Finding Empirical Formulas

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# Finding Empirical Formulas - PowerPoint PPT Presentation

Unit 1-9 Chapter 3.5. Finding Empirical Formulas. Formulas. Empirical formula : the lowest whole number ratio of atoms in a compound. molecular formula = (empirical formula) n [ n = integer] molecular formula = C 6 H 6 = (CH) 6 empirical formula = CH.

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Unit 1-9

Chapter 3.5

### Finding Empirical Formulas

Formulas

Empirical formula: the lowest whole number ratio of atoms in a compound.

• molecular formula = (empirical formula)n [n = integer]
• molecular formula = C6H6 = (CH)6
• empirical formula = CH

Molecular formula: the true number of atoms of each element in the formula of a compound.

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 compoundsMIGHTbe empirical (lowest whole number ratio).

Molecular:

C6H12O6

H2O

C12H22O11

Empirical:

H2O

CH2O

C12H22O11

Calculating Empirical Formulas by mass %

One can calculate the empirical formula from the percent composition:

• 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.
Calculating Empirical Formulas

The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol

16.00 g

1 mol

14.01 g

1 mol

1.01 g

1 mol

12.01 g

Calculating Empirical Formulas

C: = 7.005  7

H: = 6.984  7

N: = 1.000

O: = 2.001  2

5.09 mol

0.7288 mol

0.7288 mol

0.7288 mol

1.458 mol

0.7288 mol

5.105 mol

0.7288 mol

Calculating Empirical Formulas
Calculating Empirical Formulas

These are the subscripts for the empirical formula:

C7H7NO2

Combustion Analysis (finding mass %)
• Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this
• C is determined from the mass of CO2 produced
• H is determined from the mass of H2O produced
• O is determined by difference after the C and H have been determined (law of conservation of mass)
Elemental Analyses
• Compounds containing other elements are analyzed using methods analogous to those used for C, H and O
• Example: Ag2O
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

Homework
• Page 98: 3.45, 47, 49, 51, 53, 55
Chapter 3.6

### Stoichiometric Calculations

Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products

Stoichiometric Calculations

From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

Stoichiometric Calculations

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Starting with 1.00 g of C6H12O6…

we calculate the moles of C6H12O6…

use the coefficients to find the moles of H2O…

and then turn the moles of water to grams

Homework
• Page 99: 57, 59, 61, 63
Chapter 3.7

### Limiting Reactants and yield

How Many Cookies Can I Make?
• You can make cookies until you run out of one of the ingredients
• Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)
How Many Cookies Can I Make?
• In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make
Limiting Reactants/reagents
• The limiting reactant is the reactant present in the smallest stoichiometric amount
• In other words, it’s the reactant you’ll run out of first (in this case, the H2)
Limiting Reactants

In the example below, the O2 would be the excess reactant/reagent

Theoretical Yield
• The theoretical yield is the amount of product that can be made
• In other words it’s the amount of product possible as calculated through the stoichiometry problem
• This is different from the actual yield, the amount one actually produces and measures

Actual Yield

Theoretical Yield

Percent Yield = x 100

Percent Yield

A comparison of the amount actually obtained to the amount it was possible to make

Homework
• Page 100: 67, 75, 77, 79