Empirical/Molecular Formulas

Objective/Warm-Up

SWBAT calculate molar mass of compounds.

What is the molar mass of each of these elements?NaClCH

GFM= Gram Formula Mass

Find the gram formula mass of C2H6OC = 12.01 x 2 = 24.02H = 1.01 x 6 = 6.06O = 16.00 x 1 = 16.01Total = 46.09 g/mol

Keep 2 decimal places.Unit is g/mol

Molar Mass

Example: Ca(OH)2

Ca = 40.01 g/mol x 1 = 40.08 g/mol O = 16.00 g/mol x 2 = 32.00 g/mol H = 1.01 g/mol x 2 = 2.02 g/mol Total = 74.10 g/mol

Practice problems

Objective/Warm-Up

SWBAT convert using molar mass.

What is the molar mass of each of these compounds?NaClCaCl2Mg(NO3)2

Objective/Warm-Up

SWBAT calculate percent composition by mass.

SWBAT distinguish between empirical and molecular formulas.

How do you find the percent of something? For example, how would you find the percent of girls or boys in this class?

SWBAT calculate percent composition by mass.

How do you find the percent of something? For example, how would you find the percent of girls or boys in this class?

Calculating Percent Composition

Example: Na2O Find the mass of each element: Na2= 22.99 x 2 = 45.98 g/mol O = 16.00 g/mol Take the part divided by the whole: % Na = (45.98 g/mol) / (61.98 g/mol) = 74.2 % % O = (16.00 g/mol) / (61.98 g/mol) = 25.8 % The total should add up to 100 %

Practice Problems

Objectives/Warm-Up

SWBAT distinguish between empirical and molecular formulas.

SWBAT determine the empirical formula of a compound from percent composition.

Find the percent of oxygen in Ca(OH)2

oxygen %2.43%10074

32 Oxygen of %

g/mol 74 2 1) (16 40 Mass Total

Intro to Empirical Formula

http://www.chemcollective.org/stoich/empiricalformula.php

CH2O C2H4O2

CH3OCH3O

Empirical FormulaEmpirical Formula

A formula that gives the simplest whole-number ratio of the atoms of each element in a compound.

Molecular Formula Empirical Formula

H2O2 HO

C6H12O6 CH2O

Wrap-Up

Give three new examples of a molecular formula and give the corresponding empirical formula.

Why is it important to know the difference between molecular and empirical formulas?

Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca.

Steps

1. Find mole amounts.

2. Divide each mole by the smallest mole.

Steps to Determine Empirical Formula

Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca.

1. Find mole amounts.2.128 g Cl x 1 mol Cl = 0.0600 mol Cl

35.45 g Cl

1.203 g Ca x 1 mol Ca = 0.0300 mol Ca

40.08 g Ca

Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca.

2. Divide each mole by the smallest mole.

Cl = 0.0600 mol Cl = 2.00 mol Cl 0.0300

Ca = 0.0300 mol Ca = 1.00 mol Ca 0.0300

Ratio – 1 Ca: 2 Cl

Empirical Formula = CaCl2

A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?

Hint

“Percent to mass

Mass to mole

Divide by small

Multiply ‘til whole”

A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?

Percent to mass: Mg – (72.2%/100)*298.12 g = 215.24 g N – (27.8%/100)*298.12 g = 82.88 g

Mass to mole: Mg – 215.24 g * ( 1 mole ) = 8.86 mole 24.3

gN – 82.88 g * ( 1 mole ) = 5.92 mole 14.01

gDivide by small: Mg - 8.86 mole/5.92 mole = 1.50 N - 5.92 mole/5.92 mole = 1.00

Multiply ‘til whole: Mg – 1.50 x 2 = 3.00

N – 1.00 x 2 = 2.00 Mg3N2

Practice

If the problem does not give you how many grams, assume 100 grams of the sample.

http://www.chemcollective.org/stoich/ef_analysis.php

Wrap-Up

What is the difference between molecular and empirical formulas?

Label as molecular or empirical:C2H4

Na2O2

Na2SO4

Explain how to calculate the empirical formula.

Warm-Up/Objective

SWBAT calculate molecular formulas from empirical formulas or percent composition.

Label as molecular or empirical:C2H4

Na2O2

Na2SO4

What are the steps to calculate the empirical formula?

You are a Forensic Scientist             The victim in the

following case is a 35-year old white male named Tony DeMoy.  Initial investigators say they found several signs around the death site that suggest foul play.  Four possible causes of his untimely death have been suggested by his wife who has been ruled out as a suspect because of a proven alibi.  Your task is to identify who and what killed Tony DeMoy.       

Molecular FormulaThe molecular formula gives the actual number of

atoms of each element in a molecular compound.

Steps

1. Find the empirical formula.

2. Calculate the Empirical Formula Mass.

3. Divide the molar mass by the “EFM”.

4. Multiply empirical formula by factor.

Find the molecular formula for a compound whose molar mass is ~124.06 and empirical formula is CH2O3.

2. “EFM” = 62.03 g

3. 124.06/62.03 = 2

4. 2(CH2O3) = C2H4O6

Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.

Steps

1. Find the empirical formula.

2. Calculate the Empirical Formula Mass.

3. Divide the molar mass by the “EFM”.

4. Multiply empirical formula by factor.

Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.

Empirical formula.A. Find mole amounts.

4.90 g N x 1 mol N = 0.350 mol N

14.01 g N

11.2 g O x 1 mol O = 0.700 mol O

16.00 g O

B. Divide each mole by the smallest mole.

N = 0.350 = 1.00 mol N0.350

O = 0.700 = 2.00 mol O0.350

Empirical Formula = NO2

Empirical Formula Mass = 46.01 g/mol

Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.

Molecular formulaMolar Mass = 92.0 g/mol = 2.00

Emp. Formula Mass 46.01 g/mol

Molecular Formula = 2 x Emp. Formula = N2O4

Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.

Solving the Crime

With a partner, analyze each piece of evidence in the lab area.

There are 4 suspected compounds. Find the molecular formula of each

compound, then see the teacher for possible identity of those compounds.

Wrap-Up

Summarize how to find the molecular formula from the empirical formula.

A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

g C – (48.38/100)*528.39 g = 255.64 g

g H – (8.12/100)*528.39 g = 42.91 g

g O – (43.5/100)*528.39 g = 229.85 g

mole C - 255.64 g * ( 1 mole ) = 21.29 mol 12.01

gmole H – 42.91 g * ( 1 mole ) = 42.49 mol 1.01 g

mole O – 229.85 g * ( 1 mole ) = 14.37 mol16.00 g

A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O

C – 21.29/14.27 = 1.49

H – 42.49/14.27 = 2.98 (esentially 3)

O – 14.27/14.27 = 1.00

C – 1.49 x 2 = 3

H – 3 x 2 = 6

O – 1 x 2 = 2

C3H6O2

A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

From last slide: Empirical formula = C3H6O2

“EFM” = 74.09

Molar mass = 222.24 = ~3

EFM 74.09

3(C3H6O2) = C9H18O6