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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