Moles and Formula Mass

The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.022 x 1023 There are exactly 12 grams of carbon-12 in one mole of carbon-12.

I didn’t discover it. Its just named after me! Avogadro’s Number 6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855). I didn’t discover it. Its just named after me! Amadeo Avogadro

Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li 6.94 g Li = g Li 45.1 1 mol Li

Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li 1 mol Li = mol Li 2.62 6.94 g Li

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol 6.02 x 1023 atoms = atoms 2.07 x 1024 1 mol

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li 1 mol Li 6.022 x 1023 atoms Li 6.94 g Li 1 mol Li (18.2)(6.022 x 1023)/6.94 = atoms Li 1.58 x 1024

Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

Formulas Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound. molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6 empirical formula = CH

Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11

Empirical Formula Determination Base calculation on 100 grams of compound. Determine moles of each element in 100 grams of compound. Divide each value of moles by the smallest of the values. Multiply each number by an integer to obtain all whole numbers.

Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 Empirical formula: C3H5O2

Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 = C6H10O4

Conservation of Mass Law In a closed system, the total mass of the system must remain constant. The total mass of the reactants must be equal to the total mass of the products. LAB CONNECTION: This law allowed us to calculate the amount of hydrate in the mixture because the amount of water lost would be uncertain if the starting mass was not the same as the ending mass.

More Conservation Laws This law implies that matter cannot be created or destroyed. [Law of conservation of matter] This is also connected to the Law of Conservation of Atoms, which states that the number of atoms of each type of element on both sides of the equations must be the same.

What happens when a given equation does not conserve mass? # atoms for reactants ≠ # atoms of products Mass is NOT conserved Must balance the equation by adding coefficients to the elements/compounds in the reaction.

How do we balance equations? For example: Think about a grilled cheese sandwich… 2 6

Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. C2H5OH + 3O2 ® 2CO2 + 3H2O reactants products When the equation is balanced it has quantitative significance: 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

How do we balance equations? Example Steps __ Zn +  _ HCl →  __ H2 + __ ZnCl2 1. Draw a T chart under the equation. Label the left side reactants and the right side products. Write the symbol and number of atoms for each element present in the reaction on its own line on both sides. No, mass is not conserved as the equation is written. 2. Does each element start and end with the same number of atoms?  Yes → Mass is conserved and the equation is balanced  No → Mass is not conserved and you need to balance the equation. (move on to step 3)

Example Steps __ Zn +  2 HCl →  __ H2 + __ ZnCl2 3. If mass is not conserved, you will balance the equation by adding coefficients before molecules. *Do not change any subscripts.* Start with an element that only appears once on each side. On the side that has fewer atoms of that element, add a coefficient before the molecule containing that element, so that the number of atoms of that element on both sides are equal. In the T chart, multiply the coefficient by the subscript on each element in the molecule to get the new numbers of atoms.

Example Steps Final answer: Zn + 2 HCl → H2 + ZnCl2 Mass is now conserved so the equation is balanced properly. 4. Again, does each element start and end with the same number of atoms?   Yes → Mass is conserved and the equation is balanced  No → Mass is not conserved and you need to change another coefficient. Repeat step 3

Concept of Limiting Reagent (Reactant) Assume 1 piece of cheese and 2 slices of bread make one sandwich. 2 ? 3 8 ?

Concept of Limiting Reagent (Reactant) 3 8 3 What is the limiting reagent? Which reactant is in excess (left over)?

Concept of Limiting Reagent (Reactant) How many sandwiches can you make with each component? Bread: 8 slices of bread x 1 sandwich =4 sandwiches 2 slices of bread Cheese: 3 slices of cheese x 1 sandwich =3 sandwiches 1 slice of cheese

Sandwiches  molecules

Another perspective…

Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.

Calculating Masses of Reactants and Products

Calculating Masses of Reactants and Products

Working a Stoichiometry Problem 25.0 kilograms of nitrogen reacts with 5.00 kg hydrogen. How many grams of ammonia are formed. 1.) Identify reactants and products and write the balanced equation. N2 (g) + H2 (g) NH3(g) a. Every reaction needs a yield sign! b. What are the reactants? c. What are the products? d. What are the balanced coefficients?

Working a Stoichiometry Problem 25.0 kilograms of nitrogen reacts with 5.00 kg hydrogen. How many grams of ammonia are formed. 2.) Convert known masses to moles. N2 (g) + 3 H2 (g) 2 NH3(g)

Working a Stoichiometry Problem 25.0 kilograms of nitrogen reacts with 5.00 kg hydrogen. How many grams of ammonia are formed. 2.) Convert known masses to moles. N2 (g) + 3 H2 (g) 2 NH3(g)

Working a Stoichiometry Problem 25.0 kilograms of nitrogen reacts with 5.00 kg hydrogen. How many grams of ammonia are formed. 3.) Determine the limiting reactant. N2 (g) + 3 H2 (g) 2 NH3(g)

Working a Stoichiometry Problem 25.0 kilograms of nitrogen reacts with 5.00 kg hydrogen. How many grams of ammonia are formed. 3.) Determine the limiting reactant. N2 (g) + 3 H2 (g) 2 NH3(g)

Working a Stoichiometry Problem 25.0 kilograms of nitrogen reacts with 5.00 kg hydrogen. How many grams of ammonia are formed. 4.) Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product. N2 (g) + 3 H2 (g) 2 NH3(g)

Working a Stoichiometry Problem 25.0 kilograms of nitrogen reacts with 5.00 kg hydrogen. How many grams of ammonia are formed. 4.) Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product. N2 (g) + 3 H2 (g) 2 NH3(g)

Working a Stoichiometry Problem 25.0 kilograms of nitrogen reacts with 5.00 kg hydrogen. How many grams of ammonia are formed. 5.) Convert from moles to grams, using the molar mass. N2 (g) + 3 H2 (g) 2 NH3(g)

Working a Stoichiometry Problem 25.0 kilograms of nitrogen reacts with 5.00 kg hydrogen. How many grams of ammonia are formed. 5.) Convert from moles to grams, using the molar mass. N2 (g) + 3 H2 (g) 2 NH3(g)

Theoretical and Percent Yield Theoretical yield = amount of a product formed when the limiting reactant is completely consumed Percent yield = Represents the actual yield (measured result) as a percentage. The formula is shown below.

Practice Problems

Practice Problems

Practice Problems