g = 18g
THERE IS 1 GRAM PER 1 MOLE SO...
Molar Mass: 18 g/mole
**masses in example were rounded from the periodic table
The percent yield has two parts: actual yield and theoretical yield. Actual yield is the amount of product produced by a chemical reaction. Theoretical yield is the maximum amount of a product that can be produced in a balanced reaction. The actual yield is usually less than the theoretical yield.
%Yield = (Actual yield/ Theoretical yield) x 100
Example: When potassium chromate is added to a solution of .500g silver nitrate, .455g of solid silver chromate is formed. What is the percent yield?
K2CrO4 + 2AgNO3 → Ag2CrO4 + 2KNO3
.500g AgNO3 x 1 mol AgNO3 x 1 mol Ag2CrO4 x 336.8g Ag2CrO4 =
169.9g AgNO3 2 mol AgNO3 1 mol Ag2CrO4
.488g Ag2CrO4 produced yet only produced .455g Ag2CrO4
.455g Ag2CrO4 x 100 = 93.2 % yield
You find the percent composition by finding the mass of the element divided by the total mass of the compound. This decimal multiplied by 100 is the percent composition.
Percent composition of MgO:
Molar mass: 24g (Mg) + 16g (O) = 40g
% Mg = 24/40 x 100 = 60 %
% O = 16/40 x 100 = 40 %
**Masses are rounded off periodic table
Empirical and Molecular Formula
Empirical Formula: The simplest positive integer ratio of atoms of each element present in a compound.
Molecular Formula: Identifies the number of each type of atom in a molecule.
Reaction stoichiometry is determining the amount of substance that is consumed or produced by a reaction.
CaCO3 + 2 HCl → CaCl2 + CO2 + H2O
Calculate moles of CO2 produced.
54.6g CO2 x (1 mol CO2/ 44.01g) = 1.24 mol CO2
Calculate moles of HCl consumed and moles of CaCl2 produced using stoichiometric factors derived from the balanced chemical equation.
1.24 mol CO2 x (2 mol HCl/1 mol CO2) = 2.48 mol HCl
1.24 mol CO2 x (1 mol CaCl2/1 mol CO2) = 1.24 mol CaCl2
Calculate mass (in grams) of CaCl2 produced.
1.24 mol CaCl2 x (111.0 g/1 mol CaCl2) = 138 g CaCl2
The reactant in a chemical reaction that limits the amount of product that can be produced.