# Empirical and Molecular Formulas

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Empirical and Molecular Formulas

We can determine its percent composition
If we are given an unknown substance how could we determine its chemical formula? We can determine its percent composition Then you can determine its chemical formula There are two types of chemical formula: Empirical Formula Molecular Formula

Empirical Formula Empirical formula: the simplest whole-number ratio of atoms in a compound This is the true formula for ionic compounds, but not for necessarily for molecular

Molecular Formula Empirical Formula C2H2 CH C4H4 C6H6 The empirical formula gives the combining ratio in its simplest form The molecular formula gives the same ratio but with the actual number of atoms Note: The empirical formula and the molecular formula can be the same in some cases. E.g. H2O, CH3, CO2

How to determine the Empirical Formula
You are given the % composition of a compound Ex: % Carbon % Hydrogen Assume that you have a g sample of the compound: Ex: g Carbon 7.7 g Hydrogen

3. Use the atomic mass of each element to determine the number of moles of each element
Covert mass to moles nC = g 12.0 g/mol nC = 7.69 mol C Covert mass to moles nH = 7.7 g 1.01 g/mol nH = 7.70 mol H 4. To obtain the simplest molar ratio (empirical formula), divide both by the smaller number of moles present C:H mol C : 7.70 mol H 7.69 1 mol C : 1 mol H, therefore C1H1 or CH is the empirical formula

Calculating Empirical Formula
E.g. What is the empirical formula of a substance that is 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen by mass? Given: Assume 100 g mC = 40.0 g mH = 6.70 g mO = 53.3 g Divide each element by the lowest quantity 3.33 mol C = mol H = mol O = 1 Mole ratio of C:H:O = 1:2:1 Covert mass to moles nC = 40.0 g 12.0 g/mol nC = 3.33 mol nO = 53.3 g 16.0 g/mol nO = 3.33 mol  The empirical formula of the substance is CH2O nH = 6.70 g 1.01 g/mol nH = 6.63 mol

Calculating Molecular Formula
Molecular formula: shows the actual number of atoms of each element in a molecule A true chemical formula for molecular compounds There are two different ways to find a molecular formula, depending on what type of information you are given…

Type 1: Given empirical formula and molar mass
Given: Empirical formula: CH2O molar mass: 180 g/mol Calculate the empirical formula molar mass M of CH2O = g/mol + 2(1.01 g/mol) g/mol M of CH2O = g/mol 2. Calculate the number of formula units # of formula units = molar mass (given) empirical molar mass (step 1) # of formula units = g/mol = 6 formula units g/mol

Molecular formula = # of formula units x empirical formula
6 x (CH2O) = C6H12O6 Therefore the molecular formula is C6H12O6 (glucose)

 The molecular formula of the compound is H2O2
E.g. What is the molecular formula of a compound that has a molar mass of 34 g/mol and the empirical formula HO? Given: Empirical Formula: HO MMcmpd = 34 g/mol MMHO = = g/mol Molecular Formula = MMcmpd MMHO = 34 g/mol 17.01 g/mol = 2  The molecular formula of the compound is H2O2

 The molecular formula of the compound is C3H9
E.g. What is the molecular formula of a compound that has a molar mass of g/mol and the empirical formula CH3? Given: Empirical Formula: CH3 MMcmpd = g/mol MMCH3 = (1.01) = g/mol Molecular Formula = MMcmpd MMCH3 = g/mol 15.04 g/mol = 3  The molecular formula of the compound is C3H9

Type 2: Given the % Composition and the Molar Mass
Step 1: Using the % composition, find the empirical formula 21.9 g Na g C 1.9 g H g O 22.99g/mol g/mol g/mol g/mol = 0.95 mol Na = 3.81 mol C =1.88 mol H =1.91 mol O 0.95 mol Na mol C mol H mol O = 1 mol Na = 4 mol C = 2 mol H = 2 mol O Given: % Na = 21.9 % % C = 45.7% % H = 1.9% % O = 30.5 % MM = g/mol

Empirical formula: NaC4H2O2
Step 2: Continue to solve for the molecular formula as for Type 1 M of NaC4H2O2 = g/mol + 4(12.01g/mol) + 2(1.01 g/mol) + 2(16.00g/mol) M of NaC4H2O2 = g/mol # of formula units = g/mol g/mol # of formula units = 2 formula units Molecular Formula = 2 x (NaC4H2O2) Molecular Formula = Na2C8H4O4

 The molecular formula is C4H10
E.g. What is the molecular formula of a compound with a molar mass of 58.0 g/mol and a percent composition of 82.5% C and 17.5% H? Given: % C = 82.5% % H = 17.5% MM = 58.0 g/mol m = assume 100 g 6.87 mol C 17.3 mol H 6.87 mol mol = 1 mol C = 2.5 *** multiply both by 2 to get full numbers Empirical formula = C2H5 M of C2H5 = 2(12.01g/mol) + 5(1.01) M of C2H5 =29.07g/mol mc = 82.5 g nc = g 12.01 g = 6.87 mol mH = 17.5 g nH = 17.5 g 1.01 g = 17.3 mol # of formula units = g/mol g/mol # of formula units = 2 formula units Molecular formula = 2 x (C2H5)  The molecular formula is C4H10

Homework P. 211 # P. 214 # 1-3 P. 218 # 17-20