# Mole Concept PBL

## Introduction

In order to exhibit a complete understanding of the mole concept and create a study guide for other students, we created a video with a brief note summary.
The Mole Concept PBL

## Molar Mass

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

Example:

H2O
H( 2 atoms) = 1g x 2

O= 16g

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

## Percent Yield

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.

Formula:

%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

SO...

.455g Ag2CrO4 x 100 = 93.2 % yield

.488g Ag2CrO4

## Percent Composition

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.

Example:

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.

Example: H6O6
H6O6
^ Molecular

HO

^Empirical

## Reaction Stoichiometry

Reaction stoichiometry is determining the amount of substance that is consumed or produced by a reaction.

Example:

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

## Limiting Reactants

The reactant in a chemical reaction that limits the amount of product that can be produced.